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~
dt  dt tan
d(
dy ) d ( if ) dt di = dt ;; = "a·cttllat s!i
Jrot
fix  xif xz
1 (
1
f
(x) = t , soy= g(t) = g(f 1 (x)) = F(x).
dy ) _ 1 [ d ( dy ) ] dy dy 1dt i1 dx  1 + (dyjdx)2 dt dx · But dx = dx/dt = ± ~
d¢> 1 ( fl±  xfl ) xy  xif ' . . · dt = 1 + (iJ/x) 2 x2 = xz +iF. Usmg the Cham Rule, and tne
~
(d:r;)2+(!!11..)2dt dt dt
r
~
13
d .• 'Jt 
~
y /(d:r;)2+(!!11..)2 dt dt = ('2+ x y•2)1/2 , we bave t hat
0
d¢> _ d¢>/dt _ ds  ds/dt 
(xii xfl ) :i;2 + y2
(b) x = x_and y = !(x )
~
_ I 1 1(±2xyxy 1 l±ii  xiJI + y2)3/ 2  (±2 + y2)3/2 ·
1 _ xii  xiJ . d¢> (±2 + y2)1/2  (~2 + 1;2)3/2 · So"'  ds 
dy .. d2y . 1 .. 0 d . x = 'x = an y = dx' y = dx2'
 II· (d~y/dx2 )  0 . (dyjcb:) i (1 +
So"' 71. X = 8  sin8
~ 2
"' =
icosB  cos [(1  cos B) 2
:i;
ld2 y/ dx2 1
(1 + (dyjdx)2] 312 •
(dyjdx)2 )31 2

= 1' cosB ~
X = sin8, and y = 1  cos8
e sin el 2
+ sin2 B)S/ 2
=
~
iJ =sine
~
y=
cos8. Therefore,
icosB  (cos 8 + sin 8)1 JcosB  II . . = . The top of the arch 1s (1  2 cos + cos2 + sm2 8)312 (2  2 cos 8)3/2 2
e
2
e
characterized by a horizontal tangent, and from Example 2(b) in Section 10.2, the tangent is horizontal when B = (2n  l )1r, d b · B · ti . fi sotaken=1 an su stttute =1rmto 1e ex~resston or ~~;: ~t=
(
2
Jcos 1r 11 11 11 _ cos1r) 3 / 2 = _ ( ))3/ 2 2 12 2 1
1 =4·
73. The coordinates of T are .(r cosB, r sin B). Since TP was unwound from arc T A, T P has length rB. Also LPTQ = LPTR  LQTR
= t 1r 
8,
soP has coordinates x = r cos B + r8 cos(~1r ~B) = r(cos B + Bsin 8),
y = rsin B  rBsin (~1r  8) = r(sin8 8cos9). X
10.3
Polar Coordinates
1. (a) (2, i)
By adding 21r to i, we obtain the point ( 2, opposite
i
is
4 3.,.,
7 ;).
The direction
so ( 2, ~ ) is a point that satisfies the r < 0
· requirement.
©
2012 Cengage ~ing. All Rights Reserved. May not be scanned, copied, or duplicated, or posted too publicly accessible wcbsitc, in whole or in part.
14
0
(b)
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
(1, 
3 ;)
+ 21r) = (1, 5; )
r
> 0: (1, 
r
< 0 : ( 1, 3r + 7r) =
r
> 0: ( ( 1), 1J +1r) = (1, 3; )
r
< 0 : (1 , ~ + 21r) = (
3 ;
(1,~)
0 _ 311' 4
(c) (  1, 2J)
,.
1,
5 2,.)
2
0
(t,f)
3. (a)
X=
1COS7r = 1( 1) = 1 and
y = 1 sin 1r = 1(0) =
0 give us
'TT
the Cartesian coordin~tes ( 1, 0).
(l, 'TT)
0
(b)
x
= 2cos( 23"' ) = 2(~) = 
y
= 2sin( 
2 ;)
give us ( 1,
y =  2sin
= 2(  ~) ~ ~VS
V3).
x = 2cos 3;
(c)
1 and
3 4"
= 2(4) = J2 and =
~( 4) =  J2
gives us ( J2, J2).
5. (a) x
= 2 andy =
 2 => r = J2 2
+ (2)2 = 2 J2 and 8 =
quadrant, the polar coordinates are (i) (2 J2, (b) x =  1 andy =
V3 =>
r
7 ;)
tan 1 (
and (ii) ( 2 J2,
:n := ~ . Since (2, 2) is in the fourth
3 ;).
= J< 1) 2 + (v'3) 2 = 2 and 8 = tan~1 ( :1) =
quadrant, the polar coordinates are (i)
2 :;.
Since_( 1, VS) is in the second
(2, ~)and (ii) (  2, 6; ) .
'·
© 2012 Cengnge Learning. All Rights Reserved. May no1 be scnnned, copied, or dupll
SECTION 10.3
7. r
~
1.
The curve r = 1 represents a circle with center
0 and radius 1. So r
~
POLAR COORDINATES
0
15
9. r ~ 0, 7r/4 S, B S, 37r/ 4 . .
1 represents the region on or
8 = IC represents a line through 0.
outside the circle. Note that 8 can take on any value.
0= 37T 4
11· 2
0='!!.. 4
< B < r1r < r < 3 •s.,.. aa 0  11< 3 r=2
o s1f3 13. ·Converting the p~lar coordinates (2, 1r/ 3) and (4, 21r/ 3) to Cartesian coordinates gives us (2 cos f, 2 sin f) = (1 , J3) and
(4 cos 2;
4 sin 2; ) = ( 2, 2 J3). Now use the distance !ormula.
,
15. r 2 = 5
x2 + y 2 = 5, a circle of radius .j5 centered at the origin.
<*
~
17.r =2 cos8
r 2 = 2rcos8
<*
2
x 2 + y =2x <* x 2 2x+ 1 +y2 = 1 <*
radius 1 centered at (1, 0). The first two equations are actually equivalent since r 2 = 2r cos()
r = 0 or
r=
(x1) 2 +y2 = 1,acircleof ~
r(r  2 cos 8) = 0
2 cos 8. But r = 2 cos 8 gives the point r = 0 (the pole) when 8 = 0. Thus, the equation r = 2 cos 8 is
equivalent to the compound condition ( r = 0 or r = 2 cos 8).
the origin with foci on the x axis. 21. y = 2
<*
23.y=1 +3x
2 r=
¢?
r = '2csc8
rsin8 = 1 +3rcos8
¢?
rsin83rcosB = 1
r s in B = 2
¢?
¢?
sin8
¢?
r(sin83cos8)=1
<*
1 r = sin::8::: 3=co_s_e=
25. x 2 +y2 =2cx
¢?
r = 0 is included in r
r 2 =2crcos8
¢?
r 2 2crcos8=0 <*
r(r2ccos8) = 0
¢?
r = Oorr=2ccos8.
= 2c cos 8 when 8 = ~ + n1r, so the curve is represented by the single equation r
~
2c cors ().
® 2012 Cc:ng.1gc Lcoming.·Ail Rights Rescn·cd. Moy not be scanned, copied, orduplicored, or posted to a publicly accc..'isiblc websil c* in whole or in pan.
~
16
D
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
27. (a) The description leads jmmediately to the polar equation 9 = ~.and the Cartesian equation
y,; tan (~) x = 7:i xis
slightly more difficult to derive.
(b) The.easier description here is the Cartesian equation x = 3.
29. r =  2 sin9 r
2
31. r
(2, 3'11'/2)
= 2(1 +cos B)
(4,0)
2'11' 0
33. r
= 9,
9;::: 0
r
(J
35. r = 4sin 39
r
./ o=f
4
,..:/ 2 (4.f) (}
4
37. r = 2cos49
r
© 2012 Ccngage Leamintt. All Rights Rc:scn·c:d. May not be SC<1J111ed, copi«<. or duplicated, or post ct.! ton publicly occc~iblc Y.•cbsih:, in whole or In Pun.
SECTION 10.3
39. r
POLAR COORDINATES
= 12sin8 0=!!. __________..• 6
41. r 2 = 9sin28 3
2..
0
3
43. r = 2 + sin 38 r
(3, ..j6)
3 2
2..
r~~+ t~
3..
6 45. r
7..
11..
6
0
6
= 1 + 2cos28
IJ
47. For(}=:= 0, 7T, and 27T, r has its minimum ;value of about 0.5. For 0 =~and We see that the graph has a similar shape for 0 . r
~
(}
~
7T and 7T
~
3 ", 2
r attains its maximum va lue of2.
8 ~ 27T. 2
® 2012 c~ngage Learning. All Rights 'Rescn·..d. May not be scanned. copic.•tl, or dupticntcd, or posted lo a publicly uccc:s.siblc website, ;n whole or in part.
0
17
18
0
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
=>
49. x = r cos 8 = (4 + 2 sec 8) cos 8 = 4 cos 8 + 2. Now, r + oo
(4 + 2sec8)+ 00 => 8+ (~)  or 8 + e2"' )+ [since we need only consider 0 $ 8 < 271'], so lim x = .
T
r oo
(6, 0)
lim (4 cos 8 + 2) = 2. Also,
8 + rr/2 
=> (4 + 2 sec B) + oo => 8 + ( ~) + or 8 + ( 3; )  , so
+ oo lim x =
=> x
lim ( 4 cos 8 + 2) = 2. Therefore, lim x = 2
r oo
8+rr/2+
T"+±oo
= 2 is a vertical asymptote.
51 . To show that x = 1 is an asymptote we must prove lim x = 1. r  ±oo
x = (r) cosO= (sin 8 tan 8) cos 8 = sin 2 ().Now; r + oo 8 + (f)  ,so lim x= r +oo
e+ (~) +,so lim X = . r ......  oo
=> sinO tan8+ oo
sin2 8 = l.Aiso,r+oo
lim 0+'Tr/2
.
lim
0+1fj 2+
x= l
=> sinBtanB>oo =>
sin2 {;I,;, 1. Therefore, lim X= 1 . ,·±oo
a vertical asymptote. Also notice that X =
sin2 e ~ 0 for all
=>
e, and X
=>
X=
1 is
= sin2 {;I $ 1 for all e. And
defined at odd multiples of ~. Therefore, the curve lies entirely within the vertical strip 0 $ x
Xi=
1, since the curve is not
< 1.
53. (a) We see that the curve r = 1 + csin 8 crosses itself at the origi n, where r = 0 (in fact the inner loop corresponds to negative rvalues,) so we solve the equation of the lima~on for r = 0
<=>
c sin 8 =  1
lei < 1, then this equation has no solution and hence there is no inner loop. But if c < the equation has the two solutions 8 = sin 1 ( 1 /c) and 8 = (} =
1r
1r
+ sin  1 (1/c) and ()= 21r sin 1 (1(c). Tn each case, r
(b) For 0 < c
sin
1
(
1, then on the interval (0, 27T)
1/c), and if c > 1, the solutions are
< 0 for (} between the two so lutions, indicating a loop.
< 1, the dimple (if it exists) is characterized by the fact that y has a local maximum at 8 =
determ.ine for what cvalu~s
So we
2
~
d2 y
dy
=> dB = cos8 + 2csinfJ cos8 = cosfJ + csin 28 =>. dB~., =  sin8 + 2ccos 28.
At 8 = 32,.., this is equal to  (  1) + 2c( 1) = 1  2c, which is negative only for c for  1 < c < 0, y only has a local minimum at e = ~ (indicating a dimple) for c
=>
3 ; .
d ~ is negative at(} = 32~, since by the Second Derivative Test. tllis indicates a.maximum:
y = rsin8 =sinO + c sin 2 (}
55. r = 2 sin (}
<=> sin 8 =  1/ c. Now if
x = r cos(} = 2 sin fJ cos 8 = sin 28, y = 1· sin 8
= 2 sin 2 B
> %. A similar argument shows that
< ~ . .
=>
dy = dyjd(J = 2 · 2 sin fJ cosO= sin 28 = tan 2B dxjdB cos2fJ·2 cos28
dx
When 8 = 7T , dy d.'t =tan ( 2 · 7r) =tan 7T = vrr; 3.
6
57. r= l / 8
=>
6
3
· 3.] [Another method: Use Equation
x=rcos8 = (cosfJ) j B, y=rsin8=(s infJ) / 8
=>
dy dyjd8 sin.O(  1/ 82 ) + (1/ 8) cos 8 8 2  sin (}+(} cosO dx = dx j dO = cos8( 1/ fJ 2 )  (1/ 8) sinB · (} 2 =  cosO  Osin B .dy  0 + 7r( 1) When 8 = 1r, dx = (1) 1r(O) =
 'Tr
l
= 1r.
® 2012 Cengoge loaming. All Rights Resef'.d. Moy not be: ~ned. copl<d, ordupl iuted, or po5led to a publicly accessible web•itc, in whole or in part. '
SECTION 10 .~
POLAR COORDINATES
0
=> x = r cos() = cos 2() cos(}, y = r sin() = cos 2(} !lin(} =>
59. r = cos 2()
dy dy / df) cos 2(} cos(}+ sin(} ( 2 sin 2()) dx = dx / d() = cos 2() ( sin(})+ cos() ( 2 sin 2()) ,0,!(J2:2:/27)+_(':v2 :::2/,.!2)~(_2;,_)  _ J2_2  1 O(J2/ 2) + (J2/ 2}(2)  J2  .
dy
1f
4 , dx
When() =
=> x = rcosO = 3cos0 cosO,_ y = rsinO = 3cos0 sinO => ,
61 . r = 3cos0
2
~ = 3 sin 9 +3cos 2 ,() = 3cos20 = 0 => 2() = ~ or 3; <=> 0 = %or So the tangent is horizontal at (
~, %)
~~ =  6 sin() cos~= 3 sin 20 = 0 63. r = 1 + cos 0 ·
and (
32,
3 ;)
[same as (
3 ;.
32,:a:)] .
=> 29 = 0 or 1r <=> 0 = 0 or~· So the tangent is vertical at (3, 0) and (o, ~).
=> x = r cos 9 = cos() (1 +cos 9), y = r sin 0 = sin 0 (1 + cos 9) =>
·.~ = (1 +cosO) cosO sin2 0 = 2cos 2 0 + cos9  1 = (2cos9  1) (cos(J + 1)
() =
'i· 1r, or 5;
=> horizontal tangent at(~ , 'i), (0, tr), and(~ , 5;
~~ =  ( 1 + cos 8) sin (}  cos ()sin() =  sin (} ( 1 + 2 cos 9) = 0
() = 0, 1r,
2 ;,
or
4 ;
x
2

bx
2
2
ay + ( ~a) = (~b) + (ta?
) .
=> sin 8 = 0 or cos 9 = 
~
=>
e : j::
= 0.
=>
(x  ~b)
2
+ (y  ~ a? = Ha2 + b2 ) , and this is a circle
(!b, ~a) and radius. ~.Ja 2 + b2 •
67. r = 1 + 2 sin(8 / 2): The parameter inteival is [0, 47r].
69. r
= e•inB 
2cos(40).
The parameter interval is [0, 21l'j.
71. r = 1 + cos999 8. The parameter interval is 1.1
[0, 2tr). /
 1.1
©
=>
=> r 2 = ar sin() + br cos 8 '=> x 2 + y 2 = ay + bx =>
+ (~b? + y 2 
with center
cos()=~ or  1
=> vertical tangent ~t (2, 0), (!, 2; ) , and (!, t;).
Note that the ~gent is horizontal, not vertical when 8 = 7r, since 65. r = a sin() + b cos 9
= 0 =>
20 12 <:engage Learning. All Rights Reserved. May not be SCIInncd, copied, or duplicated, or posted too publicly accessible website, in whole or in port.
19
20
D
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
73. It appears that the graph of r the graph of r origin by
= 1 + sin(~  "if) is the same shape as r = 1 + sin ll
= 1 + sine, but ro,tated counterclockwise about the
"if. Similarly, the graph of r =
1 + sin(
r = l +s~(o  ~_)
e. . ,. ~) is rotated by·
r=
1+ sin(ll  ~)
i . In general, the graph of r = f (e  a) is the same shape as that of r = f (e), but rotated counterclockwise through a about the origin. That is, for any point (ro, eo) on the curve r = j(e), the point (ro, Bo +a) is on the curve r
0.9
= j(e a), since ro = f(Oa) = f((eo +a) a).
= 1 + c cos e, where c is a real number. If c = 0, we get a circle of radius 1 centered at the pole. For 0 ~ c s 0.5, the curve gets slightly larger, moves right, and flattens out a bit on the left side. For 0.5 < c < 1, the left side has a dimple shape. For c = 1, ,the dimple becomes a cusp. For c > 1, there is an internal loop. For c 2: 0, the
75. Consider curves with polar equation r
rightmost point on the cu;ve is (1
+ c, 0). For c < 0, the curves are reflections through the vertical axis of the curves
with c > 0.
c = 0.25
77. tan'lj! = tan(¢ e)=
_
c
= 0.75
tan ¢  tane r/> e 1 + tan tan
dy   tane dx dy 1 + dx ta11e
~  ~ tan e ~ (~ ~in e + r cos e) 
 : + :~ tane
 (:;.cos
c= 1
tan e (
c= 2
dy/dB dxjde  tane dyjde
1 + dx j de tane
~ cos B
r sine)
e r sine) + tan B( :; sine + r cos e)
sin2 e rcosB+r ·  case dr dr sin2 e de case+ de. case
_ rcos2 B +rsin2 e r 2 iJ dr . 2 e = dr I dO  dr de cos . + dB sm
10.4 Areas and Lengths in Polar Coordinates
@ 2012 Censagc Learning. All Rights Reserved. May not be scanned: copied, or duplicated, or posted to 0: publicly ncccssiblc website, in whole or in part.
SECTION 10.4 A~~S AND LENGTHS IN POLAR COORDINATES
3. r 2 = 9sin28, r ~ 0, ·a~ 8::; 7rj2.
A 
7.
1'
1
~/2
0
2 lr d8 2
1~/2
.
= 4 + 3sin8,
0
1(9sin2B)d8 l!. 2  2 [ lcos2B)~ 2 0
/2
l!.(4
1  1)=l!.2
%::; () ::; %·
1
= ~
~/2
2
(16 + 9sin 8) dO
[by Theorem 4.5.6(b) [ET 5.5.7(b)]]
rr/2
r /2
= ~·2}
0
[16 + 9 ·~(1 cos28))d~
1 (";  ~cos28) rr / 2
=
9. The area is bounded by r
'
d8 =
[~1 8  tsin2BJ; 12 = ( 4 ~"  O) (0 _
0) =
4~"
= 2 sin 8 for 8 = 0 to 8 = 7r.
~r 2 d8 := ~ 1 rr (2 sinfJ) 2 dB= ~
A= 1"'
[by Theorem 4.5.6(a) [ET 5.5.7(a)]] .
1"'
(2, 'TT/ 2)
2
4sin Bd8
=21" ~(1  cos2fJ)dB ~ [e  ~sin2eJ: =1r Also, note that th is is a circle with radius 1, so its area is 1r(1) 2 =
1 ~r d8 = l ~1 2
2
11. A =
"
2
1r.
fo rr(9+12.cos8+4cos 8)d8 2
2
rr H3 + 2cos8) d8=!
2
2
=
.,..
= ~
[9 + 12cos8 + 4 · t(1 .+ cos 28)] dB
{2rr
lo
(11 + 12 cos 8 + 2 cos 28) d8 =
Hne +
_(_l,_'TT:)H=

':..._:..1+
12 sin O+ sin 28) ~71'
= H227r) = 111r
13. A = =
r 71' ~r2 d8=
Jo
~ fo
= ~
"
1
~ r"(4+4si~4fJ+sin2 4fJ)d8
· Jo
2
2"
0
r rr t{2 + sin4fJ) 2 dfJ =
lo
I4 + 4sin48 +
3
~(1 cos88)] dB
(~ + 4 sin48  ~ cos88) dB=~ [~B cos4fJ
= ~[(97r 1)  ( 1))
·
.
fs sin8B] ~"'
= ~7('
® 20 12 Ccngagc Learning. All RighiS R=:rvod. May not be sconncd, copied, or duplicotcd, or posted to a publicly accessible website, in whole or in pan.
0
21
22
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
D
1 (39 2 2
=
. + 2o1 sm 109] o21r =
1( ) 2 3'1T
3 27T
=
1.4
17. The curve passes through the pole when r
9=
=0
4 cos 30
==?
=0
==?
cos 39 = 0
39 = ~
==?
+ 1rn
=?
%+·in. The part of the shaded loop above the polar axis is traced out for
9 = 0 to 9 = 7T /6,' so we'll use
19. r = 0
sin49 = 0
==?
7T / 6
48
==?
r
and 7T /6 as our limits of integration.
= 1rn
9=
==?
= 4 cos 30
r=sin 40
~n.
r'4 Hsin48) dfJ. = t Jor'4sin 48d8 = t Jor '4t(1cos89)d8 0
=4
2
2
A = Jo
.
1 (9
.
· ]"'I 4 B SlD 89 0 1

0
=4
0
1 (,.)
.1 Tii7T
=
4
This is a
21.
lima~on,
with inner loop traced
r = 1 + 2 sin 8 (rect.) 3
out between 8 =
7 ;
and 1 ~" [found by
solving r = 0].
17T ·
0=6 31r/2
A = 2[ 71r /6
=
ll l'l'
0= 6
77r /6
=> cos 8 =
~

=(
[8+ sin 28)~ 13 = i +
4
C; + 2 v'3  4) =
9 ;)
12 ) dB=
= J;; a { 4 [~ (1 + cos 28)] 
37r/2 71r /6
=> 8 = :g. or
A = 2 f 0,. 13 ~[(2 cos8) 2
=
··••,
. [37f/2 [ !(1 + 2sin8? d8 = (1 + 4sin8 + 4sin2 8) d8 =
[9  4cos0 + 20 ~n28J~:~~
23. 2 cos B = 1
.
5 ;
7T 
[1 +4sin8 + 4· t{1 cos29)] d9 .
.
¥
.
J0" 13 (4cos2 8  l) d8
1} d8 =
J; /S ( 1 + 2 cos 28) d8 r = 2cos0
@) 2012 Cengogc Lcaming. All Rights Reserved. Moy not be oamned, copied, or duplicated, or posted to a publicly accessible website, in whole or in prut.
SECTION 10.4
AREAS AND LENGTHS IN POLAR COORDINATES
0
25. To find the area inside the l~miniscate r 2 = 8 cos 29 and outside the circle r = 2, r=2
we first note that the two curves intersect when r 2 = 8 cos 29 and ~ = 2, .
I
that is, when cos2B = ~For 1r < B ::; 1r, cos2B = ~
{::} 2B = ±7r/ 3
{::} B = ±1r/ 6 or ±57r/ 6. The figure shows that the desired area is
or ±511' /3
4 times the area between the curves from 0 to 1r/6. Thus,
A = 4 J0" /. 6 [~(8 cos2B) ~(2) 2 ] dB = 8 J0" 16 (2 cos 2B  1) dB
]
. =8 [ sin2B  B
rr~
0
.
.=8(VS/27r/6) =4\1'3411'/3
27. 3cosB=1+cosB {::} cosB=~
=> B=t ori.
8 =1!. I 3 ,l'
A= 2 f 0" 13 ~ [(3 cos 8? (1 +cos 0) 2 ) dB =
J0"' 13 (8 cos2 8' 2 cos B 1) dB = f 0" 13 [4(1 +cos 2B) 
2 cos B  1) dB
= J0.,.13 (3 + 4cos28 2cosB) dB= [3B + 2sin2B 2sinf!] ~ 1 3
= 1T +v'3.J3=1T
= sinB. => J3 =
29. v'3cosB
A Jo  f "'13 12 (sin B) 2 dB +
1'"' 12
1
hr/ 3 . 2
(J3 cos B)2
3
 Jo f"' / 1. 1(1  cos28) dB+ 2 2
 1 4 [B

=> tanB = J3 => B =
sinBB cos
i·
d()
f rr/3 "/ 2 12 · 3 · 1(1 + cos2B) dB 2
12 12 sm · 28]"o13 +!!4 [B +.!2 sm · 28]1f7f/3
=
t [(i  4)  o] + ~ [ (~ + o) ( f + 4)]
_ 
.l!.. _ 12
_il 16
+. .!!:8 _
31. sin 2B = cos 2B
=>
1fl _ 16

5rr _
24
2 sin B = 1 cos2B
r =
J3cos8
fl 4
=> tan 28 = 1 => 28 = ~ =>
=>
B=. f
A= 8 · 2 f0"/8'~ sin 2 2B dB= 8 f 0"'18 H1 cos 48) dB = 4[9
t sin48]~
18
= 4(f 
33. sin2B = cos2B => tan2B = 1 =>
i
·1) =
2B =
~1
~
A= 4 f 0"' 18 ~sin 2B dB [since r 2 = sin 2B) =
f 0"'18 2 sin2B dB= [; cos 28] ; 18
= ~
v'2 ( 
1)
=> (} =
f
....···B=i
= 1 ~ Y2.
© 2012 Ccngage learning. All Riglus Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessibk website, in whole or in part.
·
23
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
24 • 0
35. The darker shaded region (from 0 = 0 to 0 = 27T / 3) represents ~ of the desired area plus ~ of the area of the inner loop. From this area, we' ll subtract~ of the area of the inner loop (the lighter shaded region from 0 = 27r /3 to 0
= 7r), and then
double that difference to obtain the desired ·area.
A = 2 [J;.,. =
13
H~ +~os0) 2
dO
g.,.13 H~ +cos 0) 2 dB]
£02.,.13 (~+cosO+ cos2 0) dB f2:
.
2 13 (~ + cos 0"+ cos 0) dB
.
= J0211'/S [~+cos 0 +
~(1 + cos28)) dO
 I2:/3
= [~ + 4
~ + sin 20 ]
sin 8 +
2
4
13
2 "
_
a + cosO+ ~(1 + cos28)) d(J
[~ + 4
0
sin 8 +
~ + sin 20 1.,. 2
4
211'/ 3
.:.8::2)  (!!:4 + !!:) 2 + (1!: 6 + .il 2 + 1!: 3  .:.8::2)
 (1!: + .il 2 + 1!: 3 
6
= ~ + ~ J3 = H 7T+3v'3) 37. The pole is a point of intersection. 1 + sin 8 = 3 sin 0 (J
=~ or
=> 1 =
sin 0 = ~
=>
2 sin (J
=>
5 ;.
The other two ~oints of intersec~ion ~e ( ~,
i) and
(
f,
5 6").
r=l + sinO
39. 2 sin 20
=1
sin 20
=?
=
~
=?
o 1!:6' &.,.6 '
2 =
13.,. 6 '
or
11" 6 °
By symmetry, the eight points of intersection are given by r = 2sin20 11' 5,. 13rr 1771" (1, 8) , where (} _ 12 , an d , 12 , 12 , and 12
i;, 1t;, 1: 2" ; and 2[ ; .
(  1, 8), where 0 =
[There are many ways to describe these points.]
41. T he pole is a point of intersection. sin (J sinO (1  2 cos 9)
=0
8 = 0, 7T,
t. or  t
and ( ~,
2 ; )
©
= sin 20 = 2 sin(} cos 0
<=> sinO= 0 or coslJ =
t
<=>
=>
=> the other intersectio~ points"are ( ~. t )
[by symmen:'].
r =sin20
2012 Ccngage Learning. All R ights Reserved. May not be scotuled, copied, or dupHcuted, or posted to a publicly accessible website, in whole or in part.
SECTION 10.4 . 3.4
43.
3
r=
AREAS AND LENGTHS IN POLAR COORDINATES
0
25
y = 2x
2(}
3
r = 1 +sinO 1.4
1.4
3
 0.3
From th~ first graph, we see that the pole .is one point of intersection. By zooming in or usin~ the cursor, we find the 0values of the intersection points to be a:~ 0.88786 estimated by plotting y = 1 +sin x andy
~
0.89 and
:r  a ~ 2.25. (The first of these values may be more easily
= 2x in rectangular coordinates; see the second graph.) By symmetry, the total
area contained is twice the area contained in the first quadrant; that is,
A= 21o. ~(20) 2 dO+ =
45. L =
2[1r/Z ~(1 +sin 0.)
2
d8 = 12
(~0 3 ] ~ + [8  2cosB + (~0  t sin2B)J:
1"' =
402 .d8 +
l1r/~ [1 + 2sinU +
H1 cos 2B)] d8
~a 3 + [(~+~)(a 2cosa + ~a:  t sin2o:)) ~ 3.4645
1b
Jr2+(d7·jd0) 2 dB= fo7r }(2cos0)2 +(2sinB.)2d8
= 17r
V
4(cos2 B + sin2 B) dB =
1"' ..f4d8 =
[2B];
= 21r
As a check, note that the curve is a circle of radius 1, so its circumference is 27r(1 ) = 21r.
=
1 Je 2
"'
2
(02 + 4)d0 =
1 2
1f
BV02 +4d0
Now let u = 02 + 4, so that du = 20d8 [Ode=
49. The curve r
t du]
= cos 4 (0 / 4) is completely traced with 0 :::; () :::; 47r.
r 2 + (drfd8) 2
= [cos4 (B/4)] 2 + = cos8.(B/4) +
[4cos3 (B/ 4) · (sin(B/4)) · t ]
2
cos6 (8 /4) sin2 (B / 4)
= cos6 (8/4)[cos2 (B/4) +sin2 (0/4)] = L
and
cos6 (B/4)
= f041r Jcos6 (0/4) dO = J04"' jcos3 .(0/4) j rj,B . = 2 f~" cos8 (B/4)dB [sincecos3 (B/4) ~ 0 for O :S B $ 27r]
= 8 J;12(1 
sin 2 u) cosudu = 8 f 01 (1 x 2 ) dx
= 8J;12 cos3 udu
[u = tB]
x = sinu, ·] [ dx = cosudu
= 8[x ~x3) ~ = 8(1 ~) = lf @ 2012 Ccngu~c Learning. All Rights Rcscrv~. May not be scanned, copied, or duplicaled. or posted too public ly accessible w·cbsite, in whole or in part
26
0
CHAPTER 111 PARAMETRIC EQUATIONS AND POLAR COORDINATES
51. One loop of the curve r = cos 28 is traced with  1rI 4 ~ 8 ~ 1r I 4. r2 +
(~~)
2
14
= cos2 28 + ( 2 sin 28) 2 = cos2 28 + 4sin2 28 = 1 + 3 sin2 28
L /"'
::::}
Vl + 3 sin
2
?r /4
53. The curver = sin(6sin9) is completely traced with 0
r2
~ 8 ~ 1r.
(~;y =s~n2 (6sin8)+36cos2 8cos2 (6sin0)
:
::::}
L
r = sin{6sin9)
L"
::::}
~
28 d8:::::: 2.4221.

= cos{6sin9)_ 6cos8, so
Jsin2 (6sin8)+36cos2 8pos2(6sinO)d8:::::8.0091.
55. (a) From (10.2.6),
s
=I: 21ryJ(dxld0)2 + (dyld9) 2 dO [from the derivation of Equation 10.4.5]
(b) The curve r 2 20 =
~
::::}
= cos 28 goes through the pole when cos 28 = 0 8=
~.
We'll rotate the curve from () = 0 to 8 =
::::} ~
(o. f)
and double
? = cos28 (1.0)
this value to obtain the total surface area generated. r 2 = cos28 ::::}
· dr 2r dO= 2sin28 ::::}
( dr )
dO
2
2
2
sin 28 sin 28  ~  cos20'
10.5 Conic Sections 1. x 2
= 6y and x 2 = 4py
::::}
4p = 6
::::}
p
= ~
The vertex is (0, 0), the focus is (0, ~),and the directrix
isy
= ~.
3. 2x =  y 2
::::}
y2
= 2x.
4p
= 2
::::} p '\
The vertex is (0, O),the focus is (  ~, 0), and the directrix is x = ~. y
x=4 6
X
X
y= ~
©
2012 Cengage Learning. All
Righ~
Reserved. May not be scanned, copied, ocduplieated, or posted to a publicly acccssible website, in whole or in pan.
=  ~.
SECTION 10.5
5. (x + 2) 2 = 8 (y 3). 4p = 8, sop= 2. The vertex is
7. y 2
(  2, 3), the focus is ( 2, 5), and the directrix is y = 1.
y2
CONIC SECTIONS
+ 2y + 12x + 25 = 0 =l> + 2y + 1 =  12x 24 =;.. (y + 1)2 =  12(x + 2). 4p = 
y
0
27
12, sop= 3.
The vertex is (  2, 1), the focus is (5,  1), and the directrix is x = 1.
 __ ,y_~_!_ X
0
:
~
X
Ijx=l 9. The equation has the form y 2 = 4px, where p < 0. Since the parabola passes through ( 1, 1), we have 12 = 4p(  1), so
4p =  1 and an equation is y 2 =  x or x =  y 2 • 4p =  1, sop = . ISX
i and the focus is ( ;i,0) while the directrix
1
= 4· y
2
(0, ±v'2).
ellipse is centered at {0, 0), with vertices at (0, ±2). The foci are
..fi 2
13. x
2
+ 9y2 = 9
<=>
x2
Y2
9 +l
b = Vi.= 1, c = ~ =
= 1
=l>
a=
v'9=I = J8 =
v'9 = 3,
15. 9x 2
9(x 2
2v'2.
The ellipse is centered at {0, 0), with vertices (± 3, 0).

2x
+ 1) + 4y 2 =
9(x 1) 2 + 4y2
The foci are (±2v'2, 0).
a = 3, b = 2, c = y
0
= 36
27 + 9
<=>
J5 =:
<=>
(x 1)2 4
y2
+ g = 1
center (1, 0),
vertices (1 , ± 3}, foci (1, ±J5)
I
3
18x + 4y2 = 27 <=>

3
y
X
(1, 3)
 I
3 X
(1, 3) 2
17. The center is (0, 0), a= 3, and b = 2, so an equation is
©
~
2
+ Y9
= 1. c = ~ =
J5, so the foci are (0, =;I=J5).
2012 Ceng3ge LC3JT1ing. All Rights Reser\'ed. Muy not be scuMed. copied, or duplicated, or pos1cd to a publicly access ible wc:bsite, in whole or in pan.
=}
28
CHAPTER 10 PARAMETRIC EQUATIONS AND POLARCOORDINATES
0
yz 19.
xz
9 =1

25
a = 5, b = 3, c = J25 + 9
::::?
= J34
::::?
center (0, 0), vertices (0, ±5), foci (0, ±J34), asymptotes y = ±~x. Note: It is helpful to draw a 2aby2b rectangle whose center is the center of
the hyperbola. The asymptotes are the extended diagonals of the rectang le.
21 . x 2
y2

= 100
x2
<=>
y2
100

100
c = JlOO + 100 = 10J2 => foci (±10 J2, 0), asymptotes y
=1
y
a = b = 10,
::::?
center (0, 0), vertices (±10, 0),
= .±f§x = ±x
X
,.._!",, ,~_;.......
23. 4x2 4(x
y2
2


24x  4y + 28 = 0
6x + 9)  (y
2
+ 4y + 4) =
4(x '_ 3)2  (y + 2)2 = 4 a
<=>
<=>
28 + 36 4 <=>
(x  3)2  (y + 2)2 = 1 => 1
4
= v'f = 1, b = v'4 = 2, c = Vf+4 = J5
::::?
center (3,  2), vertices (4,  2) and (2, 2), foci (3 ± asymptotes y
25. x 2
=y+1
+2=
J5, 2),
±2(x 3).
<=> x 2
= 1(y + 1). This is an equation of a parabola with 4p = 1, so p =
i. The vertex is (0, 1) and the
focus is (0,  ~). 27. x 2
= 4y 2y2
x; + (y ~ 1)
2
<=> x 2
+ 2y2 
4y
=0
<=> x 2 + 2(y2

2y + 1)
=2
<=> x 2
+ 2(y 
1)
2
=2
<=>
= l.. This is an equation of an ellipse with vertices a~ (±J2, 1) . The foci ~eat (±.,J2=1, 1)
29. y2 + iy = 4x2 + 3
<=> y 2
+ 2y + 1 = 4x 2 + 4
of a hyperbola with vertices (0,  1 ± 2)
<=>
(y
+ 1? 
4x 2
=4
<=>
= {0, 1) and (0, 3). The foci are at (0, 
1
(y: ) 1±
= (±1,
1) .
2 
J4+T)
x 2 = 1. This is an equation = (0,  1 ±
J5).
31. The parabola wi~ vertex {0, 0) and focus (1 , 0) ?pens to the right and hasp= 1, so its equation is y 2 = 4px, or y 2 = 4x. ' 33. The distance from. U1e focus ( 4, 0) to the directrix x = 2,is 2  (  4) = 6, so the distance from the focus to the vertex is ~(6) = 3 and the vertex is (  1, 0). Since the focus is to the left of the vertex, p =  3. An equation is y 2 .= 4p(x + 1)
y
2
= 12(x + 1). © 2012 CcnGogc: Lcmning. Al l Rjglns Rest.! rvcd. Mny not he scanned. copied, or duplicutccl, or posted to a publicly occcssiblc website. in whole or in p..1rt.
=>
SECTION 10.5
CONIC SECTIONS
29
0
35. A parabola with vertical axis and vertex (2, 3) has equation y 3 = a(x 2?. Since it passes through (I; 5), we have 53=a(I2) 2
'*
a =2,soanequationisy 3 =2(x2? .
37. The ellipse with foci (±2, 0) and vertices (± 5, 0) has center (0, 0) and a horizontal major axis, with a = 5 and c = 2,
so b2 = a2
c 2 = 25 4 = 21. An equation is

x2
12
+ ;1 = 25
1.
39. Since the vertices are (0, 0) and (0, 8), the ellipse has center (0, 4) with a vertical axis and a = 4. The foci at (0, 2) and (0, 6) 2
are 2 units from the center, soc= 2 and b = .,Ja2  c2 = .,j4 2

_41. An equation of an ellipse with center (1,4) and vertex ( 1,0) is (x
from the center, soc= 2. Thus, b2
+ 22
= 42
'*
. 2
2
b = 25  9 = 16, so the equation is
~
~2
!
2
1
2
)
+ (y ~24)
2 ....:. ;
2
a
b
'*
= 1. Foci (±5, 0)
2
= 1
= 1. The focus (1,6) is 2 units
+ (y 164 )
c = 5 and 32
2
= 1
.
+ b2 =52
'*
2
 r6 = 1.
45. The center of a hyperbola with vertices ( 3,  4) and ( 3, 6) is ( 3, 1), so a = 5 and a.n equation is . (x + 3) 2 · · b2 = l.Focl(  3,  7)and( 3,9)* c =8,so5 2 +b2 =82
'*
b2 =6425=39andthe
(  1} 2 {x + 3) 2 equation is y = I 25 39 . .
.
2
47. The center of a hyperbola with vertices (±3, 0) is (0, 0), so a = 3 and an equation is
Asymptotes y
= ±2x
b
'* a =
2
'*
~2
•

2
t
2
= 1.
. 2 2 b = 2(3) = 6 and the equation is ~  JL = 1. 9 36
49. In Figure 8, we see that the point on the ellipse closest to a focus is the closer vertex (which is a distance
a  c from it) while the farthest point is the other vertex (at a distance of a+ c). So for this lunar orbit, (a c)+ (a+ c) = 2a = (1728 + 110) + (1728 + 314), or a = 1940; and (a+ c) .(a  c) = 2c = 314  110, or c = 102. Thus, b2 = a
2

y2
x2
2
c = 3,753,196, and the equation is , , 3 763 600
+ 3,753, 196
=I.
51. (a) Set up the coordinate system so that A is ( 200, 0) and B is (200, 0) .
IPAI\PBI =
(1200)(980) = 1,176,000 ft =
b2  c2
3,339,375 121

2 
a 
121x 2 1,500,625
2
1~ mi = 2a
*
'*
2
1 b2 = 12 and the equation is (x + ? . , 12 2
43. An equation of a hyperbola with vertices (±3, 0) is
v'f2. An equation is (x  2 0) + (y2 4 )
22 =
a= 1 ~ 5 , and c ~ 200 so
2
121y  1 3,339,375  .
© 20 12 Ccngoge Leoming. All Rights Rescn'Cil. May not be scanned, copied. or duplicated. or posted to a publi<:ly uccusible website, in whole or in part.
30
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
0
=> x = 200 =>
(b) Due north of B
(121}(200) 2 1,500,625
2
121y 1 3,339,375 =
=>
y = 133,575 ~ 248 mi 539
53. The function whose graph is the upper branch of this hyperbola is concave upward. The function is
55. (a) If k
> 16, then k 
> 0, and
16
2
2
xk + yk = 1 is an ellipse since it is the sum of two squares on the left side.   16 2
(b) IfO
< k < 16, then k 
16
< 0, and xk
2
+ k
~
= 1 is a hypel"bola since it is the difference of two squares on the
16
left side. (c) If k
< 0, then k 
16
< 0, and there is no cw11e since the left side is the sum of two negative tenns, which cannot equal
(d) In case (a), a 2 = k, b2
b2
= 16 
k , and c2
= 4ptJ =>
57. x 2
2
y :;
= k 16, and c2 = a2 
= a2 +
2x = 4py'
= ~; (x 
b2
b2 = 16, so the foci are at (±4, 0}. Jn case (b), k  16
are
= a2 + p
y
=> y' = ; , so the tangent line at (x0 , y 0 ) is .
p
xo). This line passes through the point (a,  p) on the
4p2
a± ..ja2 + 4p2 2
k,
= 16, and so again the foci are at (±4, 0}.
. . so  p  x6p = xop (a  xo ) =>  4p 2  x 2 = 2axo  2x 2 d1rectnx, 0 0 4 2 x5  2axo  4p2 = 0 <=> x5  2axo + a 2 = a2 + 4p2 <=>
(xo a) 2
< 0, so a 2 =
I.
<=> (a,p)
y=  p
<=> x 0 =a± Ja2 + 4p2 • The slopes of the tangent lines at x = a± Ja2 + 4p2
, so the product of the two s lopes ts
a+ Ja2 +
4p2
''. 2p
a  Ja2
+ 4p2
2p
=
a2 _ . (a2 + 4p2) 4p2. =   = 1 4p2 4p2 '
showing that the tangent lines are perpendicular. 59. 9x2 + 4y 2
= 36
2
<=>
~
2
+ Y9
= 1. We use the parametrization x
= 2 cost, y = 3 s in t, 0 ~ 't ~ 21r. ·The circumference
is given by
L =I:" J(dx/dt) 2
+ (dyjdt)2 dt = ]~2 " J( 
2 s in t) 2 + (3 cost)2 dt
=I:" )4sin2 t +
9cos2 t dt
= I:" v4 + 5 cos2 t dt Now use Simpson's Rule with n = 8, 11t =
L ~ ss = ~ [J(o) + 4f(*)
+ 2f(~) +
2
1T
8.
0
= ~.and f(t) = 4
4Jen + 2J(1r) +
4/(
v'4 + 5cos2 t to get
5 ;)
+ 21en
+
4tCn
+ J(21T)J ~ 15.9.
® 2012 Ccngage leArning. AJI ftis.hts Rcscn·cd. May nol be scannc:c.l, copicd or t.luplicah!d, or posted to a publicly uccessibJc website, in whole or in pout. 9
SECTION 10.5 x2 y2 61. ;_2   2 = 1
a
b
A= 2
la a
r
=?
y2 a:2 _ a2  2 = b a2 a
0
31
b y = ± Jx2 a2. a
=?
J x 2 ·_ a 2 dx ~ 2b [~ J x2 
!!_
CONIC SECTIONS
a2 
2
42
2
In Jx +
J x2 
a2J ] c
a·
b = [cvc2  a 2 a2 ln jc + vc2  a 2 j + a 2 ln!al) a Since a 2 + b2 = c2 , c2  a 2 = b2 , and J c2  a 2 = b.
=!!.a [cb a 2 1n(c + b)+ a 2 !na) = !!_a [ch + a2 (!na = b2 c/a + abln[a/( b +c)), where c2 = a 2 + b2 . 63. 9x2
+ 4y2 =
x2
Y2
'4 + g = 1
36 <::>
top half of the ellipse is
H1rab) =
y = 1
A
1b a
=  3 · '2 87T
1 [f(x))2 dx = 12
37T
1 2
0
2
( 4  x ) dx
a= 3, b = 2. By symmetry, x = 0. By Example 2 in Section 7.3, the area o_f the
=?
+ 4y2 =
37T. Solve 9x2
9x2 +4y 2 =36 <::> 4y 2 =369x 2
1
~
36 for y to get an equation for the top half of the ell ipse:
y 2 = ~(4x 2 )
2
 2
!n(b +c)))
2
1 ( 3 V 4 x 2 ) 2 2
3 [4x = 47T
1 x 3] 3
dx=  3
87T
y = ~v'4x 2 .Now
=?
12
y
(4 x 2 )dx
 2
2 0
3 ( 16) = 4 = 47T 3 7T
so the centroid is (0, 4/ rr). 2
. . .tmp I'tctt . Iy, •x 65. Dt' ffierenttatmg a2 line at Pis 
tan a
2
= 1
+ yb2
[y
=?
=f. 0]. Thus, the slope ofthe tangent
b:x1 . The slope of F1P is ___)f];_ and of F2P is ____lt!__ By the formula from Problems Plus, we have +
a Y1
=
=
X1
2 Yl b Xt Xt_+_c + a 2_y_1 _.:........;.::.''=1_ b2 x1Y1 2 a yt (Xt +c) b2 (cxt+a2 ) CIJt(CXt + a 2 )
·
_

c
X1 
yt
a2 + b2 x 1(x1 +c) _ a 2 y1 (x1 + c)  b2X tY1 
c
a 2 b2 + b2 cx1 C2 XtY1
+ a 2CIJ1
b2
= cy1
and  a
2
y'f. 
b2 x1 ( x1 c)
 a
2 2
b
+ b2 cx1
= a 2y1 (x1 c)  b2X 1Y 1 = c2x1y1  a 2C1Jt =
b2 (ext  a 2) CIJ1 (cx1 
a 2)
b2
= CIJ1
Thus, a = {3.
© 2012 Cc:ngoge Lco.ming. All Rights Reserved: Muy not be scanned, copied, or duplicaled, or posted to a publicly ncccs.•db lc wcbsiu~. in whole or in part.
32
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
0
10.6 Conic Sections in Polar Coordinates 1. The.directrix x = 4 is to the right of the focus at the origin, so we use the form with "+ e cos(}" in the denominator.
ed l2 . 4 4 (See Theorem 6 and Figure 2.) An equation is r = (} = (} = (}. 1 2 +cos ' 1+ ecos 1+ 2 cos
3. The directrix y = 2 is above the focus at the origin, so Wf! use the form w ith "+ e sin(}" in the d enominator. An equation is ed _ 1.5(2) _ 6 1 + e sin ()  1 + 1.5 sin(}  2 + 3 sin(}·
_ r 
5. The vertex (4, 3Jr/2) is 4 units below the focus at the origin, so the directrix is 8 units below the focus ( d = 8), and we use the form with " e sin(}" in the denominator. e = 1 for a parabola, so an equation is
ed 1(8) 8 r = 1  e sin () = 1  1 sin (} = 1  sin (} · 7. The directrix r = 4 sec (} (equ ivalent to r cos(} = 4 or x· = 4) is to the right of the focus at the origin, so we w ill use the form with"+ e cos()" in the denominator. The distance from the focus to the directrix is d = 4, so an equation is
~{4)
ed
r = 1 + ecosB = 1 + ~cos(}· 4
9. r =
5  4sinB
2
2=
4 2 +cos((
5 4 5 where e = .1 and ed = .1 = / 1/5 . 1  ~sinB' 5 &
. l/
=> d = 1. y
(4 , 7r/2)
(a) Eccentricity = e = ~ (b) Since e = ~
<
1, the conic is an ellipse.
(c) Since" e sin (}" appears in the denominator, the directrix is below the focus
IFll =
at the origin, d = (d) The vertices are
11.
r=
2
1, so an equat'ion of the directrix is y =  1.
F2
(4, ¥) and(~ , 3; ) .
1/3
3 + 3 sinB · . 1 /~ =
l +
2/ 3 1
X
1 · y=1 ( 4 3'1T)
'2
sinB ' wheree=1and ed= 3
=> d= ~. y
y = 2/ 3
(a) Eccentricity= e = 1 (b) Since e
= .1, Lhe conic is a parabola.
(c) Since"+ e sin(}" appears in the denom inator, the directrix is above the focus at the origin. d =
IFll = j , so an equation o flhe directrix is y =
(d) T~e vertex is at ( ~,
i·
¥), midway between .the focus and directrix.
@) 2012 CcngJige Leon~ing. All Rights Rc.c:n..1. May not be scanned. copied, or duplieoted. or posted to a publicly accessible website, in whole or in P"fL
CONIC SECTIONS IN POLAR COORDINATES
SECTION 10.6
13. r =
:
6+
3 2 1 6 · 1 = ( , where e = cos 9 116 1+ 3 cos0
(a) Eccentricity= e =
(b) Since e =
t
~ and e d = ~
:::}
d=
~
t
< 1, the conic is an ellipse.
(c) Since " + e cos
I
9:
e" appears in the denominator, the directrix is to the right of
the focus at the origin. d =
!Fll = ~.so an equation of the directrix is
x=21
:
(f,1T)
!
x = ~(d) The vertices are ( ~, 0) and ( that is, (
t, 1r), so the center is midway between them~
fn, 1r) .
3 1/ 4 3/ 4 3 15. r = 4  8 cos 9 . 1I 4 = 1  2 cos 9' where e = 2 and ed = 4
d  1s ·
(a) Eccentricity = e = 2
(b) Since e
= 2 > 1, the conic is a hyperbola.
(c) Since " e cos 0" appears in the denominator, the directrix is to the left of the focus at the origin. d'=
IFll = ~.so an equation of the directrix is
3 .X   s·
(d) The vertices are (  ~. 0) and
(i, 1r), so the center is midway between them,
that is, (~ , 1r).
17. (a) r =
1
; .
sm 9
, where e
= 2 and ed =
1 :::}
d
= 12 . The eccentricity
e = 2 > 1, so the conic is a hyperbola. Since" e sin B" appears in the denominator, the directrix is below the focus at the origin. d =
IFll = ~.
so an equation of the directrix is y =  ~ . The vertices are (  1, ~) and 1 (3,
3 2 3 2'~~" ) , sothe center .IS m1"dway between th l'!m, t hat 1.s, ( 3 , 2" ) .
3
(b) By the discussion that precedes Example 4, the equation . IS T
1
=
0
1  2sin ( e
;)•
3
2
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33
34
0
19. Fore
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
< 1 the curve is an ellipse. It is nearly circular when e is close to 0. As e
increases, the graph is stretched out to the right, and grows larger (that is, its righthand focus moves to the right while its lefthand focus remains at the origin.) At e = 1, the curve becomes a parabola with focus at the origin. 21.I PFI= e iPLI
=> r =e[drcos(7r9)J = e(d+rcos8)
=*
ed
=> r=....,.
r(1  ecos~) = ed
1 ecosO
X
x=d
23. IPFI = ei P ll
r(1  esin 9)
y
=> r = e[d rsin(9 1r)] = e(d+r sinO) => ed => r = ""7
= ed
.. 1 esin9
X
l
•
y=  d
25. We are given e = 0.093 and a = 2.28 x 10 8 . By (7), we have
r=
a(1  e 2 ) 2.28 x 108 (1 (0.093) 2 ] = 1 + ecos 8 1 + 0.093 cosO
27. Here 2a = length of major axis = 36.18 AU
r = 18.09(1  (0.97)
2
1 + 0.97 cos 9
]
~
1.07 1 + 0.97 cos 8
~
.
2.26 x 108 1 + 0.093cos8
::::=::::
=> a = 18.09 AU and e = 0.97. By (7), the equation of the orbit is
· d"1stance from the comet to t he sun 1s · . By (&), the maxtmum
18.09(1 + 0.97) ~ 35.64 AU or about 3.314 billion miles.
29. The minimum distance is at perihelion, where 4.6 x 107 = r = a(1  e) = a(1  0.206)
= a(0.794} =>
a = 4.6 x 107/ 0.794. So the maximum distance, which is at aphelion, is
r = a(1 + e) = (4.6 x 107/ 0.794) (1.206} ~ 7.0 x 107 km. 31.
From Exercise 29, we have e = 0.206 and a(l e)= 4.6 x 10 7 km. Thus, a= 4.6 x 10 7/ 0.794. From (7), we can write the .
.
equation of Mercury's orbit as r =a
1
e2
1 + ecos 8
•
. So smce
2
dr a(1 e )esin8 = d9 (1 + ecos8) 2
=>
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CHAPTER 10 REVIEW
0
35
the length of the orbit is
1 2
L=
,.
Jr 2
+ (dr/d0)2 d(J =
a(l e2 )
1
2 +2 +ecos
2
,.
.Jl(t e
0
0
~)~sO dO:::::: 3.6 x 108 km
This seems reasonable, since ,Mercury's orbit is nearly circular, and the circumference of a circle of radius a is 21ra :::::: 3.6 x 108 km.·
10
Review CONCEPT CHECK
1. (a) A parametric curve is a set of points of the form
(x, y) = (J(t), g(t)), where f and g are continuous funct ions of a
variable t. (b) Sketching a parametric curve, like sketching the.graph of a function, is difficult to do in general. We can plot points on the curve by finding f(t) and g(t) for various values oft, either by hand or with a calculator or computer. Sometimes, when
f
and g are given by formulas, we can eliminate t from the equations x = j(t) andy
= g(t) to get a Cartesian equation
relating x andy. It may be easier to graph that equation than to work with the original formulas for x andy in terms oft.
2. (a) You can find :
as a function oft by
calculating~~ = ~~J~~
[if dxjdt
f:
0].
(b) Calculate the area as I: y dx = I: g(t) j'(t)dt [or I; g(t) !' (t)dt ifthe leftmost point is (!({3), g({J)) ~ther than (f(a),g(a))].
3. (a) L = I: J(dxjdt)2 (b)
+ (dy j dt)2 dt =I:
+[.g'(t))2 dt
vlf'(t))2
s =:: I: 27ryJ(dxfdt)2 + (dyfdt) 2 dt =I: 21rg(t) v lf'(t)F + [g'(t)J2 dt
.
4. (a) See Figure 5 in Section 10.3. (b) x
= rcosO,
y = r sinO
(c) To find a polar representation (r, 0) with r ;?. 0 and 0
s 0 < 27r, firs t calculate r =·Jx2 + y2. TI1en 0 is specified by
cosO= xjr and sinO= yjr. d
5. (a) Calculate
Jx =
!:.._ (y)
dy
~ = ~ dO
dO (x) 2
(b) calculate A= I: !r dO= I:
(c) L =I: J(dx/d8.)2
!:.._ (r ::;in 8)
=
. : : l J:;: 
d() (rcosB)
(~) sin(} + r
dr) (dO
cos(}
. ,
wherer = f(O).
cos ()  r sm (}
tlfCOW d8
+ (dy/d0) 2 dO= I:
Jr 2
+ (dr/d0) 2 dO= .r: J[f(0))2 + [!'(0)]2 dO
6. (a) A parabola is a set of points in a plane whose distances from a fixed point F (t11e focus) and a fixed line l (the 'd irectrix) are equal. (b) x 2
= 4py; y 2 = 4px ®
2012 Ccngagc L.carning. All Ri~llls Reserved. May 001 be liCUnncd. copied. or duplicated. or posted 10 a publicly ncccssible website. in whole or in part.
36
PARAMETRIC EQUATIONS AND POLAR COORDINATES
CHAPTER 10
D
7. (a) An ellipse is a set of points in a plane the sum of whose distances from two fixed points (the foci ) is a constant. 2
(b)
X
a2
2
+ _Y_ = 1. a2  c2
8. (a) A hyp~rbola is a set of points in a plane the difference of whose distances from two fixed points (the foci) is a constant.
This difference should be interpreted as the larger distance minus the smaller distance. 2
2
(b) ::__  _ Y_
c2 a2
a2
=±
(c) y
=1
Jc2 _ o,2 a
x
9. (a) Jfa conic section has focus F and corresponding directrix l, then the eccentricity e is the fixed ratio /PFI / /Pl / for points
P of the conic section. (b) e
< 1 for an ellipse; e > 1 for a hyperbola; e = 1 for a parabola.
(c) x = d: r =
ed . ed · . x =  d:.r = . y = d: 1 + e cos (} 1  e cos (}
1'
=
ed ed . . y =  d: r = :::: 1 + e sm 8 1  e sin 8 ·
TRUEFALSEQUIZ 1. False.
3. False.
= (t but its graph bas a vertical tangent when t = 1.
Consider the curve defined by x = f(t)
1) 3 and y= g(t)
= (t 1)2 : Then g'(t) =
2(t  1), so g'(1) = 0,
Note: The statement is true iff' (1) =I= 0 when l (1) = 0.
For example, iff (t) = cost and g( t) = sin t for 0 ::; t :S 4rr, then the curve is a circle of radius I, hence its length 4
is 2rr, but f 0 " J [!'(t)]2
+ [g'(t))2 dt = .{04 " J(
sin t) 2 +(cos t) 2 dt =
f04"
1 dt = 4rr, since as t increases
from 0 to 4rr, the circle is traversed twice. . 5. True.
The curve r = 1  sin 28 is unchanged if we rotate it through 180° about 0 because 1  sin 2(8 + rr) = 1  sin(28 + 2rr) = 1  sin 28. So it's unchanged if we replacer by r. (See the discussion
after Example 8 in Section 10.3.) In other words, it's the same curve as r = (1 sin 20) = sin 28  1. 7. False.
The first pair of equations gives the portion of the parabola y = x 2 with x ;::: 0, whereas the second pair of equations traces out the whole parabola y = x2 •
9. True.
By rotating and translating ~he parabola, we can assume it has an equation of the form y = cx2 , where c > 0. Tile tangent at the point (a, ca2 ) is the line y  ca2 = 2ca(x  a); i.e., y = 2cax  ca2 • This tangent meets the parabola at the points [since c
(x, cx
2
)
(x, cx2 ) where cx 2 = 2cax  ca2 • This equation is equivalent to x2 = 2a~  a 2
> 0]. But x 2 = 2ax  a 2 <=> x 2 =
2
(a, ca
).

2ax + a 2 = 0 <=>
(x a) 2 = 0 <=>
x =a <=>
This shows that eacb tangent meets the parabola at exactly one point.
© 1012 Cc:nKogc Lc.aming. All Rights Reserved. Mny nul be scnnnccl, copied, or dupllcntcd, or posted to a publicly accessible website, in whole or in part.
CHAPTER 10 REVIEW
0
37
EXERCISES 1.
+ 4t, y = 2  t,  4 ~ t ~ 1. ~ = 2  y, SO 2 2 2 X = {2  y) + 4{2 y) = 4 4y + y + B  4y = y  By+ 12 {::} x + 4 = y 2  By+ 16 = (y  4) 2 . This is part of a parabola with vertex X
y
= t2
( 4, 4), opening to the right. X
3. y
= sec() = ~ () = .! . Since 0 ~ () ~ 1r /2, 0 < x ~ 1 and y ;::: 1. COS X
y
This is part of the hyperbola y = 1/x.
\ ,1,1),00 0
5. Three different sets of parametric equations for the curve y = (i) (ii)
X X
= t, y
= t
4
,
IX are
= Vt 2 y =t
(iii) x = tan2 t, y = tan t, 0 ~ t <
1r
/2
There are many other sets of equations that also give this curve. The Cartesian coordinates are x = 4 cos
7. (a)
y = 4sin ;,. = 4(
(b)
Give~ x
=  3 and y= 3, we haver =
( 3, 3) is in the second
= 4 ( ~) = 2 and
4) = 2 J3, that is, the point (  2,2 J3).
J( 3) 2 + 32 = JIB = 3 v'2. Also, tan() = ~ '*
quadrant,(}= ' 34".
1 111") and ( 3 '2 ( 3 y '2 ~, 4 v. .L.i,
2 ;
Thus, one set of polar coordinates for ( 3, 3) is (3
tan() =
V2,
3,;'"),
3
~ , and since
and two others are
7 ,.) 4 .
9. r = 1  cosO. This cardioid is symmetric about the polar axis.
r 2
(2, 7T) 7T
®
27T 0
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38
0
PARAMETRIC EQUATIONS AND POLAR COORDINATES
CHAPTER 10
r
11. r = cos 30. This is a threeleaved rose. The curve is traced twice.
13. r
= 1 +cos 2e. The curve is
sym·metric about the pole and
axes.
1 1T
1T
2
15. r =
(2,0)
(2, 1r)
both the horizontal and vertical
3
=}
t + 2sin£1
e= 2
31T
2
> 1, so the conic is a hyperbola. ·de = 3 =>
d = ~ and the form "+2 s in(}" imply that the directrix is above the focus at the origin and bas equation y
17. x+y=2
= ~The vertices are (1, ~) and (  3, 3; ) .
rcosll+rsinll = 2
¢:>
r(cos0+sin 0)=2
¢:>
2 .
¢:>
r=   

cos(J + s in£1
19. r = (sin £1)/0. As B. ±oo, r __, 0.
As (} __, 0, r .. 1. In the first figure, there are an infinite number of x intercepts at x
= 1rn, n
a nonzero
integer. These correspond to pole points in the second figure.
0.25  0.75
21. x = lnt, y = 1 +
2
dy
dx
When t = 1, (x , y) = (0, 2) and dyjdx 23. r = e 0
=> y
1
dy
t ; t = 1. dt = 2t and dt = t' so dx
=
dy / dt
2.
= rsin£1 = e 0 s in(} and x
= T"cos O = e  0 cosO
dy dyjd£1 *sin(} +rcosO _ e 0 sin(}+ e 0 cos(}  e 0 dx = dx j d(} = ~;cos(}  rsin(}  e 0 cosB  e 0 sinO ·  e 0 When (}
dy = 1r, dx
=
0  ( 1) 1 = 1 + 0 1
=
2t
= dxjdt = 1/t = 2t 2 · => sine cosO
= cos B+sine ·
1.
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CHAPTER 10 REVIEW
25. x
= t + sin t, y = t 
d (dy)
dt dx dxjdt
cost
0
39
dy dyjdt 1 + sint dx = d.xjdt = 1 +cos t
~
{1+cost) cost {1 +sint)( sint) 2 2 _ __ _ ___:(,_1,...+.:..__c_os_ t')2 _ _ _ __ = cost+ cos t +sin t + sin t = 1 + cost + sin t l +cost (l+cost)3 (1 + cost)3 3
2 .
27. We graph the curve x = t  3t, y = t + t + 1 for 2.2 :=::; t :=::; 1.2.
By zooming in or using a cursor, we find that the lowest point is about (1.4, 0. 75). To find the exact values, we find the tvalue at which
dyjdt=2t+1 = 0
t=~~ <=> , (x,y),;,(lj,~).
¢=>
29. x ·= 2acost acos2t sint = 0 or cost = ~
y = 2asin t  a sin 2t
~
~ ~
dx = 2asin t + 2asin2t = 2asint(2cost  1) = 0 dt
¢:>
t = O,·t, 1r, orr;. dy = 2a cost  2a cos 2t = 2a(l +cost 2 cos2 t) = 2a(1 cos t)(1 + 2 cost) = 0 dt
271" 4" t = 0'3,or3.
Thus the graph has vertical tangents where t = what the slope is where t
t.
1r
and ~">3" , and horizontal tangents where t
= 0, we use !'Hospital's Ru;e to evaluate t>0 lim ddy/jddt X t t
X
0
a
0
3"
~a
3
 ~a
~a 2
7r
 3a
0
4 7r
~a
~a 2
51r
~a
_ :.::]:a
2
(3a,O)
(a,O)
2
fu :=::; 8 :=::;
{'0 , there are two petals, one with r
> 0 and one
< 0.
33. The curves intersect when 4 cos 8
for
To determine
)C
31. The curve r 2 = 9 cos 50 has 10 " petals." For instance, for with r
4 ;.
y
2rr
3
and
= 0, so there is a. horizontal tangent there.
y :.::]:a
3
= 2;
1r
:=::;
8 :=::;
1r.
=2
~
cos 8
=~
~
8 = ±i
r = 4cos 0
The points of intersection are (2 , i) a nd (2, ~).
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=}
40
D
CHAPTER 10
PARAMETRIC EQUATIONS AND POLAR COORDINATES
=>
35. The curves intersect where 2 sin () = sin() + cos () sin 0
= cos 0 => 0 = "i, and also at the origin (at which 0 = 34rr 8=f
on the second curve).
A= I 0" 14 ~(2 sin0) 2 dO+ I:;f ~(sinO+ cos 0) 2 dO 4
= Io"
14
(1 cos 20) dO+~ I:;t (1 + sin2fJ) dO 4
L =I~ .J(dxjdt)2 + (dyjdt) 2 dt
= 1~ .J(6t)2 +
{6t2)2 dt =I: V36t_2 + 36t4 dt = I~ y'36t2v'1 + t2 dt
=I~ 6\t! v1 + t 2 dt = 6 I~ t v'f+t2 dt = 6 I 1
5
J:
= 6. ~ · ~ [u312 = 2(53 12 39. L =I:"
'
r = sin 0 + cos 8
r = 2 sin 8
= [0 ~ ~ sin20]~ 14 + [~0: t cos2BJ!~~ = ~(1r 1)
u
1 2 ;
(~du)
1) = 2(5VS 1)
Jr2 + (drjdB)2 dB = I:" }(1/0)2 + (1/0 2 )2 dB=
~ [ #+! +In (B + J B2 + 1 )] 2,. = #+I ()
7r:
L7 2
dB
"'
v' 471"2 + 1 271"
+ In ( 271" + ~ )
11"
71"
+ v'71"2 + 1
s = I 14 27ry.J(dxfdt)2 t. (dyfdt)2 dt = I 14 21r{~t3 + ~c 2 ) J(2/0) 2 + (t2 t3)2 dt 4 3 + .!c 2) . l(t2 + t  3) 2 dt = 21r f 4 (.!t 5 + !! + l t 5) dt  21r [...!..t6 + l!t = 21r J1 (.!t 3 2 v 1 3 6 2 18 6
4) 4  471 ·295 1r 1c 8 1 1024
43. For all c except  1, the curve ·is asymptotic to the line x = 1. For c
< 1, the curve bulges to the right near y =
bulge becomes smaller, until at c
=
0. As c increases, the
1 the curve is the straight line x
= 1.
As c continues to increase, the curve bulges to the left, until at c = 0 there is a cusp at the origin. For c > 0, there is a loop to the left of the origin, whose size and roundness increase as c increases. Note that the xintercept of the curve is always c.
© 2012 Ccngag~ Ltaming. All Rights R.cscrvcd. Muy not be scanned, copied, or duplicated, or po!!1etl lo n publicly accessible website, in whole or in part.
CHAPTER 10 REVIEW 2
45. ~ +
9
2
.
47. 6y2
'lL = 1 is an ellipse with center (0, 0). 8
a = 3, b = 2 J2, c = 1
=?
36y
+ 55 = 0
~
6y + 9) =  (x + 1)

41
~
2
(y  3) = [;(x + 1), aparabola with vertex (1, 3),
foci (±1, 0), vertices (±3, 0).
opening to the left, p
J
y
+X 
6(y2
0
2 2
directrix x
=  f.t
=?
focus ( ;~, 3) and
= ~. y
3
3
X
( 1,3)
0
X
49. The ellipse with foci (±4, 0) and vertices (±5, 0) has center (0, 0) and a horizontal major axis, with a = 5 and c 2
so b2 = a 2
2
c2 ·= 5

2
4

= 9. An equation is ~5
+ Y
9
= 1. 2
.
The asymptote y 10b2 = 16
::::}
= 3x has slope 3, sob = l  ::::} b2

a
3
*
a= 3b and a 2 + b2
= c2 y2
= " and so a2 = 16 
~ = L52 • Thus, an equation is  " 7215
=?
xb: = 1. ?
51 . The center of a hyperbola with foci {0, ±4) is (0, 0), so c= 4 and an equation is y 2 a
= 4,
2
·
{3b? + b2
x2
 8 5 = 1
= 42
=?
5y2 5x2 1, or  72 8
= 1.
53. x 2 = (y  100) has its vertex at (0,100), so one of the vertices ofthe ellipse is (0, 100). Another form of the equation of a
= 4p(y 
parabola is x 2 found at .(0,
b2
__
a2
_
3 9
100) so 4p(y  100)
=
(y 100)
=?
4p
= 1
__
2
401  399
2
3 9
or 25 +
.
x2
(
,== 25. So the equation of th e e ll ipse is /1 + 2
(8y  399? 160,801
55. Directrix x
p = ~.Therefore the shared focus is 3 9
8
x2
::::}
~ ) so 2c = ~ 0 ::::} c = ~ and the center of the ellipse is (0, ~ ). So a= 100 3~ 9 = 4 ~ 1 and
c2
3 9
=4
=?
y  839!))2 a2
=1
=?
=1
·
d
= 4, so e =
~
=?
ed 4  ..,.. .,  1 +ecosfJ 3+cosfJ'
r 
3 57. (a) Jf (a, b) lies on the curve, then there is some parameter value t1 such that~ = a and tr ~ =b. l ft 1 = 0, 1 + tl 1 +tl the point is (0, 0), which lies on the line y 3(1/tl )
3ti
= x.
lft1 =I 0, then the point corresponding tot=
3(1/td
x = 1 + ( 1 /t ) 3 = t~ + 1 = b, Y = 1 + ( 1 /h) 3 = 1
tt3tl+ 1 = a.
So (b, a) also lies on the curve. [Another way to see
this is to do part (e) first; the result is immediate.] The curve intersects the line y = x when
t=
t
2
::::}
..!. is given by tl
~3 1+ t
2
=
3 t 1 +t3
t = 0 or. 1, so the points are (0, 0) and ( ~, ~).
::::}
42 . D
.CHAPTER 10
(b) dy
dt
= (1 +t
PARAMETRIC EQUATIONS AND POLAR COORDINATES
3
)(Gt)  3 t2(3t (1+t3)2
2
4
= Gt  3t =Dwhen6t  3t4 =3t(.2t3 )=D ·=> (1+t3)2
)
horizontal tangents at (0, D) and (
(0, D) and ( ?'4,
t=Dort= 3'2,so thereare y.r.
V2', ?'4). Using the symmetry from part (a), we see that there are vertical tangents at
.v2).
(c) Notice that as t > 1 +,we have x> oo andy . y  {x  1) = y + x + 1 =
oo. As t> 1, we have x . oo andy
>
3t+3t 2 +{1+t3)
1+ t 3
3
=
{t+1) = 1 +t3
{t+1)
>
 oo. Also
2
t 2 t+1
>
0 as t
>
1. Soy = x  1 is a
slant asymptote. .d) dx = (I+ t 3 )(3) 3t(3t 2 ) = 3  6t3 dy 6t 3t4 dy dyjdt t(2  t 3) ( dt (1 + t3)2 (1 + t3)2 and from part (b) we have dt = (1 + t3)2. So dx = dxjdt = 1 2t3 .
y
So the curve is concave upward there and has a minimum point at (D, D) and a maximum point at ( .v2,
W). Using this together with the
X
information from parts (a), (b), and (c), we sketch the curve.
3
3
(e)
3t
x? + y 3 =
( 1+ t3
)
27t3 + 27t6 .(1 + t3)3
+ ( 1 3t2 + t3 )
.
2
'y =x  l
27t3 (1 +t3 ) (1 + t3)3
3
3xy = 3C!\3 )C~t3 ) = ( 1 ~~3 ) 2 ,sox 3 +y3 =3xy. (f) We start with the equation from part (e) and substitute x = r cos 0, y = r sin 0. Then x 3 + y 3 = 3xy r 3 cos3
0+
s in3
1'3
e ~ 3r2 cos e sin e. For
) 3(_1 cosO · cosO
r
=1 0, this gives
r=
=>
.
3 ~OS e s~: Dividing numerator and denominator cos O+ sm 0
sinO
.
by cos3 0, we obtam r =
. sm3 e 1 + cos3 0
3sec0 tan B 1 + tan 3
e.
(g) The loop corresponds to 0 E (D, ~).so its area is 
{"/
2
r
A}0
2
r / (l+ sec e tan e 0  ~ {""' u du tan 0)2d 2} {1+u 2
(3sec0 tan0)  ~ 1+tan3 0 d02}0 2
.! ("/ 2
2d02}0
2
2
2
2
3
0
3)
[let u = tan OJ
= lim l!.[~(l+u3)l]b =;! b+oo 2
3
a
2
(h) By symmetry, the area between the foUum and the line y = x  1 is equal to the enclosed area in the third quadrant, plus twice the enclosed area in the fourth quadrant. The area in the third quadrant is~ . and since y = x  1
r sine =  r cos 0 1 1 /_"'/ 2  rr/2
4 [(

=> r =  . e
. () 1 () ) sm + cos
!;JU
2
1 e' the area in the fourth quadrant is +cos
()  (3sec0 tan0) 1 +tan3
2 ]
. 1 + 2 (1) = 3 . d() CAS 1.. Therefore, the total area IS 2 2 2 2
=
© 2012 Ccngage Learning. Al l Rights Reserved. M1ty not be scanned, copied. ~r duplicated, or posted too. publicly accessible wcb~itc, in whole or in part.
=>
0
PROBLEMS PLUS
1. x =
l
t
cos u   du, y =
1
dx = 0 dt
lt
tL
sin u dx cost dy sin t .   du, so by FTCl, we have dt = t and dt =  t  . Vertical tangent lines occur when
1
tL
cost = 0. The parameter value corresponding to (x, y) = {0, 0) is t = 1, so tl1e nearest vertical tangent .
{:}
occurs when t =
~.
Therefore, tl1e arc length between these points is
(dx)
7r/2
L=
1
dt
1
2
2
+ (dy) dt
dt=
11r/Z
2
cos
1
t
+
t2
. 2t
~dt = t2
j '1f/Zdt ~ [ lnt]7f12 = ln :!!. 1
t
1
2
3. ln. terms ofx andy, we have x =?'cosO= (1 + csin B)cosB =cos O+ csinBcosB =cosO + ~csin2B and 2
y = rsinB = (1 + csinO)sinB =sinO+ csin B. Now  1 $sinO $ 1
=> 1 $sinO t csin2 0 $ 1 + c $2, so
1 $ y $ 2. Furthermore, y = 2 when c = 1 and B = ~ , while y = 1 for c = 0 and 0 =
3 ; .
Therefore, we need a viewing.
rectangle with  1 $ y $ 2. To find the xvalues, iook at the equation x =cos B + ~ csin 20 and use the fact that sin 20 ~ 0 for 0 $ B $ ~ and sin 28 $ 0 for  ~ $ B $ 0. [Because r =::, 1 + csin 0 is symmetric about the yaxis, we only need to consider

~
S 8. $
~ .)
So for 
~
$ B .$ 0, x has a maximum value when c = 0 and then x = cos 0 has a maximum value
of 1 at B = 0. Thus, the maximum value of x must occur on [0, ~] with c = 1. Then x = cos 0 + ~ sin 20 ~~ = sinB +cos 2B= sinB + 12 sin2 9
[but sin B #  1 for 0 $ 9 $
H
=?
If sin 9 = ~, then B =
=>
j~ = (2sin91)(sin9+1) = 0when sin9 =  1or~
*
and
2.1
c=t
x = cos {f + ~sin 3 = ~V3. Thus, the maximum value of xis ~V3, and, by symmetry, the minimum value is  ~ V3. Therefore, the smallest viewing rectangle that contains every member of the famil y of polar curves
r = 1 + csinB, where 0 $ c $ 1, is [~V3, ~V3] x [1 , 2].
5. Without Joss of generality, assume the hyperbola has equation :: 
~: = 1. Use implicit differentiation to get
2 2 :  Yb2y' = 0, soy' = b:x. The tangent line at the point (c, d) on the hyperbola has equation y d = b:d c (x  c). a a y . a ' 2
The tangent line intersects the asymptote y = !!. x when .!!.x  d = bzdc (x c) a a a
=> abdx a 2 d 2 = b2 cx  b2 c 2 =>
© 2012 Cengnge le>ming. All Rigllls Reserved. Moy not be: seonncd, copied, or duplicated, or posted to n publicly a=ssiblc website. in who le or in pan.
43
44
D
CHAPTER 10 PROBLEMS PLUS 2 2
=>
X =
2 2
a d b c b(ad _be)
ad + bc d h . bad +be b an t e yva 1Ue IS  b
=
a
ad 'tbe
=  a
·
. .1ar1y, t he tangent 1.me mtersects . b at (be bad, aad  be ) . Th e m1.d pomt . oftl1ese mtersect1 . .on pomts . 1 .s S1m1 y = ax 
bead) (ad+be + adb e)) = (12bc 12ad) . a2b , 2;;:= ( c, d), the pomt of tangency.
1 (ad+bc 1 a( 2  b + b • 2  Note: Jf y
=·o, then at (±a, 0), the tangent line is x = ± a, and the points of intetsection are clearly equidistant from the point
of tangency.
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INFINITE SEQUENCES AND SERIES
11 . D 11.1
Sequences
1. (a) A sequence is an ordered list of numbers. It can also be defined as a function whose domain is the set of positive integers, (b) The terms a,. approach 8 as n becomes large. In fact, we can make an as close to 8 as we like by taking n sufficiently large. (c) The terms an become large as n becomes large. In fact, we can make an as large as we like by taking n sufficiently large.
2n
. {
2 4 6 8 10 } 1 + 1' 4 + 1 ' 9 + 1' 16 + 1' 25 + 1' . . .
3. an
= n2 + 1, so the sequence IS
5. an
=
(1)n l . . { 1 1 1 1 1 } 5n , so the sequence IS 51' 52' 53 ' 54' 55 '...
7. ~
=
} 1 . {1 1 1 1 1 (n + 1)!, so the sequence IS 2!'"3!' 4f•5!' 6!'...
9. a 1
= 1, a,,+1 =
a 3 = 5a2  3
5an 3.
= 5(2) 
{1
= 5' 
{11 1
= 5a3 
3 = 5(7)  3
{
4 3 8 5 } 1, 5' 5' 17' 13 ' . . . .
1 1 1 1 } · 25' 125 '  625' 3125'... .
1
1
= 2' 6' 24 , 120, 720 '...
Each term is defined in terms of the preceding term.
3 = 7. a4
=
= 32.
as
= 5a4 
}
a 2 = 5a1
. 
3 = 5(1) 3 = 2.
3 = 5(32)  3 = 157.
The sequence is {1, 2, 7, 32, 157, ... }.
Q.4
a5
= 1:
4
a
4
=:' 1 ~;/7 = ~
,The sequence is {2,
a3 = = 1 +2/52/5 1 + a3
1 2n1
an=    .
15. {  3, 2, 1, ~, ;~, ... }. The first term is 3 and each term is  ~ times the preceding one, so a,. = 3 ( ~
wegetan
2
7•
~. ~. ~' ~ •... }.
The denominator of the nth term is the nth positive odd integer, so
17. { ~, ~,%, 1: , 25~, ... }.
=
The numerator of the nth term is n 2 and its denominator is n
f 1 .
+ 1. Including the alternating signs,
= (1)n+l _n2_ _
®
n+1
2012 Ccngage Learning. All Rights Reset\'ed. Moy not be scnnned. copied, or duplicated, or posted to u publicly accessible website, in whole or in par1.
45
46
D
19.
3n an= 1+ 6n
n
'
1
0.4286
2
0.4615
3 4
0.4737
5
0.4839 0.4865
21 .
10
5
n
It appears that lim an = 0.5. n+oo
lim n+oo
~ = lim 1 + 6n
n+oo
(3n)jn (1 + 6n)jn
=
lim n+oo
= ~ = .!
3 1/n + 6
6
2
0.4909 0.4918
.....
an= 1 + (U~
n
1 2
0.5000
3
1.2500 0.8750
4
1.0625
5 6 7
0.9688 1.0156
8 9
1.0039 0.9980 1.0010
0
10
5
n
I
It appears that lim a,. = 1.
I
nco
0.9922
10
=1
.
0
0.4884 0.4898
8 9 10
.........
0.5 0.4
0.4800
6 7
23. an
INFINITE SEQUENCES AND SERIES
CHAPTER 11
lim (1+( ~f) =lim 1+ lim ( ~f=1+0 = 1 since
n oo
ntoo
noo
""
(0.2)", so lim a,. = 1  0 = 1 by (9) . Converges n  oo
3 + 5n2 (3 + 5n2)jn2 5 + 3/n2 5+ 0 25. a,, = n + n 2 = (n + n 2)/n2 = 1 + l /n, so a,, + i + 0 = 5 as n + oo.
Converges
27.. Because the natural exponential function is continuous at 0, Theorem 7 enables us to write lim an= lim e 1 fn = e1imnoo( l /nl = e0 = 1. Converges
,.,_ex:>
noo
2mr h lim' b 1' (2mr)/n lim 271' 271' 71' s· . . .,. b 29. Ifb n =  , t en n = 1m ( )/ = / =  =  . mce tan IS contmuous at 4 , y 1 + 8n n+oo n+oo 1 + 8n n n>oo 1 n + 8 8 4 Theorem 7, lim tan ( n >oo
31. an =
2 2 n1r ) n 71' ) = tan ( lim 1 + 8n n+oo 1 + 8n n2 j,;r;J
n2
vn3 + 4n = ../n3 + 4n/v'n3 =
.,jii
=tan~ 4
= 1.
Converges
Jl + 4jn2 , so an + oo as n+ oo since
lim
.,jii = oo and
n>OO
lim j l + 4/ n 2 = 1. Diverges
1\+00
33. lim
n+oo
Jan! =
lim
n+oo
I(%"I = 21 2vn
lim
n+oo
1 1 =  (0) = 0, so lim an = 0 by (6). Converges n 112 noo 2
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SECTION 11.1
35. a,. = cos( n /2).
SEQUENCES
0
47
This sequence diverges s ince the tenns don 't approach any particular real number as n> oo.
The terms take on val4es between  1 and I. 1 _ (2n  1)! = (2n  1}! · = · ___. 0 as n> oo. 37 · an  (2n + 1)! (2n + 1}(2n}(2n 1)! (2n + 1}(2n) 1+e
2n
' >
en e
41. an = n 2 e
43. 0
< 
n
2
n = . Smce en •
2
x2
•
lim 
X+00 e X
cos n < _..!:.._ · [since 0 2n  2"
. ( 1/ n ) 45. an = n sm
n
0 as n> oo because 1 + e 2 n
>
C onverges
1 and e"  en + oo. Converges
2x 11 2 H = lim  = lim  = 0, it follows from Theorem 3 that lim an= 0 . Converges XtCO eX XtOO e:t noo
82 1 :s; cos2 n :s; 1], so since n>oe> lim = 0, { 'c 0 n n } converges to 0 by the Squeeze Theorem. " ·
2
2
1.tm sin t / n n) . s·mce lim sin(1/x) / = sin(1/ = to+ x ..... oo 1 X t 1
[w here t = 1/ x ]
= 1, .tt .,•OIIows ·from Theorem 3
that {an } converges to I .
= lim x > oo
lim
!too
( 2)"' = 1+ 
X
.
lim e 1n Y = e 2 , so by Theorem 3, lim
:roo
n +oo
2 1 + 2/x
( 2)" = 1+ n
2
=2
::::}
e 2 • Converges
2
2 2 + 1/n 49. an = In(2n + 1) ln(n + 1) =In ( 2n n 2 ++ 1) = In .(2 + / n ) + In 2 as n > oo. 1 1 1 2
Converges
51. a.,, = arct an(ln n). Let f(x) = arctan(ln x). Then lim f(x) = ~since In x + oo as x> oo and arctan is continuous. x  oo
Thus, lim a ,= n+oo
lim f(n) =
n +oo
~
Converges
53. {0, 1, 0 , 0, 1, 0, 0, 0, 1, ... } diverges since the sequence takes on only two values, 0 and I , and never stays arbitrarily close to either one (or any other value) for n sufficiently large. n! 55. an= 2 n
57.
1
2
3
(n 1)
= 2 · 2 · 2 · · · · ·  2
n 1 n · 2 2': 2 · 2
2~~
(for n
> 1]
=
n
4> oo as n+ oo, so {an} diverges.
From the graph, it appears.that the sequence converges to 1. { ( 2/e)n} converges to 0 by (7), and hence {1 + ( 2/e)" }
.. . . .. . .
converges to 1 + 0
= 1.
0 '~~~~ 21
®
2012 Ccng;,gc Lc011T1ing. AU RighiS R.escn·l.d. May not be scanned, copied. or duplicated. or posted to a publicly accessible website, in whole or in part.
I
48
D
CHAPTER 11
'
INFINITE SEQUENCES AND SERIES
From the graph, it appears that the sequence converges to t.
59.
As n+ oo,
3/ n 2 +2
.. ...
8 + 1/ n
=>
so lim an =~ · n +oo
0 '~~"'./ 21 2
61.
2.~..
..
From the. graph, it appears 'that the sequence {an} ,;_ {
~ ~o~: } is
divergent, since it osci llates between 1 and 1 (approximately). To n2
, then prove this, suppose that {a,.} converges to L. If bn =  2
1+n
..
{bn} converges to 1, and lim abn n+oo
n
=!::.1 = L. But an bn
= cosn, so
lim abn does not exist. This contradiction shows that {an} diverges.
2
oon
•
From the graph, it appears that the sequence approaches 0.
63.
O < a,.= :<::;
1 n · (1) · (1) · · · · · (1) 2
So by the Squeeze Theorem, {
0~~~~~J IO
65. (a) a,. = 1000(1.06)."
1 · 3 · 5 · · · · · (2n 1) 1 3 5 2n 1 n =  ·  ·  · · .. · (2n) 2n 2n 2n 2n
=
1 n + 0 as n + oo 2
1 · 3 · 5 · · · · · (2n  1)} ( n)" converges to 0. 2
=> at = 1060, a2 = 1123.60, a3 = 1191.02, a4 = 1262.48, and as = 1338.23.
(b) lim a,. = 1000 lim (1.06)", so the sequence diverges by (9) with r = 1.06 > 1. n  oo
n+oo
67. (a) We are given that the init!al population is 5000, so Po = 5000. The number of catfish increases by 8% per month and is
decreased by 300 per month, so Pt P,.
= 1.08Pnt 
= Po + 8%Po 
69.
lim xr"'
300, and so on. Thus,
= 5000, we get P 1 = 5100, P 2 = 520.8, P 3 = 5325 (rounding any portion of a
= 5451, P5 = 5587, and P6 = 5734, which is lhe number of catfish in the pond after six monlhs.
If lrl ~ 1, then {r"} diverges by (9), so {nrn} diverges also, since lnr''l = n lr"l :C  t OO
= l.08P1 
300.
(b) Using the recursive formula with Po catfish), P,.
300 = 1.08Po  300, H
= :coo lim =..._ :lb ~ ( r:Z: X+00

In\
T T%
= lim ___!__I ."' :::: 0, so lim nr" %+00 nT noo
~
lr''l· If lrl <
1 then
= 0, and. hence {nr"} converges.
whenever lrl < 1.
® 2012 Ccngogc Lcoming. All Rights Rc:sen'Ctl. May not be S
SECTION 11.1
SEQUENCES
D
49
71. Since {a,.} is a decreasing sequence, an > an+l for all n ~ 1. Because all of its terms lie between 5 and 8, {an} is a
bounded sequence. By the Monotonic Sequence Theorem,'{an} is convergent; that is, {an} has a limit L. L must be less than 8 since {an} is decreasing, so 5 $ L
73. a,. =
n 2
1
< 8.
' d . . IS ecreasmg smce an+1 =
+3
bounded since 0
1 1 2 (n + 1) + 3 = 2n + 5
1
< 2n + 3
= a 1, "tOr each n
~ 1. The sequence is
i fo r all n ~ 1. Note that a1 = t·
75. The tenns of an = n( 1)n alternate in sign, so the sequen~e is not monotonic. The first five terms are  1, 2,  3, 4, and 5. Since lim Ia.. I = lim n n. +oo
n+oo
n
77. an= 
n2
= oo, the sequence is not bounded. .
.
( )
defines a decreasmg sequence smce fo r f x = 

+1
.
x
x
for x ~ 1. The sequence is bounded since 0
< a"
2

+.1
'( ) (x ,f x =
2
+( 1)(1))2x(2x) x2 + 1
lim a,
= n+oo lim 21  <112 ") =
21
1  x2 <0 (x2 + 1)2 
$ ~ for all n ~ 1.
79. For { v tn 2, y rnFn 2 y 2, y. I 2 y rnFn 2 v 2, .. . } , a1· _ 2 112 , a2 _ 2 3/4 , aa _ 2 7/B , ... , so an _ 2 (2"1)/2"
noo
=
= 2 1(1/2">.
= 2.
Alternate solution: Let L = lim an. (We could show the limit exists by showing that {an} is bounded and increasing.) n+oo
Then L must satisfy L = ~
81 . a 1 = 1, an.+l
=3 
.
=> L
.
2
= 2L
=> L(L  2)
= 0. L =/= 0 since the sequence increases, soL = 2.
1 . We show by induction that {an} is increasing and bounded above by 3. Let Pn be the proposition an

< an < 3. Clearly P1 is true. Assume that Pn
that an+l > an and 0
1 1   · >  .Now an+2 an+l a,., above by 3, so 1 = a 1
=3
1 1   > 3an+l a.,
= an+l
~
is true. Then an+l > an
Pn+l·
an+l
=>
an
This proves that {an} is increasing and bounded
< an < 3, that is, {an} is bound~d , and hence convergent by the Monotonic Seque11ce Theorem. => £ 2
If L = lim an, then lim an+l = L also, soL must satisfy L = 3 1/ L n+oo
=> 1 < 1
n+oo

3L + 1 = 0
=> £
_ 3±.,/5

2
•
But L > 1, soL=¥. 83. (a) Let a,. be the number of rabbit pairs in the nth montl1. Clearly a1 = 1 = a2. In the nth month, each pair that is 2 or more months old (that is, a,.2 pairs) will produce a new pair to add to the an1 pairs already present. Thus, an= qn 1 + a n2, so that {an}= {fn}, the Fibonacci sequence. ( b) an= f,.+ 1
f ,.
fn fn1 + fn  2 an1 =   = fn  1 fn 1
then L = . lim an 1 and L n+oo
= 1 + .f,. 2 = 1 + f n1
= nlim a,. 2, soL must satisfy +oo ·
1
L = 1 + L
1
1
/ ·= 1 + . If L = lim an, f n 1 fn2 an2 noo
=> L 2  L  l
=0
=> L =
[s ince L must be positive]. ® 2012 Ccngagc Lcoming. /\11 Riglns Rcsc~rvcd. May no1 be scnnncd. copied, or duplicated, or poi!.cd loa publicly occ<..ssiblc wcbsilc, in whole or in parL
¥
50
D
CHAPTER 11
IN_FINITE SEQUENCES AND SERIES
50
85. (a)
From the graph, it appears that the sequence {
~~ }
5
converges to 0, that is, lim n 1 = 0. n c::o n.
0 '~~~~L~ 10 O.Or3_ _ _ _ _ _ _ _ _ _ _ _ _"'
(b)
y= 0.1 7.5
y = 0.001
"''~~'~'
9.5 0'<:::::============~ l5.5
12.5
0
From the first graph, it seems that the smallest possible value of N corresponding to c = 0.1 is 9, since n 6 / n!
< 0.1
whenever n ~ 10, but 96 / 9! > 0.1. From the second graph, it seems that for c = 0.001, the smallest pos~ible value for N is 11 sin~e n 5 fn ! < 0.001 whenever n ~ 12. 87. Theorem 6: If lim
n +oo
lanl = 0 then n+oo lim lanl = 0, and since ian I ~an ~ Ia,. I, we have that lim an= 0 by the noo
Squeeze Theorem. 89. To Prove: If lim an = 0 and tl. tOO
Proof: Since
{bn} is bounded, then nlim (a.,b,.) = 0. + oo
{bn} is bounded, there is a posi!ive number M
all n ~ 1. Let f: > 0 be given. Since lim an n~oo
la,.bn 
Ol = lanbnl = lanl lbnl ~ lanl M
lim (anbn)
n t oo
91 . (a)
=
such that lb,.l
~
M and hence,
lanl lbnl ~ lanl M
= 0, there is an integer N such that ian  01 < Ia,. OlM < ~ · M
=
f:
for
Me if n > N . Then.
for all n > N . Since f: was arbitrary,
= 0.
First we show that a > a1 > b1 > b. a 1  b1 =
ai
b
.Jab =
~ ( ~ 2Vllb +b) = ~ ( y'a 
a a1 = aHa+ b)= Ha  ~) > 0 and b b1 way we can show that a1 ak
= b
/bf > 0 .Jab=
[since a > b]
=>
a1
>
b1 . Also
Vii( Vb  01) < 0, so a >.a1 > b1 >b. In the same
> a2 > b2 > b1 and so .the given assertion is true for n = 1. Suppose it is true for n = k, that is,
> ak+l > bk+ l > bk. Then
® 20J2 Cengoge Learning. All Rights Rescr...cd. Mny not be scanned, copied, or dupl icated. or posted to t1 publicly accessible wcbsilc, in whole or in putt.
SECTION 11.2
ak+l  a1.:+2
= ak+l Hak+l + bk+l) = ~(ak+l  bk+l) > 0, and
bk+l  bk+ 2
= bk+ 1 
so the assertion is true for n = k (b) From part (a) we have a
= ~ ( ~ Jak+l ) < 0
Jak+lbk+l
=*
a 1.:+1
SERIES
0
51
> ak+2 > bk+2 > bk+l•
+ 1. Thus, it is true for a ll n by mathematical induction.
> a,. > an+ 1 > bn+1 > b.. > b, which shows that both sequences, {an } and {bn}, are
monotonic and bounded. So they are both convergent by the Monotonic Sequence Theorem.
. b,. = {3 . Then lim an+l = lim a,. +b., (c) Let lim an =a and Itm n+oo n+oo n  oo n+oo 2
2a: =
Q
93. (a) Suppose p2
(b)
+ {3 =>
= {3.
{Pn} converges top. T hen Pn+l =. ~ a+p...
+ ap =
Pn+l =
Q
"*
bp
p(p + a  b) = 0
~ = (~);n < (!!.)Pn + 1
a + pn
(c) By part (b), P1
<
n..oo
....2::
a
a
=>
n+00
a
bp a+p
=>
p=
> 1.
y
P
< ( ~ )P2 < ( ~
y
Po, etc. In general, Pn
< (~
r
Po.
(!!.)n ·po = 0 since b < a. [sy (7), lim rn = 0 if ; 1 < r < 1. Here r = !!.a E (0, 1) .J a 11.+00
(d) Let a < b. We first show, by induction, that ifpo For n
=>
=>
p = 0 or p = b  a.
since 1 + Pn
(~)Po. P2 < ( ~ )P1 _< ( ~
so lim Pn $ lim
a: +f3
 => => a : = 2
= 0, we have Pl 
po
bpo
= a +Po
po =
Now we suppose the assertion is true for n
< b  a, then Pn < b  a arid Pn+l > Pn · Po(b  a a+po
Po) > 0 since Po < b a. So Pl > po. ·
= k, that is, Pk < b a and Pk+l > Pk · Then
a(ba  Pk)  b _ a_ ~ __ a(b  a)+ bp~.; ap~.;  bpk __ __.!.. _ ___...!~ b a  Pk+l a+pk a+pk a + pk bpk+l  Pk+l + Pk+l
Pk+l
< b
Pk+2
> ·Pk+l· Thus, the assertion is true for n = k + 1.
a. And Pk+2  Pk+l =
a
A similar proof by induction shows that ifpo
= Pk+l(b a 
Pk+l) a+ Pk+l
> 0 because p~; <
b  a. So
.
> 0 smce Pk+l < b  a. Therefore,
It is therefore true for all n by mathematical induction.
> b a, then Pn > b a and {Pn}
is decreasing.
In either case the sequence {Pn } is bounded and monotonic, so it is convergent by the Monotonic Sequence Theorem.
rt then follows from part (a) that lim Pn = b a. noo
11.2 Series 1. (a) A sequence is an ordered list of numbers whereas a series is the sum of a list of numbers. (b) A series is convergent if the sequence of partial s ums is a convergent sequence. A series is divergent if it is not convergent. (i) 2012 Ccngagc L<.1ming. All Rights Rcscm..1. May nol he scanned. CO)Jied. or duplicated. or poslcd loa publicly accessible wcbsilc. in whole or in p:111.
52
0
CHAPTER 11
INFINITESEQUENCES AND SERIES
00
3.
I: an = noo lim s, =
n=l
00
5. For
1
lim [2 3(0.8)n)
n.oo
1
I: 3 • a,, = 3· n=l n n
s~ = 83 +a~ ~
ss = sr + as
~
81 = a1
1.1777, ss =
84
=
= ntoo lim 2 
3 lim (0.8)" = 2  3(0) n+oo
1
=2
1
13 = 1, 82 = s1 + a2 = 1 + 23 = 1.125, sa =
+as~
1.1857, so = ss + ao
~
1.1903, s 7 =
8G
82
+ aa ~ 1.1620,
+ a 7 ~ 1.1932, and
1.1952. It appears that the series is convergent.
' 7. For s3
n
oo
n
I: ~·a,, = r.:;· n=l 1 + yn 1 + yn s2 +as~
=
2.4265,
sr = so+ ar:::::: 8.9644,
S4
St
= a1
= sa+ a4
ss =
::::::
87 +as~
1 i;= 

3.7598,
8::,
1+
J1
=
= 0.5 82
'
84
+ a5
= St + a2 =
::::::
5.3049,
86
2 0.5 + ~ ~ 1.3284, 1 + v2 =
s5
+
a 6 ::::::
7.0443,
11.0540. It appears that the series is divergent.
9.
n
8n
1
Ot  +;++..+i ll
 2.40000
2
 1.92000
3
 2.01600
4
 1.99680
5
 2.00064
6
 1.99987
7
2.00003
8
1.99999
9
 2.00000
10
2.00000
3
From the graph and the table, it seems that the series converges to  2. In fact, it is a geometric · ·th 2 d I · · . ~ sen es w1 a=  .4 an r =&·so tis sum IS 6 . n=l
(
12 )n 5
=
2.4 2.4 ( 1) =  =  2. 1  & 1.2
Note that the dot corresponding to n = 1 is part of both {an} and {Sn}.
TI86 Note: To graph {an} and { sn }, set your calculator to Param mode and DrawDot mode. (Draw Dot is under GRAPH, MORE, FORMT (F3).) Now under E (t) = make the assignments: xt1=t, yt1=12/ ( 5) t, xt 2=t , yt2=surn seq ( ytl, t, 1, t , 1) . (sum and seq are under LIST, OPS (F5), MORE.) Under WIND use 1 , 10, 1, 0, 1 0, 1,  3, 1, 1 toobtain a graph s im ilar to the one above. Then use TRACE (F4) to see the values.
10/~
11.
n
Sn.
1
0.44721
2
1.15432
3
1.98637
4
2.88080
5 6 7
3.80927 4.75796 5.71948
8
6.68962
9
7.66581
10
8.64639
0....___....___
The series
f:
n=l
__.._ _..__
__..__ _..__, 11
~ diverges, since its terms do not approach 0. n2 + 4
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SECTION 11.2
SERIES
0
13.
s.,
n
0.29289
1 2
0.42265
3
0.50000 0.55279
4
0.59175
5 6
0.62204 0.64645
7
o'~~~~~~~~~L~~
From the graph and the table, it seems that the series converges.
0.66667
8 9
0.68377 0.69849
10
so
00 (1
15. (a) lim an = lim noo
noo
(b) Since lim an = 1 ;
64 9

(
1) = ·
1   
1
y'kTI
~ = ~.so the seqitence {an} is convergent by (11.1.1). 3n + 1 3 ~ =I 0, the series
n +oo
17. 3  4 +
1)
I:      = lim Vn v'n + 1 k oo
n=l
f: an is divergent by the Test for Divergence.
n= l
+ ·· · is a geometric series with ratio r = l Since lrl = ~ > 1, the series diverges.
19. 10  2 + 0.4  0.08 + · · · is a geometric series with ratio 
21.
u
2 10
=
i· Since lr\ = k < 1, the series converges to
f: 6(0.9)nl is a geometric series with first term a = 6 and ratio r = 0.9. Since lrl = 0.9 < 1, the series converges to n=l
a 6 6   =   =  = 60. 1 7· 10.9 0.1 23.
f: (~ 3~nl 4
n=l
converges to
25.
oo I: n=O
27. 1 3
1
=
1
~ f:· (~) n . The latter series is geometric w.ith a= 1 and ratio r =~.Since lrl = ~ < 1, it 4
(
4
n= l
1
/ ) =
 3 4
1 oo (?T) n·
?Tn
= 
n+l
3
3
I: 
n= O
3
IS
~.
Thus, tbe given series converges to
9
12
15
=
t.
. . . ?T . . . ageometnc. senes w1th rat1o r =  . Smce \rl > 1, the senes dtverges. 3
+ 1 + 1 + 1 + 1 + · · · = 6
(!) (~)
~ 1 1 ~ 1 Th'IS ts . a constant mu It1p . Ie o f the d'tvergent harmoruc . senes, . so L   =  L  . 3 n= l n 3n
n= l
it diverges.
1 . by the "'.est ".or o·1vergence smce . lim. an 29. ~ L., n    dtverges n=l
3n  1
noo
lim n  1 = 1 r = noo 3n  1 3
....£
0.
© 2012 Ccnll"lle l earning. All RighiS Reserved. May not be scanned. copied. or duplicotcd. or posted loa publicly a<:<:essiblc website, in wbolc or in part.
53
54
0
CHAPTER 11
INFINITE SEQUENCES AND SERIES
31. Converges. 00
00
00
1 + 2" ( 2) "] L:=L: 1 + 2") =L: 1 ) " +n=1 3" n=1 3n 3" . n=1 3 3 (
~
 ! 2~ +~ 1  2/3  2 +  2
 1  1/ 3 00
33.
L: o/2 = 2 + v'2 + ?12 +
[sum of two convergent geometric series]
[ (
~+
· · · diverges by the Test for Divergence since
n=l
= noo lim o/2 = lim 21 / n = 2° = 1 =/= 0. noo
lim a,.
n.oo
35.
37.
E In ( 2nn ++11 2 2
n=l
)
.
diverges by the Test for Divergence since
E (~)" is a geometric series with ratio r = I ~ 1.047. It diverges because lrl ~ 1.
k=O 00
39.
41 .
L: arctan n diverges by the Test for Divergence sincen.+oo lim an = lim arctan n = %=/= 0. n=l n+oo ~ 6
n= l
1 e
~
=~ 6
n=1
1/e 1  11e
1 (  )" 1·s a geometric · senes · w1t · h fi rst term a e
e e
1/e 1  11e
1 e 1
to   =   ·  = .By Example 7, oo
(1
n~1 en +
1
·)
n(n + 1)
00
1
""
· 1
1 1 · r =  . s·mce Ir I =  < 1, the senes · converges = 1 and ratio e
oo
L: (
e
1
· = 1. Thus, by Theorem 8(ii),
)
n=1 n n + 1
·
1
1
= n~l en + n~l n(n + 1) = e 1 + 1 = e 
43. Using partial fractions, the partial sums of the series '
e
e 1 1 + e 1
e
= e
1·
E+ are n 1
n =2
n L: . (t 
s,. =
i= 2
=
(1
2. 1)(t + 1)
!) + (! !) + (! !) + ... + ( ' 1 3 2 4 3 5 n3
. senes . an d sn . sum 1.s a te Iescopmg Th 1s Thus,
E i:= n  1
n=2
45. For the series
oo :E
n=1
l~
n+oo
Sn
3
n(n+3)
n ( 1 ) = i=2 L: .t 11 . t +1
=
1 = 1 + 12 ~ n 1
lim (1 n+00
t(1 + 3)
i=1
~
3 l'tm Bn = l'1m n ( n + 3) = noo n oo
(1 + 21 + 31 
)
[using partial fractions). The latter sum is
t+3
\
= 6 n= l
n
+!2  n1 1  .!.) = ~n 2
+ (l2 ....: l) + (13  l) + (l4  l) + ... + (1  l) 4 5 6 7
TIms,
1 ·
n 3 n (1 , s,. = :E . . = L: :  . 1
i =1
.!.)
_ 1) + ( 1 n 1 n2 n
(1  l) + (1  _1 ) + (1  _1 ) + (.!.  4) n3
n
n 2
1 n+1 1 1 + 21 + 31
 +1 n
1
 + 2 1l

n + 1
1
n+ 2 
n1
1
n+3
[e t l . senes ' ] escopmg
1 ) = 1 + 21 + 31 = 11 . 6
..,+ n 3
n+2
converges
© 2012 Cengagc Learning. All Righ!S Reserved. Muy not be scanned. copied. or duplicotcd, or po>led lo n publicly occcs:Jiblc wobsitc. In whole or ip part.
n
n+3
SECTION 11.2
47.
1
F~r the series ..~
(
0
55
e1 1"  e 1/ (n+l)) ,
= i~ ( el/i _ el/(i+I))
Sn
SERIES
=
(el _ el / 2) + (el / 2 _ el/ 3) + ... + ( el /n _ el / (n+l))
=e_
el /(n+l) [telescoping series]
f: ( el/n 
Thus,
el/ (n+l))
n =l
=
lim
49. (a) Many people would guess that x 9 = 0.99999 .. . = 10 +
(b) x
r
Sn
n+oo
e0
= e 1.
Converges
< 1, but note that x consists of an infinite number of9s.
9 9 + 100 1000
0.9 . _ 0. = = 0.1. Its sum IS 1 1
=e
= lim ( e  e1 / (n+l) )
noo
0.9 0. 9
~ + 10,9000 + · · · = .~
9
l"h" . . . 1s a geo~etnc senes w1th a 1
1 10.. , w 11c
= 0.9 and
= 1, that 1.s, x = 1.
(c) The number 1 has two decimal representations, 1.00000 . .. and 0.99999 . . . . (d) Except for 0, all rational nurnbe.rs that have a terminating decimal representation can be written in more than one way. For example, 0.5 can be written as 0.49999 . . . as well as 0.50000 ... . 
51. 0.8
=
. .tl 8 d + · ·· .IS a geometriC. senes WI 1 a = an r = 10
8 8 + 102 10
1 l a . t converges to  10 1  1"
=
8/10 / 1  1 10
516 516 "516 516 . . . . 516 53. 2.516 = 2 + + + + · · · IS a geometnc senes With a= and r + · · · . Now 106 103 106 103 103 3 3 _a_ = 516/ 10 = 516/ 10 = 516 . Thus. 2 .516 = 2 516 = 2514 = 838 1 .:... r 1  1/ 103 999/ 103 999 ' + 999 999 333 · 
42 42 104 + 106 +
55. 1.5342 = 1.53 +
a It converges to 1 
1·
=
42/ 104 1  1/ 102
42 Thus, 1.5342 = 1.53 + 9900
57.
00
00
n=l
n=l
2..: (5)"xn = 2..: ( 5x )" JxJ
00
n=O
153
42/ 104 99/ 102
= 100 +
42 9900
=
=
. . . . 42 + · · · 1s a geometnc senes w1th a = and r =
1 . 102
104
42 9900 · 15,147 42 15,189 5063 9900 + 9900 = 9900 ~r 3300 ·
is a geometric series with r =  5x, so ilie series converges
lrl < 1
¢:?
~
..
n=O
a
 1
1
1 r =
~ 2"
L.. ;; n=O X
2
< Jxl
< t, that is, t < X<
X
)

5 x
1 + 5x ·
00
3
¢:?
61. '
=
1 . It converges to 103
r. In that case, the sum of the series is _1 a_r = 1  ( 5 5x = 'E (x  2)" = E (X   2 ) " is a geometric series with r = X2, so the series converges Jrl < 3 ~
J 5xJ < 1
59.
42 42 · · · . Now 104 + 106
=
= 89 :
1 =
_ x 2 3
~
3
x 2 3
<
<1
 3
1
3
3  (x  2)
5 x ·
< x 2 < 3
~
 1
3
X
x
~
< x < 5. In that case, the sum of the series is
~ (2)". . . senes . Wlw .... r = :• 2 sothe senes . converges L..IS a geometric n=O
1
_
> 2 or x < 2. In that case, the sum of the series is  a 1
~
X
1 r
=
Jrl <
__!__ / 1 2
X
=
1
~ . X  2
© 2012Ccngage !.earning. All RighiS Rcscn "Cd. May nol be scaru>Cd. copic:d. or duplicated. or posted lo a publicly aecC>Sible wcbsilc. in whole or in part.
56
0
INFINITE SEQUENCES AND SERIES
63.
:L: en:z' = :L: (e"T is a geometric series with r = e"', so the series converges
CHAPTER 11
00
00
n =O
n=O
 1 < e"' < 1
lrl < 1 {::} lex! < 1 {::}
{::}
1 l ex
<=> 0 < e"' < 1 {::} x < 0. In that case, the sum of the series is _a_ =   . 1 r
65. After defining J, We use convert (f , parfrac ); in Maple, Apart in Mathematica, or Expand Rational and .
. 0 . fi d th h I . 3n2 + 3n + 1 enve to n at t e genera term IS (n 2 + n)3
.
s~mpl~fy 10
Sn
=
1 ( ) . So the nth partial sum is . n+1 3
= n13
n(1
~.:'f1
1
1 ) ( 1) (1 1) (1 1 ) k3  (k + 1)3 = 1  23 + 23  33 + " . + n3  (n + 1)3 = . 1  (n + 1)3
= 1. This can be confirmed by directly computing the sum using
The series converges to lim s,. noo
s um (f , n=l.. i nfinity) ; (in Maple), Sum [f , {n , 1 , In f inity}] (in Mathematica), or Calculus Sum (from 1 to oo) and Simplify (in Derive). 67. For n
= 1, a1 = 0 since s1 = 0. For n > . an
A Iso,
~
=
I'
nL;:l a,.,. = n~~ Sn
Sn 
1,
n 1
Sn 1
(n  1)  1
= n _ _+ _l  (n 1) + 1 =
(n  l )n (n + l)(n 2) 2 (n + l)n = n '( n +:1=)
1' 1  1/n 1 = n~ 1 + 1/n = .
69. (a) The quantity of the drug in the body after the first tablet is 150 m g. After the second tablet, there is 150 mg plus 5%
of the first 150mg tablet, that is, (150 + 150(0.05)] mg. After the third tablet, the quantity is (150 + 150(0.05) + 150(0.05) 2 ]
= 157.875 mg. After n tablets, the quantity (in n
1
.
.
mg) is
150(1  0 .05") _ 0.0 1 5
.
150 + 150 ( 0.05) + ... + 150 ( 0.05 )  . ·We can use Formula 3 to wnte this as
_(b) The number of milligrams remaining in the body in the long run is lim ( 3~go (1  0.05")] = n..oo
.
3
3000 ( =W 1
") 0.05 ..
~go (1 0) ~ 157.895,
only 0.02 mg more than the amount after 3 tablets.
71. (a) The first step in the chain occurs when the local government spends D dollars. The people who receive it spend a fraction e of those D dollars, that is, D e dollars. Those who receive the De dollars spend a fraction c of it, that is, D c 2 dollars. Continuing in U1is way, we see U1at the total spending after n transactions is
Sn = D + De + De2 + · · · + Dc"  1 = (b) lim Sn n.oo
=
1
lim D(  c" ) 1 C
n.+oo
[since c + s
= D 8
If c = 0.8, then s 73.
f
=1 
= __..!2_ 1
D(1 en) by (3). 1 c lim (1  c")
C n oo
= 1] = kD
[since k
= __..!2_ 1 C
[s ince 0
=>
lim c"
n.oo
= o] .
= 1/ s]
c = 0.2 and the multiplier is k
= 1/s =
(1 + c) n is a geometric series with a ,;, (1 + c) 2 and r
= (1 +
5. c) t, so the series converges when
n=2 1
IC1+ c) 1 < 1
{::}
l1 +el > 1
{::}
1 +e>1or1+e<  1
{::}
c > Oore <
7
2.Wecalculatethesumofthe
© 2012 Ccnsogc Learning. All Rights Rcscrvcd, Muy not be SC4JU1cd, copied, or duplicated, or posccd to a publicly ncccssiblc website, in whole or in pan.
SERIES
SECTION 11.2
.
2c2 + 2c  1 = 0 Soc= :02
1
1 (1
)
+c
1
c =  2~ m =
<:::}
57
2
(1+ c)  2
.
scn es and set 1t equal to 2:
D
=2
<:::}
±'1
1
.
1 ) (1+ c
2  2 ( 1 ) 1 +c
<:::}
1 = 2(1 + c) 2
2(1 +c)

<
However, the negative root is inadmissible because 2
<:::}
q 1 < 0.
.
1
1
1
75. e"" = e1+2+3+··+;; = e 1e 1 12e 113 · · · e 1/ n
.
.
~) (1 + ~) · · · (1
> (1 + 1) (1 +
+ ~)
[e"' > 1 + x]
= ~~~ ... n + 1 =n + 1 123 n
> n + 1 and n+ lim e"" = oo. Since { sn} is increasing, lim s,. = oo, implying that the harmonic series is oo n + oo
Thus, e·'" divergent.
77. Let dn be the diameter of Cn. We draw lines from the centers of the C; to the center of D (or C), and using the Pythagorean Theorem, we can write 2
2
1 2 + (1  ~d1) = (1 + ~d1)
¢:}
1 = (1 + ~d 1 )  (1 ~d1 ) = 2d 1 2
2
==> d1
[differenceofsquares]
= ~·
Sim ilarly,
1 = (l + ~d2)  (1  d1  ~d2) = 2d2 + 2d1 di  d1d2 2
2
= (2  dl)(d1 + d2) 1
d2 = 2 _ d 1
d,.+1
1

1 12
¢:}
(1
dd , 1 = (1 + 2d3 1 )2  (1 
2 _ d1
¥ ~) . 2
= ( =2
=
d1
1
i=1
lfwe actually calculate d2 and
'
1 respect1·veIy, so we suspect that .m general , d n 3·4
 = ly
< n dk 
'
=
1 k (k + 1)
.
T
= k1 
formula for dn+l. we get dn+l
1  .Then k+ 1 .
n
2:= d; =
;= 1
_ _ n ] [ 1 n+1
=
(
2
n

)

n+1
2
1 )2
d1  d2  2ds
<:::}

d3 
+ d2)) 2 _ (dt + d )
[1 (dl
2
,
2
d~ from the formulas above, we find that they are.!.6 = 2 1. 3 and 1
= (
. duct1on: . • 0 prove th'IS, we use m Assume that for all ) . .,.,
n n+ 1 1
1  n+ 1
1 2 (n+1) n +2
= n +n 1
[telescoping sum). Substituting this into o ur
1
= (n + 1)(n + 2 ), and the induction is complete.
n+1
Now, we observe that the partial s ums 2:=~= 1 d; of the diameters of the circles approach 1 as n+ oo; that is, 00
00
n=1
n =1
:2:= an = :2:= (
1
n n
) = 1, which is what we wanted to PJOVe. +1
79. The series 1  1 + 1 1 + 1  1 + · · · diverges (geometric series with r = 1) so we cannot say that
0=1 1 +1  1 + 11+· · · . 81.
2:=""= can = n 1
.
and m general,
lim l:=~=l ca; = nlim c 2:='~ a ; = c nlim 2:=~~ 1 a; = c 2:::,'= 1 a,., which exists by hypothesis.\ oo ' 1 + oo
n+ oo
® 201 2 Cc:ngagc l camins All RighLtt Resen:~.d. ~~fay no1 be scunrx.d. copied. or duplicated. or posted to a publicly accessible website, in whole or in part.
58
0
CHAPTER 11
INFINITE SEQUENCES AND SERIES
83. Suppose on the contrary that l:(an Theorem 8(iii), I: [(a,.
+ bn) converges. Then l:(an + bn) and I: an are convergent series. So by
+ bn)  a,.] would also b~ convergent. But I: [(an + bn) an] = I: bn, a contradiction, since
I: bn is given to be divergent. 85. The partial sums {sn} form an increasing sequence, since s,.  Sn 1 =an > 0 for all n. Also, the sequence {sn} is bounded since Sn
::::;
1000 for all n. So by the Monotonic Sequence Theorem, the sequence of partial sums converges, that is, the series
I: an is convergent. 87. (a) At the first step, only the interval ( ~ , ~) (length ~) is removed. At the second step, we remove the intervals ( ( ~ ' ~).which have a total length of2 ·
Ut At the third step we remove 2 1
2
intervals, each of length (~ )
at the nth step we remove 2n 1 intervals, each oflength ( ~) n, for a length of 2" 1 length of all removed intervals is
f ~ (t)
n
= ~ ( ~) n
1
•
•
In general,
Thus, the total
= 1 :./;13 = 1 [geometric se~ies with a = t and 7' = t] . Notice. that at
1
n =l
the n th step, the leftmost interval that is removed is ( ( the rightmost interval removed is (1 
· ( ~)"'
3
i, ~) and
1)" , (j )"),so we never remov~ 0, and 0 is in the Cantor set. Also,
(j)", 1  (tf'), so 1 is never removed. Some other numbers in the Cantor set
dB
12127
are 3• 3•·9• 9• 9• an 9· (b) The area removed at the first step is removed at t.he nth step is {8)"00
n>;1
1(8)"9 9
1
1/ 9 8/ 9
=1
n
oo
23
1ar
2
· (
i) 3 • In general, the area
=
H~rl. so the totat area of all removed squares is
=
2+ ~
= 1. 1
1
89. (a) For n>;1 (n + 1)!' 81
~; at the second step, 8 · ( ~) ; at tbe third step, {8) 2
= 1. 2 = 2' S2
4
1
119
5
2
=
6' sa
5
=
6+
3 1 . 2. 3. 4
23 24,
.
s4 =  + = .The denommators are (n + 1}!, so a guess would be sn = 24 1 · 2 · 3 · 4 · 5 120 (b) For n
1
= 1, 8 1 = 2 = Sk+l
=
21  1
~·so the formula holds for n = 1. Assume 8k =
(k + 1}!  1 (k + 1)!
k+1 + (k + 2)!
=
(k + 1}!  1 (k + 1)!
(k
k +1 + (k + l)!(k + 2)
+ 1)! 
(k + )! 1
=
1
(n + 1}!1 ) ( n + 1 .1
. Then
(k + 2) !  (k + 2) (k + 2)!
+k +1
 (k + 2)! 1 
(It+ 2)!
Thus, the formula is true for n
.
(c) lim
noo
8n
·
.
= lim
,._00
= k + 1. So by induction, the guess is correct.
(n + 1)!  1 . [ 1 ] oo n (n + 1)I. = nhm 1  ( L: (n + 1)! = 1. oo n + 1)'. = 1 and so "=1
® 2012 Ccngagc LearninH,. All Rights Reserved. Muy not be scanned, copied, or duplicated, or poste<.llo a publicly accessible websile, in whole or In part.
SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS
0
59
11.3 The Integral :rest and Estimates of Sums
=
1. The picture sh?WS that a 2
1 a 3 = "1.3
<
.3
131
.
1 2
1. 3
r
<1
2
·1 x 1.3
1
"1.3 dx, and so on, so X
2
y
dx,
00 1 [ 00 1 2:: 1':3 < "1.3 dx. The n. 1 X
n= 2
integral converges by (7.8.2) withp = 1.3
> 1, so the series converges. 0
2
3
X
4
3. The function f(x) = 1/ ~ = x 1/ 5 is continuous, positive, and decreasing on [1 , oo ), so the Integral Test applies. 4 15 ] t = lim (~t4 1 5  .§.) = oo, so E = t.oo lim J' x 1 / 5 dx = lim [2x .1/ ifrf diverges. 1 t.oo 4 1 t.oo ·. 4 n= 1 1
f
oo 1
x  1 15 dx . .
5. The function f(x) = ( x ~ ) 3 is continuous, 2 1
.1roo (2x +1 1)3 dx ~ lim 1r (2x +1 1) t+oo
1
1
3
dx
positi~e, and decreasing on [1, oo), so the lf1tegral Test applies .
=
lim t.oo.
Since this improper integral isconvergent, the series .
7. The function f(x)
}
roo ~ + 1 'dx = 1
X
[.!4 (2x + 1)2 1
E (2n +1 1)
n=l
3
]t
=
1
1 lim ( t oo 4(2t + 1)2
+ _.!_) = _.!__ 36
36
is also convergent by the Integral Test.
=/positive, and decreasing on [1, oo), so the futegral Test applies. + 1 is continuous, . . X
lim
r /+ 1 dx =
t .oo } 1
X
1 2.
1
2
lim [ ln( x + 1)] t =  lim [ln( t> oo
.2 t+oo
1
e + 1) ·
ln 2]
= oo. Sine~ this improper
E~ is also divergent by the Integral Test. +1
integral is divergent, the series
n=1 . n
9.
E n ~ is a pseries with p = /2 > 1, so it converges by (1 ). . .
n=1
1 Thi . 1 1 'th p = 3 .. ++ ···= ~ s tsapseneswL 6 3 · 64 125 n=l n
1 1 11.1 +++ 8
27
1 3
1 5
1 7
1 9
1
00
13. 1 + + +  +  + · · · =
L: . 2n  1
n =l
The function f( x) =
. by ( I. ) > 1,so1tconverges
1
x _ · is 2 1
continuous, positive, and decreasing on [1, oo), so the· Integral Test applies.
1 · dx = lim 2x 1 t +oo
r oo 
}1
ft2 1 1
X
1
dx
= t lim + oo
a1n 12x  111 ~ = ~
lim (1n(2t  1)  0)
t>oo
r: 
= oo: so the series n=l 2n 1
diverges.
15.
oo
L:
n =1
vfn+ 4 n
E n4
n=l
'oo
2
2
=4 .
= nL: = l
( vfn  2 n
oo 1 oo 4 + 2n4 ) = nL: 3 / 2 + L: 2• =l n n =1 n
~
6 n= l
1 .IS a convergent psenes . w1'th p 3!2
n
E ~n is a constant multiple of a convergent pseries with
p
3
= 2
> 1..
= 2 > 1, so it converges. The sum of two
n= l
convergent series_isconvergent, so the original series is conv~rgent.
©
201 2 Ccognse Learning. All Righls RCSCJV
1
60
0
INFINITE SEQUENCES AND SERIES
CI:IAPTER 11
17. The function f(x)
+4
=~ X
l 1 ·= . oo
1
  dx x2 + 4
lim
t +00
is continuous, positive, and decreasing on [1, oo), so·we can apply the Integral Test.
.
1
~t  2
x + 4
1
· . [1 _x]t =· ·1 . [· _(t)
dx = lim tan t + oo 2
1
2
2
1
hm tan
1
2
t+ oa
 tan
_(1)] 1
2
=.!. [~  tan(.!.)] . 2 1
. 2 2 Therefore, the series
f: ~ converges. n +
n=l
ill n = . ~ ln n smce . . f(. x ) = ln x 1s . contmuous . · · on [2, oo ) . 19 . ~ ~ ~ ln 1 = 0 . Tbe funchon and postttve 3 3 3 n= l n n =t n 1 X I
3
2
2
2
f'( x) = x (1/x)(ln x)(3x )=x 3x lnx=1  3lnx
.
(x3)2
x > e113 ~ 1.4,' so f
1 2
x4
1 _ 3 lnx < O {::}
ln x> 1
{::}
3
is decre~ing on [2, oo), and the Integral Test applies.
1t
00
Inx dx = lim x3
x6
ln x dx
t oo
2
x3
~
1 lim [ lnx2   ] t 2x 4x2 1
t+oo
=
.!.] (~) 41 , so the series f:
lim [_!._2 (2 ln t + 1) + 4t 4
t+oo
n= 2
In~ n
converges. (*): u = hl x, dv = x  3 dx
I
ln x d x
~
(**):lim t+oo
=  21 x 2 1n x  .;· (~2lnt+ 1) 2
= (1 /x) dx, v = ~x 2 , so
=> du 1
2x
2 (
1/ x ) dx
=
1
2x
21n ·
x
+ 21 ; · x 3dx =
J:!:_lim 2/ t = l4 lim ]:_ =0. t + oa 8t t+oo t2
4t
21. f (x) = + is continuous and positive on [2, oo), and also decreasing since J'(x) =
x= x
.
use the lntegra1Test.1 ."" .
2
+. C.
 21 x 2 1nx 41 x 2
X
1 ln dx= lim [ln(lnx)J; = lim X
t+oo
t+oo
1
x2
Tlnln)2 < 0 for x > 2, so we can'
x
[ln(lnt)~ln(ln2)]= oo,sotheseries n=2 f n 11ndiverges. n .
23. The function f(x )'= e1 1"jx 2 is continuous, positive, and decreasing on [1, oo), so the Integral Test applies. [g(x) = e 1 1~ is decreasing and dividing by x 2 does~'t change that fact.]
l
~t e1frz;
oo
f(x) dx = lim
too
1
1
·
X
t too
.
t
x] = 
 2 dx = lim [e 11
1
· . lim (e1/ t  e) = (1  e) = e 1, so the series t +CXJ
00
el /n
2:: n 2
n=1
converges. . 25. The funct10n f(x) = 

1

q; 2 +x 3
1 = 2 x

1 1 +   [by partial fractions] is continuous, positive and decreasing on [1, oo), x x+ l

so the Integral Test applies. 00
1 1
f(x)dx = lim t + oo
It(
1 + 1) dx = lim [  1  ln x +ln(x +:t) . 21  X
1
X
X
+1
t+oo
X
J
t
l
. [ 1 + ln t +1 + 1ln2] =0 + 0 + 1ln 2 = hm t+oo t t ' I converges, so the senes . ~ The .mtegra ~ n=l
1

n 2 +n3
converges. . I
® 2012 Cengnge Learning. All Rights Reserved. Mny not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in prut.
, SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS
27. T he function f(x)
.
COS7Tn
· oo
L:
~. y n
n=l
1
00
29. We have already shown (in Exercise 2 1) that when p = 1 the series .
(
X
61
= coFxx is neither positive nor decreasing on (1, oo), so the hypotheses of the Integral Test are not
satisfied for the senes
f (x) =
0
E (l ) n=2 n n n p
1 ) is continuous and positive on [2, oo), and f' (x) = ln x P ·
X
diverges, so assume that p =1 L
~~ X~'P: 1 < 0 ifx > e  P, so that f
is eventually
decreasing and we can use the Integral Test.
roo ::1 : d l' [ (ln X)1p] t } 2 x (ln x)P x=t~ 1 p 2 <0
This limit exists whenever 1  p
p
¢:?
>
[forp :f 1] = lim [ (ln t) 1 p  (ln 2)1  p] h oo · 1 p 1 p
1, so the series converges for p
>L
31 . Clearly the series cannot converge if p 2': ~. because then lim n(1 + n 2)P =1 0. So assume p. < ~ . Then ·
n
f(x) = x(1 + x 2)P is continuous, positive, and eventually decreasing on [1 , oo), and we can use the Integral Test.
This limit exists and is finite
p
¢:?
+1
p
¢:?
<
1, ~o the series converges whenever p
<
1.
> 1. Unless specified otherwise, the domain of a function f
33. Since th is is a pseries with p = x, ( (x) is defined when x
is the
set of real numbers x such that the expression for f (x) makes sense and defines a real number. So, in the case of a series, it's the set of real numbers x such that the series is convergent.
(3)
4
00 35. (a) n~l
00
(b)
00
E
;;:
n=l
1
81 4 = 81
n
1
00
I:
n= 1
1
1 4
n
= 81
(7T4) 90 00
1
k~6 (k  2)" = 34 + 44 + 54 + ...
97T4
= 1o
1
= k~3 k4 =
4
. 7T 90 
37. (a) j (x ) = ...; is positive and continuous and f'( x ) = ~ X
00
X
.1
n~l n2 ~ sw Rw :::;
(b)
8 10
+
=
1
I2
1
1
1)
t .oo
X
00
is negative for x > 0, and so the Integral Test applies.
t .oo
10
1
(.!.t + 10 ) 1
=
1 , so the error is at most 0.1. 10
1\
:::; S :::; 8 10
00
21 dx :::;
4 7r 17 [subtract a I and a 2] = 90  16
1
1 lim [ ] t = lim
roo ...;dx = X
11
1
14 + 24
+ 22 + 32 + .. : + 102 ~ 1.549768.
} 10
1
(
S :::; SlO
X
1.549768 + 0 .090909
+
10
21 dx X
=?
SlQ
+
= 1.640677 :::; s :::; 1.549768 + 0.1 =
+ fa
1.649768, so we gets
~
1.64522 (the average of 1.640677
and 1.649768) with error :::; 0.005 (the maximum of 1.649768  1.64522 and 1.64522 1.640677, rounded up).
® 20 12 Cengage Lc:unint;. All RighiS Reserved. May riot be scanned, copied. or duplicntcd, or posted to n publicly accessible \\"
62
0
CHAPTER 11
INFINITE SEQUENCES 1,\ND SERIES
(c) The estimate in part (b) iss ~ 1.64522 with error ~ 0.005. The exact value given in Exercise ~4 is 1r2 / 6 ~ 1.644934. The difference is less than 0.0003. (d) Rn
39. f(x)
~ /.~ ;. dx = .!.. So Rn ,. x n
4
41.
{::}
n > 1000.
= 1/{2x + 1)6 is continuous, pos itive, and decreasing on [1, oo), so the Integral Test applies. Using (3),
1 10( n + ) 5 2 1 S4
n
~ 1 oo (2x + 1)  6 dx = t~ [ 10(2; ~ 1) 5
R,.
1 1000
< 0.001 if.!. <  
~ 1~ 6
(2n + 1)5
¢:}
1
1
= n~1 (2n + 1)6 = 36 +
1 S6
+
1 76
r
~ 20,000
l0( n\ 2
n
{::}
1
) 5 . To be correct to five decimal places, we want
~ ~ ( .e'20,000 1) ~ 3.12, ;muse n = 4.
1
+ ge ~ 0.001446 ~ 0.00145.
f: n1.oo1 = f: n 1 ~01 is a convergent p series with p = 1.001 > 1. Using (2), we get
n=1
n=1
Rn < !.oo xl.001 dx = lim [ x  o.oo1 ] t =  1000 lim [1]t =  1000( 1 )
 n
We want Rn
< 0.000 000 005
n > (2 X 10ll) 1000
= 21000 X
{::}
1000 n 0.001
1 1 1 1 f(x)=,+++ .. ·+ 1 x2 3 4 n Bn
~
ln b
= (e1" b) Inn = <
 1
10301
X
{::}
n°·001
1011.000
Jt f(x) dx, so with
n0.001
n
>
5
1000 10
x
= l.07 X
=
1000. n0.001
{::}
9
1011.301. y
1 dx= ln n. x
1
1 1 1 1 = 1+ z + '3 + 4 + .. · + ;:;: ~ 1 + ln n.
8109 ~ 1 + ln 109 ~ 21.72 bin"
X
9
x0.001
1"
(b) By part (a), s10o ~ 1 + ln 106 ~ 14.82
45.
< 5 x 10
1011,000 ~ l.07
43. (a) From the figure, a2 +as+···+ an ~
Thus,
t>oo
t>oo  0.001 "
¢:}
0
< 15 and
2
3
4 ·"
n
< 22. 1
(e1" ") In b = nln b = n _ 1n b. This is a pseries, which converges for all b such that  ln b > 1
b < e 1
{::}
¢:}
b < 1/e [with b > 0].
11.4 The Comparison Tests 1. (a) We cannot say anything about I: an. If an
> bn for all nand :L b,. is convergent, then :L an could be convergent or
divergent. (See the note after Example 2.) (b) If an < bn for all n, then I; an is convergent. [This is part (i) of the Comparison.Test.) 3.
n n+ <  n 3 2n 2 3 1
= 2n1 2
" a II n < 21 .or n
because it is a pseries with p = 2
®
~
~ 1, so L..n=l
n  converges by companson . ~ 1 , wh'1cI1 converges WI'th L..2 2n 3 + 1 n=l n

> 1.
2012 Cengage Learning. A ll Ri£bts Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 11.4
n +1 5. 
>  n
nVn
7.
= 
1
Jn
nVn
~ n for all n 2: 1, soLn=l
pseries with p =
! ~ 1.
9" 3 + 10n
=
< 9"
10"
( 9 )" for all n > 1. 10  .
+1 1 d'1verges by companson · ··h L~ w1t
n vn
n=l
1 1
vn
,
THE COMPARISON TESTS
D
63
· 1s · a wh'1ch d'Jverges because 1t
00
oo '9n n L:: (for is a convergent geometric series (lrl'= fo < 1), so n=l L:: n=l 3 + 10
converges by the Comparison Test. 1 · > 1 for all k 2: 3 ['smce ln k• > 1 "k .or .> _ 3] , so ~ LJ lnkd. k 1verges by compan·son w1t'h~ LJ k , wh'h 1c d'1verges because 1t
9. lnk k
k
k=3 •
k =3
is a pseries with p
=:1 ~ 1 (the harmonic series). Thus, f
lnkk diverges since a finite number of terms doesn't affect the
k=l
I
convergence or divergence of a series. 11 .
{fk
+ 4k + 3
Jk3
<
ifk
kl/3 1 ~ = k = k7 16 for all k 312 .v k3
· which converges because it is a pseries with p 13.
{fk
00 •
L::
k= l
3
,jk
+ 4k + 3
. . converges by compar1son w1th
1
00
L:: k=l
k
716
,
= ~ > 1.
arctann 7r/2 fior aII n 2: 1, so ~ arctann . 1 hi h . 1. < ""1.2 LJ ~, converges by comparison w1t. h 1r ~ LJ ""1.2• w c converges because 1t is a 2 n=l n n 2 . n n =l n 2 constant times apseries with p
15.
2: 1, so
4'' + n _
3
1
2
= _1 .2 > 1.
4 . 4" = 4 (4)" > 3n"" 3 for all n 2:
00
4n+l
n=l
3" 2
00
1.
00
n~l 4 (4)" 3 = 4 n~l
(4)" 3
is adivergent geometric series (lrl = ~
> 1), so
L::   diverges by the Comparison Test. 17. Use the Lirni~ Comparison Test with an lim an = lim n bn noo Jn 2 + 1
noo
=
1
= JTi2'+T and bn =
1 )1 + (1 fn2)
lim ' n
n:1
= 1 > 0. Since the harmonic series
f
n=l
.!. diverges, so does n
1
00
L::l Jn2+1' n= n +1 1 4" . . Companson . TiestwJtlan· .I 19. Use the L1m1t =  + andbn . 1 + 3n
4
= 3"n :
1 +4" . lim= . a.. . 1 +,,3" . 1 +4" ·3" . 1+ . ( 1 +1 ) · 1 =1 > 0 lim = hm  = hm 4" ·  3"  = hm n__.ob bn n__.oo !._ · noo 1 + 3" 4" n__.oo 4" 1 + 3" n.oo 4" .I_ + ~
Since the geometric series 2::: bn =
1+4" 1+4"   >  >  (4" ) 1 + 3"
3n
©
+ 3"
2 3"
~
'
1
r diverges, so does n=l f 11 ++ 34: . Alternatively, use the Comparison Test with
2::: (1
1(4)" . or use the Test for Divergence.
= 
2
3
201 2 Cengoge Le:uning. All R;ghL< Resel'\'ed, May not be SCIUUled, copied, or duplicated, or posted to a publicly acc'cssihlc website, in whole or in pM.
64
D
CHAPTER 11
INFINITE SEQUENCES AND SERIES
f+"2
21 . Use the Limit Comparison Test with an = .
3 2
. n 1 Jn+2 = n.oo hm 2n2 + n + 1
an bn
hm 
n.oo
Since
2n +n+ 1
and bn =
vn)
f: ;
[P = ~ > 1], the series
= (15+ + n2n 2 )2 3
3
and bn
4
f: f+"2
=
1
n3
n2 1 ) =   for all n n n+ 1 n+ 1 ·
2
> 0·
also converges.
:
*
l'
+ 2 . 2 0 s· ~ 1 . l )2 = > . mce n'S::l n 3 IS a convergent
= n!...~ (.l;,. +
...
~
1, so
oo
I:
../n4 + 1
n=l n 3 + n 2
diverges by comparison with
n=2n
~) n
2
e n and bn = e": lim abn = lim .(1 noo
n
noo
f: en = n=l f: ·e:. is a convergent geometric series [lrl = ~ < 1], the series n=l f: (1 + ~) n=l n 1 29. Clearly n! = n(n  1)(n  2) · · · (3)(2) > 2 · 2 · 2 · · .. · 2 · 2 = 2'' 1 so..!_ <  . . ' n!  2nl
[lr I = ~
1], so
<
an
+ ~) n
2
= 1
> 0. Since
2
en also converges.
E  1
n =1 2 n 1
is a convergent geometric
E \n . converges by the Comparison Test.
n=l
31. Use the Limit Comparison Test with an
.
1 
= f: ~.which diverges because it is a pseries with p = 1 ~ 1.
27. Use the Limit Comparison Test with an = (1 +
series
= /1 = 2
. L.J ~ ( 5 + 2n) a lso converges. = 3 > 1], t he senes n.=l 1 + n 2 2
2(
f: 
n.oo 2 +
4
li n (5 + 2n) l' 5n + 2n 1ln = n.~ ( 1 + n2)2 = n~ (1 + n2)2 · l l (n2 )2
1 n.=ln+1
) 1 +2ln 1ln + 1l n 2
= lim
. n= l 2n + n + 1
23. Use the Limit Compa'rison Test with a,.
. [p psenes
.
(n 3 12 Jn+2)l(n 3 12 n.oo (2n2 + n + 1)ln 2 .
= lim
is a convergent pseries n=l n 12
lim an n .oo bn.
; : n 12
.
sin(1ln)
I1m  = 1rm n.oo bn n.oo
1ln
== sin ( ; )
and bn
= ; . Then I: an and 2: bn are series with positive terms and
. sinO . ~ . . . . _= hm   = 1 > 0. Smce L.J bn IS the divergent harmontc senes, 0 +D 0 n=l
00
2: sin (1 l n) also diverges. n= 1
[Note that we could also use !' Hospital's Rule to ev~luate the limit:
·
. sin(ll x) H • cos(l/x) · ( l/x 2 ) 1 . hm I = %oo lim I = x_____. hmoo cos X =cos O= 1.) 1 X 1 X 2
x  oo
lO
1
33. n~l ~ =
1
00
Rw ~ Tw ~
1
1
.J2 + ..ff7 + v'82 +
1 10
1 1 · · · + .JI0,05I ~ 1.24856. Now ~ <
1 ] t = lim 21 dx = lim [ X
t .oo
X
10
© . 20 12 Ccngage Leaming . All Ri(;hts Reserved.
t .oo
(
1 
t
1) +10
1
..JTiJ =
1
.
n 2 , so the error IS
= 1 = 0.1.
10
Mny no t be scanned, copied, or duplicated, or posted ton publicly accessible website, in who le or ln pan.
SECTION 11.5
35.
cos2 1 = ~
10
s ncos2 n n =l I;
5
l
R 10 ~ T10 ~
oo
10
2
2
2
cos 2 cos 3 ++ + .. · + cos51010 52 53
1
.
dx= lim
t~oo
5"'
j·t 5 10
x
~
ALTERNATING SERIES
0
65
cos2 n 1 . 0.07393. Now   <  so the error IS 5"  5n'
. [ 5 "' ] ' . ( 5 t dx= lim  ln = lim  5 10 t~ oo 1n 5 t~oo
. ~ .s a convergent . . senes . <  9 for each n, and smce L..,  9 1 geometnc 10"  10" n=1 10n
37. Since d,.
+
1o) = 5101ln
5 ln S
(I r I =
1
10
5
< 6.4 x 10
_
8.
< 1) , o.d 1 d 2 d 3 . .. = ~ L..,  dn n= l
10"
will always converge by the Comparison Test. 39. Since I: a.. converges, lim an = 0, so there exists N such that ian n>oo
all n
01 < 1 for all n > N => .
0 ~ ar,. < 1 for
=> 0 ~ a~, ~ an. Since I; an converges, so does I; a~ by the Comparison Test.
>N
41 . (a) S ince lim an = oo, there is an integer N such that abn > 1 whenever n > N. (Take M = 1 in Definition 11.1.5 .) oo~
Then an
n
.
> bn whenever n > Nand since I: b,. is divergent, I; a"
is also divergent by the Comparison Test.
. If an =  1 and bn = 1 fior n > X (b) (1) _ .2, t hen I'1m an b = lim. . n = lim ln =II lim  1
n
Inn
so by part (a),
~
L.., .
n=2
.. If an (11)
1
ln n
.
IS
n oo
n ........ oo
1n n
X
X+ 00
xoo
1
1 X
= lim x = oo, xoo
. d1vergent.
1 th ln n and bn = , = en ~ L..,
n
n
n
n=l
bn 1·s the d'1vergent h armomc · senes · an d I'1m b an n+oo
n.
lim Inn = l im lnx = oo,
,,_ oo
xoo
00
so
I: an.diverges by part (a). n=l
43. lim nan = lim a/,. , so we apply the Limit Comparison Test w ith bn n+oo n  oo 1 n series converge or both series diverge, and we also know that .
f
n =l
= .!n . Since ,,_00 lim nan > 0 we know that either both
.! diverges n
[pserics with p = 1]. Therefore, I; an must be
divergent. 45. Yes. Since
I: an is a convergent series with positive terms,
lim an= 0 by Theorem 11.2.6, and I: bn
n ~oo
=I: sin( an) is a
series with positive terms (for large enough n). We have lim bn = lim sin(an) = 1 > 0 by Tbeor~m 2.4.2 noo
an
n +oo
an
[ET Theorem 3.3.2]. Thus, l::bn is also convergent by the Limit Comparison Test.
11.5
Alternating Series
1. (a) An alternating series is a series whose terms are alternately positive and negative.
(b) An alternating series
f; an= n=l f; (l)"
n=l
1
bn, where bn =ian I, converges if O < b,.+l ~ b,. for all nand lim b,,
= 0.
n+oo
(This is the Alternating Series Test.) (c) The error involved in using the partial sum s.,. as an approximation to the total sum s is the remainder R ,. = s  sn and the size of the error is smaller than bn+l; that is, IR~,I ~ bn+l· (This is the Alternating Series Estimation Theor~m.) ©
2012 Cc ngace Learning. AU Rights Reserved. Mny not be scanned, copied, orduplicntcd, o r posted loa publicly i1Cccssiblc website, in whole or In p:.rt
66
0
INFINITE SEQUENCES AND SERIES
CHAPTER 11
00 2 4 6 8 10 n 2n 2n . 2 2 . 3.   +   +   +···= :E{1)   .Now limb,. = l i m   = lun / =  =fO.Smce 8 7 5 6 9 n=l n+ 4 noo noo n + 4 ntoo 1 + 4 n 1
lim
n+oo
an =I 0 (in fact the limit does not exist), the series diverges by the Test for Divergence.
00
00
n= l
n=l
5. :E a,. = :E (1f 1
2n
1 +
00
1
n=l
2n + 1
= :E (1)" 1 b,.. Now b,. = 
1
> 0, {bn} is decreasing, and
lim b,. = 0, so the noo
series converges by the Alternating Series Test.
7.
E a,.= E( 1)" 32nn + 11 = f (1)"b,.. Now lim b,. =
n=l
n=l
n=l
lim
n+oo
n+oo
~+ 11jnn =
3 2
=f 0. Since
lim n+00
a,. =f 0
(in fact the limit does not exist), the series diverges by the Test for Divergence. 00
9. :E n= l
a,.=
00
00
n=l
n =l
:E ( 1)"e" = :E (1)"bn. Now b,. =
1
n > 0, {bn} is decreasing, and e
lim b,. = 0, so the series converges n+oo
by the Alternating Series Test. n2
11. b,. = n 3
+ 4 > 0 for n 2:: (x 3
1. {bn} is decreasing for n 2:: 2 since
+ 4)(2x) 
x 2 (3x 2 )
(x3 + 4 )2
lim b,. = lim n +00
13. lim
n+oo
nop
_
+ 8 3x3 )
x(2x 3
(x3 + 4 ) 2

x(8 x 3 )
_
 (x 3 + 4 )2 < 0 for x > 2. Also,
1 = 0. Thus, the series 1)"+1 ~converges by the Alternating Series Test. / n/ 1 + 4 n3 n= l n3 + 4
E(
b~ = lim e2 fn = e0 = 1, so ra.oo
lim ( l)n 1 e21" does not exist. Thus, the series
n+oo
f: (1)"
n=l
1 2
e 1" diverges by the
Test for Divergence. 15. a,. = oo
:E
1
+ .,fti
fon
sin(n +.! )1r 1+
n=O
17.
sin(n + ~)1r
(1)"
=
1
+ .,fii,"
Now b,. =
1
1
+ Vn > 0 for n 2:: q, {b,.} is decreasing, and ..~~~ b,. =
0, so the series
converges by the Alternating Series Test.
n~l (  1)" sin(~J
bn
=sin(~) > 0 for n 2:: 2 and sin(~)
2::
sin(
n : 1 ).
and ,.~ sin(~) =
sinO = 0, so the
series converges by the AJternating Series Test.
n"
19. n!
= n·n···· ·n > n => 1· 2 · ·· · · n 
n" n!
l i m  =oo n.+oo
=>
lim n ..... oo
( 1)" " 
n.1
n
·
oo
by the Test for Divergence. 21.
The graph gives us an estimate for the sum of the series
E (O.~)" of 0.55. n.
n=l
bs
(0 8)" =~ : : : 0.000 004, so
1
©
n
does not exist. So the series :E ( 1}" ~, diverges n=l n.
20U ~ngoge Learning. All Rights Reserved. Muy no1 bo scanned, copied, or duplicated. or po!lttd too publicly accessible website, In \\tlole or in port.
SECTION 11.5
ALTERNATING SERIES
D
67
~ ( 0.8t :.._  ,;.., (  0.8t ' ~57 LJ '
LJ n=1
n.
n.
n=1
~ 0.8 + o.32  o.oss3 + o.oi 706 o.oo2 731 + o.ooo 364 o.ooo 042 ~ o.sso7
Adding b8 to s 7 does t:~ot change the fourth decimal place of 57, so the sum_of the series, correct to four decimal places, is  0.5507. 1 f: ( 1n);+satisfies (i) of the Alternating Series Test because ( ) · n +1 1
23. The series
n=l
=
series is convergent. Now bs
1 56
= 0.000064 > 0.00005 and b6' =
1 66
~
6
0.00002
<
in and (ii)' lim in = 0, so the n + oo
< 0.00005, so by the Alternating Series
Estimation Theorem, n = 5. (That is, since the 6th term is less than the desired error, we need to afld the first 5 terms to get the sum to the desired accuracy.) 25. The series
f: 10" ( 1)",·n. satisfies (i) of the Alternating Series Test because 10."+ltn + 1 )'. < 10"1 n 1 and (ii)
n =O
=
so the series !s convergent. Now bs
!
10 31
~ 0.000 167 > 0.000 005 and b4 =
!
10 41
1 lim   = 0, 10" n !
nioo
= 0.000 004 < 0.000 005, so by
the Alternating Series Estimation Theorem, n = 4 (since the series starts with n = 0, not n
= 1). (That is, since the 5th tenn
is less than the desired error, we need to add the first 4 terms to get the sum to the desired accuracy.) 1
1
27. b4 = I =   :::; o.ooo 025, so 8. 40,320 00
n~1
( 1)" (2n)!
3
~ 53 = n~l
(1)" ' (2n)!
1
1
= 2 +
1
24  . 720
~ ~0 .459722
Adding b4 to S3 does not change the fourth decinlal place of 5s , so by th~ Alternating Series Estimation Theorem, the sum of . the series, correct to four decimal places, is  0.4597. 72 29. b7 = 107

= 0.000 004 9, so 00
(
"'
n~
1
1)n ln2 10" :::; 56 =
6 (1)n ln2 10" =
"' n7;: 1
1 10 
4 100
+
9 1000 
16 10,000
+
2s 100,000 
36 1,000,000
· = 0·067 614
Adding b7 to 56 does not change the fourth decimal place of 56, so by the Alternating Series Estimation Theorem, the sum of the series, correct to four decimal places, is 0.0676. I
00
31.
I:
n= 1
(
1)nl n
=
1 1 1 1 1 1 1 1  +    + · · · +    +    + · · · . The 50th partial sum of this series is an 2 3 4 49 50 51 52
underestimate, since "~
1
(
 1~nl
= 5so + (
1 1 1 ) + ( ; ) + ; · ·, and the te~s in parentheses are all positive. ' 52 54 51 3
The result can be seen geometrically in Figure 1. 1 33. Clearly bn =   is decreasing and eventually positive and lim bn = 0 f~r any p. So the Series converges (by the n+p n + oo Alternating Series Test) for any P. for which every bn is defined, that is, n + p :/: 0 for n
2 1, or p is not a negative integer.
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CHAPTER 11
INFINITE SEQUENCES AND SERIES
68
D
35.
2: b2 n = 2: 1/(2n)2 clearly converges (by comparison with ~e pseries for p = 2). So suppose that 2: (1rI bn .
.
converges. Then by Th~orem 11.2.8(ii), so does 2: [(1
t. 1 bn + bn] ::: 2( 1 + l + i + · · · ) = 22: 2n 11 . But thi~
diverges by comparison with the harmonic series, a contradiction. Therefore, 2: (....:1)" 1 b,. must diverge. The Alternating Series Test does not apply since {bn} is not.decreasing.
11.6
Absolute Convergence and the Ratio and Root Tests
1. (a) Siilce lim Ian+l I = 8 noo
an.
(b) Since lim I an+l I n.CX>
an
> 1, part (b) of the Ratio Test tells us that the series 2: an is divergent. .
.
= 0.8 < 1: part (a) of the Ratio Test tells us that the series 2: an is absolutely convergent (and
therefore convergent). (c) Since lim I an+l I = 1, the Ratio Test fails and the series 2: an might converge or it might diverge. ~
3. lim ntoo
an
.
.
1 5 1 1 1 lan+l l = noo lim In+ · " 1 = lim 1.!. · n+ 1 =.!.lim + /n = .!. (1) = .!. < 1,sotheseries E ~is an 5n+l n ntoo 5 n 5 ntoo 1 5 5 n=l 5n
absolutely co.nvergent by the Ratio Test.
> 0 for n 2::' 0, {bn} is decreasing for n ~ 0, and lim
1 5. bn = 5n + 1
n>oo
bn = 0, so
.
E 5n( 1+t 1 converges by the Alternating
n=O
Series Test. To determine absolute convergence, choose an = ~ to get n l'rm' 5n + 1 = 5 > 0 , so ~ 1 .d . b .h L ' . C . ,.. . h the = n+oo ~   1verges y t e mut ompar1son . est w1t n n=l 5n + 1 ' .
an = 1'tm / ( 1/n ) lim b n n~ 1 5n + 1
n~~
hannonic series. Thus, the series
=
7. lim lak+Il .
k~oo
ak
E k ( ~)
E ( 1)" is conditionally convergent.
n=O 5n+ 1 1
(~_)'"+ ] =
1 lim [(k+: ) koo k(~)k
lim k + k too k
1(~) 3
1
= ~3 k~oo lim (1+ .!..) = ~'(1) = k 3
'
'
13 < ·1,sotheseries
is apsolutely convergent by the Ratio Test. since the terms of this series are positive, absolute convergence is the
k
n=1
same as convergence.
4 1 9. lim 1an+ I I = lim [ (l.l)n+l4 . ~J = lim (1.1)n 4 ·= (1.1) lim · n 
so the series
= (1.1)
lim ( n o oo
1+
~~
n
)4
> 1,
f: ( 1)n (l.~)n diverges by the Ratio Test. n
n= l
el/n 3 ~
11 . Since 0 ~ 
n
oo
( 1)"el/n.
n=l
n3
2::
e
3
n
(
1)
= e 3 n
and
oo
2:
n=l
· ' 1 is a convergent pseries [p = 3 3 n
00
e l /n n3
> 1], L: n= l
converges, and so
is absolutely convergent.
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SECTION 11.6 13.
.
lan+l l
.
[
2
w n+ l
Ji_~ ~ = J2...~ (n + 2) 42n+3 .
{n+1)4 "+ . 10"
1
]
,
ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS
D
69
. (10 n + 1) 5 . . 00 10" = n~ 42 . n + 2 = 8 < 1, so the senes n~l (n + 1)42~+1
is absolutely convergent by the Ratio Test. Since the.terms of this series are positive, absolute convergence is the same as convergence. 15.
(1)" arctann\ rr/ 2 . oo rr/ 2 rr <  2 , so smce 2:::  2 = 2 2 n n n = l n \
00
2:::
1 2 converges (p = 2
n =l
n
. . oo > 1), the giVen senes 2::: n=l
(1)" arctann
n2
converges absolutely by the Comparison Test. 17.
f: (Inn1)" converges by the Alte~nating Series Test s.ince
..= 2
 1
Inn
>
.!, and since n
1 lim = 0 and 1nn
n oo
{ln1n } is decreasing. Now Inn < n, so
f: .! is the divergent (partial) harmonic series, f: 1 1 diverges by the Comparison Test. Thus, n=2.n nn n= 2
~ (I1n )".IS condi' w twnaIIy convergent. n=2 n 19.
00 . . 00 cos{nrr/ 3) :5 11 and 2::: 11 converges (use the Ratto Test), so the senes 2::: converges absolutely by the n. n= l n. n=l n.1
icos{nrr/ 3)1
n.1
Comparison Test.
. 21 . lim noo
y/i(lJ =
2 2 00 1 < 1, so the senes . 2::: . n + 1 = hm . 1+ + 1/n / = lim 2 1 n 2 n+oo 2n 2 + 1 n oo 2 n =l
2
(
n + 1 ) " IS . absolutely convergent by the 2n 2 + 1
Root Test. 2
23. lim
n..oo
y'jaJ =
lim " (1 +
n... oo
1)"
.!)" n
= lim (1 + n~ oo
.!)" = e>1 n
[by Equation 7.4.9 (or 7.4*.9) [ ET 3.6.6] ],
2
00 so the series L;:
n =l
(
1+ n
diverges by the Root Test.
.
25.
.
100 10 . Ian+ll . I(n + 1) C!1QO"+l · n! . I lim . 100 1) hm   Inn  (n +10 a,. n co (n + 1)! n 0100" .noo n+ 1 n 
noo
. 100 lim  ( 1 + 1 )100 nco n + 1 n
= 0 ·1 = 0 < 1 so the series
oo
nl00100"
n=l
n1
2:::
. is absolutely convergent by the Ratio Test.
27. Use the Ratio Test with the series 1·3 ·5 1 ·3 ·5 ·7 n  1 1 · 3 · 5 · · · · · (2n  1) _ co ,._ 1 1 · 3 · 5 · · · · · (2n  1) 1 ·3 1  3 ! +  5! 7! + .. · +( 1) {2n  1)! +···n~l(  1 ) · (2n  1)! (2n1)! I lim l an+ll = lim ~ (  1)"~1·3·5 .... ·(2n 1)[2(n +1)l]_ a,. noo (2(n+1)  1]! ( 1)" 1 ·1·3·5·····(2n1)
n  oo
=lim ~ (1)(2n+l)(2n 1)! 1 = lim _.!_=0 < 1, n oo (2n + 1)(2n)(2n  1)! noo 2n.. so the given series is ahsolutely convergent and therefore convergent.
® 2012 Ccngagc Lc.lm..ing. All Rights Resen·cd. May not be scanned. copiLd. or dupllcaled. or posted to a publicly ncccssiblc: \\'Cbsilc. in whole or in pan.
INFINITE SEQUENCES AND SERIES
CHAPTER 11
70
D
29 ·
n~l
00
00
2 · 4 · 6 · · · · · (2n) n!
(2 · 1) · (2 · 2) · (2 · 3) · · · ·. (2. n) nl
= n~l
oo
.2nn!
oo
.
.
= .~1 ;;;[ = n~l 2n, wh1ch diverges by the Test for
Divergence since lim 2n = oo. n oo
I a,. I =
.. lim an+l . de fi mt1on, 31 . B y the recurs1ve noo
00
5 > .1, so the series diverges by the Ratio Test.
4
00
n
n=l
a n+l l . lirn  = 11m
I
15n + 11 = 4n + 3
bn I: b"" cosn1r = I: (1 )" 2':., where bn
•
33. The series
n+oo
lim
n400
an
noo
I( l)n+lb~+l + · ( n
> 0 for n ~ 1 and lim bn
n
n=l
1
n ) b  1 n ~
n oo
I= 1.
1m
n ~o oo
b · n n n +1
1 2
= .
. ~ = 12 ( 1) = 12. < 1, so the senes LJ n= l
b~ cos n 1r iS . n
absolutely convergent by the Ratio Test. 3 1 1 3 35. (a) lim 1 / (n + ) 1 = lim n 3 n _,oo 1/n3 n +oo (n + 1) (b) lim
I(n2 ++ 1) · 2n" I=
(c) lim
Iv~ ·( n+1  3
n  oo
n .... oo
(d) lim n .... oo
lim n + 2n
1
n
V:
1
n t oo
n 
1
=
1 (1 + 1/ n) 3
lim n >oo
1 = nlim ( + +oo 2
1 = 3 lim ·;
n
n>oo
n = + 1
= 1.
= 1 .
1  ) 2n
2
3 lim n + oo
Inconclusive
Conclusive (convergent)
v~/ I+Vn
= 3. Conclusive (divergent)
I
.Jn + 1 · 1 +n21 1/n2 + 1 ]  = lim. [ R +1  · 1 + (n + 1) 2 Vn n>oo n 1jn2 + (1 + 1/n)2
I= lim I(nxn+~)' + . · n! I =
37. (a) lim \ an+l
a,.
n >oo
X"
n >oo
lim 1 x1 n +1
n>oo
. = 1. rnconc lus1ve
1 = lx l nlim   = lxl · 0 = 0 < 1, so by the Ratio Test the > oo n + 1
00 x n series I; I converges for all x .
n =O
n.
(b) Since the series of part (a) always convergt<s, we must have lim xn, n. oo
0
39. (a)
Ss
1
= n~l n2" =
1
1 . 1
2 + 8 + 24 +
1 64
+
= 2 (n: n+1
1'n+l 
rn =
n
2 (n+ 2)  2 (n+ 1)
in using s s is Rs ~
a6
1
_ lim
Tn
=
n>oo
1
=
1 160
661
= 960
1
n
2
< 0.00005
(11 + 1 )2 11
¢?
= 0 by Theorem 11.2.6.
. :::::: 0.68854. Now the ratiOS
) form an increasing sequence, since
(n + 1) 2  n(n + 2) 2 (n + 1)(n + 2 )
1/(6 · 2 6 ) _ 1 112
=
1 2(n + 1)(n+ 2) > 0. SobyExercise34(b), theerror
1 :::::: 0.00521. 192
(b) The error in using s ,. as an approximation to the sum is Rn =
(n + 1)
n.
an+~
1 2
= (
2 ) n+I . We want Rn < 0.00005 n+12
(n + 1)2n > 20,000. To find such an n we can use trial and error or a graph. We calculate
= 24,576, so su =
£; n 12"
:::::: 0.693109 is within 0.00005 ofthe actual sum.
n= l
©
¢?
20l2 Ccngo.gc Learning. All Rights Reserved. May nol be scanned, copied, or duplicttted, or posted to tt publicly accessible website, in whole or in p.1r1.
SECTION 11.6
41. (i) Following the hint, we get that Iani the series 2::=N
ABSOLUTE CONVERGENCE AND THE RATIO AND ROOTTESTS
0
71
< rn for n 2: N, and ~o since the geometric series 2:::'= 1 r" converges [0 < r < 1],
lanl converges as well by the Comparison Test, and hence so does 2:::'=1 lanl, so 2::_:'=1 an is absolutely
convergent. (ii) If
lim . y'j'aJ = L > 1, then there is an integer N
such that
ntoo
lim
an
n~oo
(iii) Consider
y'j'aJ >.1 for all n 2: N, so lanl > 1 for n 2: N. Thus,
"# 0, so 2:::'=1 an diverges by the Test for Divergence:
f
n =l
~
[diverges] and
n
f
n=l
~
y'j'aJ =
[converges]. For each sum, lim
n
1, so the Root Test is inconclusive·.
n+oo
2:: an is absolutely convergent, and since la;t I ~ !ani and Ia~ I ~. I ani (because a;t' and a~ each equal
43. (a) Since
either an or 0), we conclude by the Comparison Test that both 2:: a;t and
2:: a;; must be absolutely convergent.
Or: Use Theorem 11 .2.8.
2:: a;t and 2:: a~
(b) We will show by contradiction that both would
l::(a;t' ~an)
by Theorem 11.2.8. But l::(a;t
must diverge. For suppose t hat 2:: a;t converged. Then so ' ~an) = 2:: [ ~(an + !ani) ~an] = ~ 2:: Ian i, which

diverges because 2: a,. is only conditionally convergent. Hence, 2: a;t can't converge. Similarly, neither can 45. Suppose that
2:: an is conditionally convergent.
2:: n 2 an is divergent: Suppose 2:: n 2 a,. converges. Then
(a)
•
'
2:: a~.
integer N > 0 such that n > N
=}
n
comparison with the convergent p series .
assumption that
2
lanl < 1.
2
lim n an '1. + 00
= 0 by Theorem 6 in Section 11.2, so there is an ;
1 For n > N, we have !ani < 2 , so
.
n
2:
!ani converges by
n>N
I;: .;.. In other words, 2:: an converges absoluiely, contradicting the n>N
n
2: an is conditionally convergent. This contradiction shows that 2:: n 2 an diverges .
Remark: The same argument shows that 2: n'~' a .. diverges for any p > 1. (b)
f (n lnl)nn is conditionally convergent. It converges by the Alternating Series Test, but does not converge absolutely .
n =2
·[by the Integral Test, since the function .
00
1 2
dx  x 1n x
00
n~ nan
2
= 00
= n~
lim
t ~oo
2
(
1t
dxx 1n x

2
=
1 f (x) = X ln X
[ t lim ln{ln x )]
t ~ oo
2
00
1
(
= oo]
. Setting an
= (l}n I n for n ~ 2, we find that n n
1)" ln n converges by the Alternating Series Test.
It is easy to find conditionally convergent series "~
is c~ntinuous, positive, and decreasing on [2, oo) and
00
(1t1
n= l
n
2:: an such·that ~na n diverges. Tw~ examples are 2::
1)"1
and
fo , both of which converge by the Alternating Series Test and fai l to converge absolutely because 2:: Ian i is a
pseries with p ©
~
1. In both cases,
2:: nan diverges by the Test for Divergence.
20l2 Ccngage Le(1111 ing. All Rights Reserved. May not be scanned, copied, or duplicated. or posted to a publ ic ly occcssiblc website, in whole or in part.
72
D
INFINITE SEQUENCES AND SERIES
CHAPTER 11
11.7 Strategy for Testing Series 1. 1 < 1 = (1)n for all n
n+3n
3n
f
> 1.
3

n=l
(~)n is a convergent geometric series [lrl = t
< 1), so
.
f
n=l
~ n
+3
converges by the Comparison Test.
f
3. lim la..l = lim _2:_ = 1, so lim a n= lim (1}" _2:_ does not exist. Thus, the series ( 1)n __.!:_diverges by n  oo noo n + 2 n<00 n 
5. lim Jan+II = lim noo 00
an
.
1 2 I(n 5)n+l + ?
n
n
•
(
(
.!)
5)" I = lim 2(n+ 1) 2 = ~ (1 + n  oo 5n2 5 noo n
llm
n 2 2 nl
2
=
~(1) = ~ < 5 · 5
1, so the series
n2 2nl
n~l ( _ )" converges by the Ratio Test.
5
7. Let f( x) = Since
J
~ Then f
xvlnx
1°/x,] = ~ dx [d..,:' 1 x u  ex
given series
f
•
2: k 2 ek
=
lim t .... oo
X
d~ =
lim [2J!nX] t = lim ·(2 villt 2
y lnx
2
t_.oo
too
00
2:
k2
.
.
""'"'k· Usmg the Rat1o Test, we get e
= 12
11.
v1Ii2) = oo. Since the integral diverges, the
n vlnn
k=I
k=l
r
J2
~ diverges.
n=2
00
2
j u 112 du = 2u11 + C = 2 JinX+ C, we find
xvlnx
f oo ~= J2 X y lnx
9.
is positive, continuous, and decreasing on [2, oo), so we can apply the In..tegral Test.
.
.e! .e! =
< 1, so the series converges.
f: (~ + ~, ) ~ E ~ + E (.!)n ·The first series converges since it is a pseries with p = 3 > 1 and the second 3 n n 3
n=l
n= l
n=l
series converges since it is geometric with lrl = ~ < 1. The sum of two convergent series is convergent. 13. lim la"+ 1 1= lim noo
an . .
n<00
J
3n+l (n+ 1) (n + 1)!
2
~1= 3nn2
2 3 2 1 lim 3 (n + 1) 2 =3lim n~ =0 < 1,sotheseries n~ n<00 (n + 1)n n <00 n . ' n=l n.
f
converges by the Ratio Test.
00
k~l
(6)k 2k13k+l = k~l 23(6)k k converges. It follows from Theorem 8(i) in Section 11.2 that the given series, k~l k , 00
00
kk
also converges.
© 2012 <:engage Lcnming. All Rights Rese"·ed. Moy not be scanned, eopied, or dupuented, or posted to • publicly accessible website. in whole oc in p:lrt.
SECTION 11.7 STRATEGY FOR TESTING SERIES
17
lim
· noo
I I= an
I
n+oo
= ~ < 1,
lim 2 + 1/ n 3 + 2/ n
=
3
n+oo
=
19. Let f(x)
73
lim 11 · 3 · 5 .. · · · (2n 1)(2n + 1) . 2 · 5 · 8 · .... (3n  1) = lim 2n + 1 2 · 5 · 8 · · · · · (3n 1)(3n + 2) 1 · 3 · 5 · · · · · (2n 1) n+oo 3n + 2
an+ l
so the series
0
oo
1 . 3. 5 .. · .. (2n  1)
n= l
2 · 5 · 8 · · · · · 3n  1
2:::
(
ln x r . Then f 1 (x)
=
vx
. I's RuIe, lim Inrn By I'Hosptta n+oo vn
) converges by the Ratio Test.
.
2  ln x
=
I'tm
n+oo
lrin.
?
< 0 when lnx > 2 or x > e, so r
x 2 312.
.
vn
(1/n ) 1/ 2 Vn
=
1'tm
n  oo
2r vn
.
ts decreasmg for n
> e2 .
=,
0 sothe senes . n~ (  1)"Inn r;;, converges by the 61 vn
Alternating Series Test. 21 . lim lan. l = lim IC l )ncos(l/n2 )1= lim icos(1 /n noo
2
).1=cos O=
n 'oo
noo
1, so the series .
E(l)'' cos(l/ n
n =l
2
)
diverges by the
Test for Divergence. 23. Using the Limit Comparison Test with
an
= tan(.!.) and b,. = .!., we have
n
lim an = lim tan(1/ n) = lim tan(l/ x) bn n +oo 1/ n :z:+oo 1/x
noo 00
00
n= l
n.=l
M:
.
n
lim sec2(1 /x). ( 1/x2) = lim sec2(1/ x) = 12 = 1 > 0. Since 1/x 2 :z:+oo
:r oo
2::: bn is the divergent harmonic series, 2::: a,.. is also divergent. · ~ li an+l 1' I (n + 1)! c" 1 . (n + 1)n! · e" . n+ 1 25. Use t I1e Ratto . est. f"LH>O m  an = .,._.00 un < +t)~ ·  1 = 1rm = 11m ~+t = 0 2 en n. ft.  00 en + 2n+ln! r&.oc e n
I
I.
2
2
oo n! < 1, so n:L :r = l en
converges.
27.
roo ln:
dx = lim [ ln X t  oo
X
./2
k ln k
 ,3
(k + l)
29.
f
an =
n =l
X
k ln k
ln k

.!.] t . [using integration by parts] X
.
.
oo
<  k 3 = k? , the gtven senes 2::: ..
M: 1. So
E ln: converges by the Integral Test, and since n
n= l
1
·
k In A:
k=t(k+ l)
3
• converges by the Companson Test.
1 f (1)" ___!_h = f ( 1)" b,,. Now b,. =    > 0, {b,..} is decreasing, and cos n cos1 n
n=l
"·=1
1
lim b, = 0, so the series n oo
converges by the Alternating Series Test. . 0 r: W rtte
1
2 = e" +2 e" < e"
co~lt n
~
Comparison Test. So
1 '. . 1 an d ~ ts a convergent geometrt.c senes, so 6~   .ts convergent by t he 6 n = l e" n = l cos11 n
E( 1) " cos ___!_h is absolutely convergent and therefore convergent. n
n=l
a~.; k+oo
31. lim
Thus,
= lim
koo
oo 2:::
3
s"
k
~
k = 1, 3
+4
5k
+ 4k =
(divide by 4")
diverges by the T<;st for Divergence.
'
74
CHAPTER 11 INFINITE SEQUENCES AND SERIES
0
33. n+oo lim
~ = n+00 lim (~ ) n + 1
1
"2 jn
= lim [( n+00
n
+ 1) I n
t
1 li m (1· + n 11
=
1
r
oo
I: e < 1,. so the series n=1
= 
n>oo
n2
(·n) n
+ 1
converges by the Root Test. 35.
an =
n 1; 1/n =
n .
~ 1 /n, so let bn =
[see Exercise 4.4.6 1], so .
th~ series
; and use the Limit Comparison Test.
11.8
n+oo n1/n
f: 1
,. diverges by comparison with the divergent harmonic series. n=l n 1 11
37. lim ~ = lim (2 1/"  1) = 1  1 = 0 ntoo
n~oo
1 l .un a, = lim . = 1 > 0
n+oo bn
.
00
< 1, so the series I: ( y/2  1)"' converges by the Root Test. n=1
.
Power Series
1. A power series is a series of the form :L:::'=o enx"
= co +c1x + c2x 2 + c3x3 +
· ··,where x is a variable and the en's are
constants called the coefficients of the series. More generally, a series of the form :L:::'=o c,,(x  a)" =eo+ c1 (x a)+ c2(x a) 2 + · · · is called a power series in
(x  a) or a power series centered at a or a power series about a, where a is a constant.
lim n+00
I
a n+1 = lim ~ Un ntOO
I
1
(1)"+ I= ( (n+1)x"H )  1
71.
nxn
I
n+1 I = lim [(1 + 1) JxJ] = Jx J. By the Ratio Test, the lim ( 1)x
n
nH)O
n
n00
00
series
I: (1)"nxn converges when !x i < 1, so the radius of convergence R =
'
.
1. Now we' II check !he endpoints, that is,
n= l
x
= ± 1. Both series n=l f: (
1)"n(±1)"
the interval of convergence is I
·
= n=l f: (=F1)"n diverge by the Test for Divergence since
·oo
= oo. Thus,
= (1 , 1).
x", then' lim . Ia,.+1 . I x"+l · 2n 11 5. If an =   1 = hm n+oo an n+oo 2n + 1 x" 2n 1 the Ratio Test, the series
lim J(=F1)"nl TI.+OO
. (2n . (2 1/n / lxJ) = JxJ. By = n+oo !1m   1 Jx l) = hm n+oo 2 + 1 n 2n + 1
~
I: 2  converges when Jx! < 1, so R =
n=l n  1
oo
1. When x
= 1, the series I: 2
1
n=l n  1
· diverges by
. . 1  > 1 and 1 ~ 1 d'1verges smce . .It .ts a constant mu It .ipIe o fth e harmoruc . senes. . companson Wit. I1 6~  1 srnce 6 n=1 2n 2n  1 . 2n 2 n=l n
When x =  1, the series
7. If an=
lim
Ian+1 I= .
f: (
x",, then lim lan+l l n. n + oo an
So, by the Ratio Test, R
n  oo
1 ) n converges by t~e Alternating Series Test. Thus, the interval of converge~ce is [1, 1). n=1 2n 1
an
©
=lim I(x"+\; x 1 = !xi + ·. I=lim 1 + n+oo
n
X"
n +oo
n
1
1 lim  
n+oo
n
+1
=lxl·
0 = 0 < 1 for all real x.
= oo and I= ( oo, oo).
~~ =n.oo lim Ix(n + 1)2 1 =lim [JxJ 2n2 noo 2
lim J (n + 1)2 x"+l . 2 2n+l n x"·
nCXJ
n!
1 .
.
(1 +
.n!.)2] = EJ.(1)2 = ~ JxJ. By the 2 .
2012 Cengagc Learning. All Rishts Reserved. May not be scanned. copied. or duplicated. or posted to u publicly occcssiblc website:, in whole or in part.
SECTION 11.8 POWER SERIES
n2xn ,. converges when 2
00
2:: (1)'" .
Ratio Test, the series
n=l
t lx l < 1
¢>
0
75
lx l < 2, so the radius of convergence is R = 2.
00
2:: (=F1)"'n 2 diverge by the Test for Divergence since n= l
lim
I
n oo
(:fl) n n 2 1 = oo. Thus, the interval of convergence is I = ( 2, 2).
11 . If an =
(  3)"x" n 312 , then . 11m
n•oo
a,.+t
Ia., I=
( 3) n+l x n+l n I I (n + 1)3/2 · (  3)nx n = 3/2
1'lffi
n~oo
1'Im
n~oo
I 3 (. n )3/21= 3 1 I 1'tm. ( X

X
n+1
n+oo
1 1 + 1/n
)3/2 . I
= 3lx l (1) = 3 lxl 00
By the Ratio Test, the series
2:: ,.=
1
00
· I: n=l
3)"
(
'
nyn
x "' converges when 3 lxl < 1
( 1)"
.
~ converges by the Alternating Series Test. When x =
n
( p = ~ > 1). Thus, the interval of convergence is [
13.Ifan = (1)" ~ . then 4" ln n
xn
00
x =  4, 2:(1)" n ln 4
1
% , the series n=l 2:: 00
n
312
S· the series
% When x =
is a convergent pseries
S, t].
lim l a n+t l = lim I xn+l . _4" ln n i= E11im 1n n =1.::1 · 1 an n+oo 4n+l ln(n + 1) x" 4 n~oo ln(n + 1) 4
n~oo
00
n
%,so R =
lx l <
= 1: 1. By the Ratio Test, the series converges when 1:1
[by !' Hospital 's Rule]
n=2
¢>
2:: n= 2
=
< 1
lx l < 4, so R = 4. When
¢:>
00 00 [(1) ( 4)t 1 . 1 1 1 = 2:  . Smce lnn < n for n ~2.  >  and 2::  is tbe " ln 4 1n n n n = 2 n n n =2 1n n
divergent harmonic series (without then = 1 term),
f 1n1n ' is divergent by the Comparison Test. When x = 4, .
n= 2 00 xn 2:: (  1)n n ln 4 n =2 n
15. If an =
1
00
= 2:: (1)" ln n=2
n
I
, which converges by the Alternating Series Test. Thus, I = (  4, 4] .
I . ·1
2
(x  2)"' , then lim . a n+t (x  2)"+1 n + )1 )   = lim ( · ( n2 + 1 n ~oo an n+ oo n +1 2 +1 X 2 "
Ratio Test, the series
oo
2::
n=O
x = 1, the series
(x 2)" n2
+1
converges when lx 21 < 1
I= lx 21 lim .
[R = 1]
n~ oo
¢:>
2
(
n + 1 ) = n +1 2+1
1 < x 2 < 1
n =O
comparison with the p series
f: n~.
[p = 2 > 1]. Thus, the interval of convergence is I = [1, 3].
1
2

 converges by
+1
·
n= l
00
1
By the Ratio Test, the series .~
3"'(x + 4)"
Vn
converges when 3 lx
+ 41
< 1
¢:>
lx
+ 41 <
1 [R =
21. By the
1 < x < 3. When
¢:>
f: ( 1)" + converges by the Alternating Series Test; when x = 3, the series f: n +1 n
n=O
lx 
t]
¢:>
® 2012 Ccngoge l.e:~ming. All Rights Rcscm :d. M•y no1be scanned, copied, ur duplicated, or posted lo • poblicly accessible wcbsile, in whole or in pnrt.
l
76
CHAPTER 11
0
INFINITE SEQUENCES AND SERIES
t < x + 4 < t
 ¥ <; x < 
<*
When
f: (1)" vn~ converges by the Alternating Series
~; =  ¥, the series
n=1
¥, the series 'f: ~ diverges [P = ~ ~ 1]. Thus, the interval of convergence is I
Test; when x =
n =l
r, then
2 nn
19. If an = (x 
11 3 •
lim
n+oo
vn
.yljO:j =
2
lim lx n
1
ntoo
= [
¥,  ¥).
= 0, so the series converges for all x (by the Root Test).
R = ooand I= ( oo, oo). 21.
an=; (x  a)", where b > 0. 1
lim lan+ll =lim (n+ 1 ) lx a.J''+
an
n>oo
b"
.
n lx
bn+l
n>oo
By the Ratio Test, the series converges when lx
a b < x < a+ b. When lx  al
23. lfa,
= n! (2x 
for all x :f;
n~oo
~ al < 1
I
<*
= b, lim ian = lim HtOO
'UtOO
n +oo
u.oo
an
(n +
I
lx . a l =
lx  ai_ b
b
71,
ix a l < b
 b<
[so R = b] <*
x a < b
<*
n = oo, so the series diverges. Thus, I= (a b, a+ b).
1 ~~~(X2x ) )"+ 1 =.lirr~ (n + 1) l2x 11+ oo as n+ oo 1n
n.
t. Since the series diverges for all :c # t, R = ,
n3
U.H X)
•
0 and I = { ~} .
then
I a,. I= lim I(5x4)"+ (n + 1)3
CLn+l . I1m 
n>oo
(1 +~)
1
1)", then lim jan+! I = lim
(5x 4)"
25. If a, =
=lim
n
a!
3
1
I = lim l5x n (5 x  4)" n> oo
n>oo
41 ( _ n _ n+1
)3=
lim l5x  41 n >oo
(1 +11/n)3
= l5x  41· 1 = l5x  41 ~
By the Ratio Test,
oo
(5x 4)"
n =l
n3
L
~ < x < 1, so R = t·
converges when l5x
When x = 1, the series
41 < 1
<*
lx
tI < t
<*
~
< x
'f: n~1 is a convergent pseries (p = 3 > 1). When x =
n=1
t
<~
<*
~.the series
f: (~"' converges by the Alternating Series Test. Thus, the interval of convergence is I= [~, 1]. n
n =l
x 'n
27. If a, = 1 . 3 . 5 . .. . . ( 2 n _ 1 ) , then · lim lan+l l n>oc
I
lim
x"+l
~ = n~oo 7173:::5·  _;(72n,1")"(2::ncf::17)
the Rat1o Test, the senes L •
•
~
00
n= l
29. (a) We are given that the power series
4
<x
©
~
~0)2
(·
1 · 3 · 5 · · · · · 2n  1
_1·3·5· . . . ·(2n  l)l_
4. ln particular, it converges when x =  2; that is,
f.. 1 ~y
_ lx_l _ _ 0<1 Tl b . lUS, y
) converges for all rea l x and we haveR= oo and
I:::'=o CnX" is convergent for x =
Ccngogc Lcnming. All Ri ghts Rcscrn:J.
r
 n~oo 2n + 1 
:c"
I= (  oo, oo) .
4. So by Theorem 3, it must converge for at least
L~o c,( 2)"
is convergent.
nol be scanned, copic:d. or Jupliculcd, or posted lit u puhl icly nccC'ssihlc. wc:bsit~. in whole or in fl:trt.
SECTION 11.8 POWER SERIES
0
n
(b) It does not follow that I:::'=o en(  4)" is necessarily convergent. [See the comments after Theorem 3 about convergence at
t /(n4").]
the endpoint of an interval. An example is Cn = (  1 (n!)k n 31. If a,.=( ) ' x , then kn.
I
lim a n+ll a.,
=
....... oo
lim [(n + 1)!]k (kn)! lxl = lim (n + 1)k lxl (n!)k [k(n + 1)]! ,.....,CX> (kn + k)(kn + k 1) · · · (kn + 2)(kn + 1)
n ..... ex>
.
= J~ 
(n+1)] [ (n+1) (n+1) (kn + 1) (kn + 2) ... (kn + k) lxl
n+1 ] lim [ n + 1] l.t.m [ 
n+ex>
= (
~
kn + 1 ,...... ex> kn + 2
J
lxl
<1
lxl
¢:}
···
. [n+1 ] 1liD noo kn + k
IX I
< kk for convergence, and the radius of convergence is R =
kk.
33. No. If a power series is centered at a, its interval of convergence is symmetric about a. lf a power series has an infinite radius of convergence, then its interval of convergence must be (  oo, oo), not [0, oo).
35. (a) If a,. =
r
,.:.,~
(1t X2n+l ) n+l, then
I(
n. n+ 1 1. 22
a
n+l li a,: = n .....~
I I
x2n+3
I
.
(n + 1)!(n + 2)! 22n+3 .
nl(n l . + 1)'22n+l . x2n+l
=
(x)2
lim
2 ,. ._..... oo
1
(n + 1)(n + 2)
= 0 for all x.
So J 1 ( x) converges for all x and its domain is (  oo, oo). (b), (c) The initial terms of J1(x) upton = 5 are ao =
and a(j
=
~·
xu , , . The partial sums seem to 176 947 200
approximate J1 ( x) well near the origin, but as lx l increases, we need to take a large number of terms to get a good approximation. 2 3 4 37. S2nl = 1 + 2x + x + 2x + x
+ 2x5 + · · · + x 2" 2 + 2x2" 1 = 1(1 + 2x) + x 2 (1 + 2x) + x 4 (1 + 2x) + · · · + x 2"  2(1 + 2x) = (1 + 2x)(1 + x 2 + x 4 + ... + x 2 ~ 2 )  x 2" 1 + 2x = (1 + 2:c)1 [by (11.2.3) with r = x 2] +    as n+ oo by (11.2.4), when lx l < 1. 1  X2
1  X2
A Iso
S2n
= S2n1 + x 2n 1 + 2x
approach   as n . 1 x2
+
+
1 +2x. _ x 2 smce x 2n 1
+
0"•Or Ix I < 1Th &: . ere,ore,
Sn +
1+2x . _ x 2 smce S2n .and S2n1 both 1
. . 1 + 2x oo. Thus, the mterval of convergence IS (  1, 1) and f (x) =   2 . 1 x
© 201 2 Cengagc Lcoming. All Rights Reserved. May not be scnnncd, copied, or dupliCD.tcd: or posted to a publicly acccs!tiblc website. in whole or in part.
78
D
CHAPTER 11.
INFINITE SEQUENCES AND SERIES
y'jCnT =
39o We use the Root Test on the series I:; c..x" We need lim \/lc,x"l =I xi lim 0
n. oo
lxl
ntoo
c !x! < 1 for convergence, or
< 1/c, so R = 1/c.
41 . For 2
< x < 3, I:; cnx" diverges and I:; dnx" .convergeso By Exercise 11.2.69, I.:;( en+ dn) x" diverges. Since both series
converge for Ixi < 2, the radius of convergence ofi.:;(en + dn) x" is 2.
11.9 Representations of Functions as Power Series 00
00
n=O
n= l
10 If f(x) = I:; c..xn has radius of convergence 10, then!' (x) = I:; nc..x" 1 also has radius of convergence 10 by ·
Theorem 20 1
3. Our goal is to write the function in the form  · , and then use Equation ( I) to represent the function as a sum of a power 1 r 1
1
00
00
series. f(x) =   = ( ) = I:; (xr' = I:; (1)"x" with 1xl < 1 1 +X 1  X n =O n= O
2 = 32( 1) x/ = 32
00
5o f(x) =
3
_x
that is, when Ixi
1
_
< 3, so R
n~o
3
oo
0
lxl < 1, so R = 1 and I= (  1, 1).
The series converges when
lxl 3 < 1,
= 3 and I = ( 3, 3).
~o [ (~)
The geometric series ...
R
(x)" 1 x" 3 or, equivalently, 2 fo 3,.+1
0
¢:?
conver~es when 1 (~) 1< 1
2
2
2 ] "
¢:?
~~ 1
<1
2
¢:?
lxl < 9
¢:?
lxl < 3, so
= 3and I= (3,3).
The series converges when lxl < 1, so R ·
A second approach:
1 +X f (x ) :::; = 1
X
= 1 and I
= (  1, 1).
(1  x) + 2 =  1 + 2 ( 1 ) 1 X 1 X
~ x n = 1 + 2 LJ ~ x n. = 1 + 2 nLJ = O n=l
A third approach:
f(x)=
1
1
+x =(l +x)( ) = (l +x)(l +x+x2 +x3 + ·o·) 1 x 1 x
= (1 + x + x 2 + x 3 + .. o) + (x + x 2 + x 3 + x 4 + ooo) = I+ 2x + 2x2 + 2x3 + o.. = 1 + 2
f:
Xno
n=l
© 2012 Cengoge Learning. All Rights Reserved. Mo.y not be scanned. copied, or duplicated, or posted toR publicly accessible website. in whole or in p:ut.
SECTION 11.9 11. f (x)
x
3
3
= x 2  x 2
..,....,,,"7
(x 2)(x + 1)
A
B
= +   => x 2 x+ 1 ·
REPRESENTATIONS OF FUNCTIONS AS POWER SERIES
D
3 = A(x + 1) + B(x  2). Let x = 2 to get A
1 and
=
79
=  1 to get B =  1. Thus
3 _ _1_ _ _ 1_=~( 1 )1 x2 x  2  X  2 X+ 1  2 1 (x/2) 1  ( x) =
f:
n=O
E (=)" _n=O E(x)n
=~
2 n=O 2
[~2 (1:)" 1(1)"']xn = E [( 1t+l n~l]x" 2 n=O 2
We represented f as the sum of two geometric series; the first converges for x E ( 2, 2) and the second converges for ( 1, ~) . Thus, the sum converges for x E (  1, 1) =I.
=
13. (a) f(x )
1
(1+x) 2
d [ L..... ~ (  1)n x " ] =,  dx n=O
= ·d ( 1) dx
1+x
~
=
L (1)"+1nx" 1
n=l
[from Exercise 3]
~
[from Theorem 2(i)] =
L (1)"(n +
n=O
l)x" with R = ~ .
In the last step, note that we decreased the initial value of the summation variable n by 1, and then increased each occurrence ofn in the term by 1 [also note that (1)"+2 = ( 1)" ].
(b)f(x)=
1 1 =1:j_[ ] =1:j_[E(1)"(n+ 1)x"] (1+x) 3 2dx (1+x) 2 2dx n=O ~
[frompart(a)]
~
= ~ L (1)"(n + 1)nxn 1 = ~ L ( 1)n(n + 2)(n + 1)x" with R = 1. n=l n=O x2 1 2 2 1 ~ (c)f(x) = ( +x) 3 =x · (l+x) 3 = x · 2nJ; { 1)''(n+2)(n+1)x" 1 0
= 1: E (1)"{n+2)(n+ 2 n=O
[frompart(b)]
l)x"+~
To write the power series with x " rather ~an ·x"+ 2 , we will decrease each occurrence of n in the term by 2 and increase 1 .CX>
the initial value of the summation variable by 2. This gives us
15. f(x) = ln(5  x) =
~~ = 5x
Putting x = 0, we get C 1 17. We know that 1 + 4X
= ln 5.
=
1  4X
f(x) = {1+x4x)2 = · 4x. for J4xJ
< 1 <=>
Jxl
~~ = 1  x/5
2
1:1[f: (::\"] 5
The series converges for Jx/51
1
(
_1: 5
2 ..~ ( 1)"(n)(n 1)x,. with R =
n=O 5}
dx = C
1.
~
< 1 <=> Jx J < 5, so R = 5.
00
) =
4 2 (1 + 4x)
f:
E xn+l = ·C..:... x" 5n=o5"(n+ 1) n=tn5n
L (4x)". Differentiating, we get
n=O
f;{  4)n+1 (n+1)xn = f;(1)n4n(n+1)xn+l 4 n=O n=O
=X
< ;}, so R = ;} .
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80
0
CHAPTER 11
INFINITE SEQUENCES AND SERIES
1
19. By Example 5, (
1 X
)2 =
00
L: (n + 1}xn. Thus,
n =O
1+ X _ 1 X _ ~ ( } ,n ~( n+l !(x ) _ (1x)2  (1 x)2 + (1 x)2  ~o n + 1 x + n~o n + 1}x 00
00
= 2::(n+1}xn+
E
nxn
[make the starting values equal]
n =l
n=O
= 1 + f; [(n + 1) + n ]xn = 1 + f; (2n + 1}xn = f; (2n + 1}xn n= l
The series converges when lx2/ 16l
82
= s1 
x3 162, 83
n= l
with R
= 1.
n=O
< 1 <=> x 2 < 16 <=> lxl < 4, so R = 4. The partial sums are 9 1 = ~. 16
5
x = 82 + 163 , 84 = 83 
x1 164 , 8s
= S4 +
x9 , .... Note that 8 1 corresponds to the first term of the infinite 165
sum, regardless of the value of the summation variable and the value of the exponent.
As n increases, 8n(x) approximates
x)
23. f(x) = In ( 1 + 1 X
better on the interval of convergence, which is (  4, 4}.
= ln(1 + x)  ln(1  x) =
=I L~0(1)nxn + n~o =1
f
(2 + 2x2 + 2x4
xn] dx
+ ·· ·) dx
=1
I
 dx 
1 +X
+
I
dx = 1 X
4 2 3 [(1 x + x  x +x
=I f;
2x2 " dx
=C +
n=O
,
But /(0) = ln t = 0, soC = 0 and we have f(x) =
oo
E

• .. )
1
dx (
X
)+
I
+ (1 + x + x
2
 dx 1 X
+ x 3 +x4 + ... )] dx
2 2 X n+l n=O 2n + 1
f;
zx2n+l
with R n = O 2n + 1
which both diverge by the Limit Comparison Test with b,. =
I
= 1. If x =
1
oo
±1, then f(x) = ± 2
E 2
n =O
.!.n .
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n
+1
,
SECTION 11.9
0
81
2x 5
2x 3
2x
l ' s2 = s1 + 3 ' S3 =
The partial sums are s1 =
REPRESENTATIONS OF FUNCTIONS AS POWER SERIES
s2
+ 5 ' ·· · ·
As n increases, s,.(x) approximates f better on the interval of convergence, which is {1, 1).
25.
=t
_ t_ B 1 t
when
I
. _1_ 8
1
lt8 l < 1
t
= t f: (tB )"
f:
=
n=O
. t tB·n+2  converges j 1   dt = C + I: 8n. + 2 The series for 1 t 00
=>
t8n+l
n =O
t8
1
8
n =O
It I < 1, so R = 1 for that series and also the series for t/{1 
¢}
t 8 ). By Theorem 2, the series for
_!_a dt also has R = 1. 1 t
x" oo x"+2 2 for lx l < 1, so x ln{1 + x) = I: { 1t  1  and n n= 1 n
oo
27. From Example 6,ln{ 1 + x) = E (  1)"  1 n =1
I
x 2 ln(1 + x) dx = C
x 2 ln{1
1
1
0.2
£
. 0
.
R = 1 for the series for ln{1 + x), so R = 1 for the series representing
n n+ 3
I
2
;· oo
oo
1
[
x6
x 
6
x lt
.
] 0.2
+ 11  ·· ·
,&n+l
00
0 .1
So
1 0
"'
(3x?"+l 2 n 1 1
. Thus,
{0.2)6
{0.2) u
6
11
0
31. We substitute 3x for x in Example 7, and find that
n= O
+1
dx=
I
109 . So I ~ 0.2 
00
2:: ( 1)
2n
+1
+
00 .
1 9 x 103 5 105
..
dx = C+ E( 1) ( n =O
~
32n+ l x2n+3 2n 1 1
0 .1
0
243 2187 35 X 107  63 X 1Q9 2187 x 63 109
so if we use
{0.2) 6/6 ~ 0.199. 989 to six decimal places.
[ 3 3 ] 33 G 35 7 37 !) x arctan (3x) dx = .....:::.__ ~ +~__E._+ ··· 1 ·3 3·5 5·7 7·9
• •IS a 1ternatmg, . The senes so 1'f we use three terms, t11e error •IS at most
0
.
.'
".32 .. +1 x2n+2
n=O
1 9 = 103  5 X 10°
Jr ·l x arctan (3x) dx ~
.
= 0.2   +   · ··. The senes IS alternating
the fi rst two terms, the error is at most (0.2) 11/11 ~ 1.9 x
. I j xarctan(3:c)d:v = x 2:(1)
= 1.
=>
2:{ l )"'x5 " dx = C + 2:(1)" x 5n ,.=o n=O
  5 dx = 1+x
+ x) dx also has R
x ln{l
5 )"=f:{1)"xu" = nf;(x =O n= o
1  ( x·5 )
 .. dx = 1 + x~
I=
n=1
x"+3 ( )'
+ x) as well. By Theorem 2, the series for
29 · 1 +1x 5 =
I
oo
+ I: (1)"'
+ .. ..
3 .5 x 108 . so
x + 35 243 ~ 0.000 983 to six decimal places. 107
<[) 10 12 Ccngagc Leming. 1\ ll Rip;hts Rcscn·cd. May not be scrumcd, copied, or duplicated, or poslt.'ll to a publicJy nccc.ssible website, in wbolc or in parL
)(
2n j 3
)
82
0
CHAPTER 11
INFINITE SEQUENCES AND SERIES
x3
33 . By Example 7' arctan x = x   3
5
1
The senes . 1s . aIternatmg, . so 1'f we use t11ree terms, t 11e error ·ts at most (0.2) 7
(0 2? Thus, to five decimal places, arctan0.2 ~ 0.2  · 
3
2
,
1
X Jo (x)
+ XJo(X) +X
2
00
Jo(x)
= n~l 00
= n~l _

00
r~l
(0 2? + · 
( 1)"2n(2n 1)x2" 22"(nl)2
2
J
1;
+ 3~0
= n~O nf ::}
f (x)
00
0.000 002.
0.197 40.
(1t2nx2 "' 22"(n!)2
00
+ 1~1
( 1)"2nx2'' 22"(n!J2
oo
+ ,~1
( 1t2nx2" 22"(n!)2
00
+ n~l
]
+ n~O oo
+ ,~1 00
+ ,~1
( 1)"x 2"+2 22" (n!)2 ( 1t 1 x 2" 22n 2 [(n 1)!] 2 (1)"(1 ) 1 22n 2x 2" 2 2 "(n!)2
x 2" ,
2
 2n + 2n 4n 22" (n.1)2
00
2n
]
X
=0
~ 0.000062, it follows from The Alternating Series Estimation Theorem that, correct to three decimal places,
Jo (x) dx ~ 1 
37. (a) f(x)
~
2 2
Ln=l
.i
7
[2n(2n  1) + 2n  2 n 22 "(n!)2
= ~ ( 1)" [4n
16 28
~
( 1)"2n(2n1)x2 " 22"(n!) 2
n~l
1 0
5
( 1)"2n(2n 1)x2 " 22n(n!)2
= ~ ( 1)"
Since
(0.2? (0.2) 5 (0.2) 7 3 + 5  7 + ... .
+ x5  x7 + · ·' · so arctan 0. 2 =. 02
Xn
I
~ 0.920. 00
nxn l
00
oo
x.,."
n=O
n
Xn1
I:  1 =
= •~1 ~ = •~1 (n 1)!
f( x)
(b) By Theorem 9.4.2, the only solution to the differential equation df(x)/d.'C = f(:c) is f(x) = Ke" , but f(O) K
= 1 and f(x) = e".
Or: We could solve the equation df(x) / dx
39. If a,..
= 1, so
= nXn2' then by the Ratio Test, n.oo lim
convergence, so R
x
= 1. When = ± 1,
= f(x.) as a separable differential equation.
Ian+ I= I(X+ an
I
lim
noo
f Ix: I= f n
n= l
" =1
n
n+l )2
1
21 .~ xn
=. lxl n+oo lim ( ....2::_1 ) 2 = lx l < 1. for n+
~which is a convergentpseries (p = 2 > 1), so the interval of n
convergence for f is [1, 1]. By Theorem 2, the radii of convergence off' and f" are both 1, so we need only check the 00
endpoints. f(x)
x"
= I; 2 n=l
n
00
nxn 1
00
x"
=> . f'( x ) = I:  2  = I;   , and this series diverges for x = 1 (harmonic series) n=l n n=O n + 1
@ 2012 Cengage Lc:tminiJ. All Rights Reserved. May not be sc:~.Mcd, copied, or duplicated, or posted w n publicly accessible wcbsile, in whole or in part.
SECTION 11.10
TAYLOR AND MACLAURIN SERIES 00
=  1 (Alternating Series Test), so the interval of convergence is [1, 1). !" (x) = I:
and conve rges for x
n=l
at both 1 and  1 (Test for Divergence) since lim
n.oo
•
41. By Example 7, tan 1 x =
6 = tan
have
1)
 1(
7r=~f
(1)"
y'3 n = O (2n + 1)3"
11.10
(1/ v'3) n +l ~( 2n + 1 = n~O 2
1)n
= n~O ( 
y3
=
2n+ 1
00
=2v'3f n=O
1 )"
n+ 1
1, 1).
r.;• we
y3
(1)" 1
3
n.1
1
y3 2n + 1' SO
(1)" . (2n + 1)3"
Taylor and Maclaurin Series J<">(a)
oo
j (5)   . 81
= n!' so ba = 
1. Using Theorem 5 with n~o b.. (x  5)", bn
3. Since j <">(o) = (n + 1)!, Equation 7 gives the Maclaurin series oo
(n + 1) 1·
n=O
n.1
oo J (n ) (O)
L:  1x" = L n.
n=O
. I
I . I
.
• x"
=
oo
L: (n + 1)xn. Applying the Ratio Test with an= (n + 1)xn gives us
n=O
hm an+l  = lim (n( + .2)x"+ll ) =
noo
an
n.oo
radius of convergence R
n +1
xn
IxI hm . n +2 = IxI · 1 = IxI. For convergence, we must have lxl < 1, so the n+oo n + 1
= 1. 4
5.
f(n) (:c)
n
j<">(o)
x) 2
0
{1
1
2(1 x) 3
2
2
6(1  x) 4
6
3
24(1  x) 5 120(1 x)
4
{1  x) 2 = f(O) + f' (O)x + !"(O) x 2 + f"'(O) x 3 + j< >(o) x4 + ... 2! 3! 41
= 1 + 2x + ¥x2 + ¥x3 + l224ox4 + ...
1
24
6
120
= 1 + 2x + 3x + 4x3 + 5x 4 + · · · = 2
f: (n + 1)xn
n=O
I a,.
lim a,.+l l :;= lim
n+00
noo
I(n(n++2)xn+ll = lxl lim n + 2 = lxl (1) = lxl < 1 1)xn n+1 n .oo
for convergence, so R = 1.
sin 7rX
7.
J
n
/ {n ) (0)
0
sin 1rx
0
1
7rCOS 7rX
7r
2
sin 1rx
2
1r
3
 7r 3 COS 7rX
4
s in 1rx
0
7r° COS7rX
5
1r
5
4
0 7r3
7f
= f(O) + !' (O)x + !" (O) x 2 + !"' (O) x 3 2!
/ ( 4 ) (0)
lim
a
n+l
= nlim . oo
3!
!(5) (0)
5
+ 4! x +  5! x +·· · 7ra 3!
7ro 5 x + ... 5!.
= 0 + 1rx + 0   x 3 + 0 + 
= 1rx 
7r 3 3 x 31
oo
= L (n= O
n.oo I a,. I
~
7r/j
+ 5! x
5
7r7 7   x 71
+ ·· ·
2n+l 7r x2n+l (2n+ 1)! ·
1)"
7r2n+3 x2n+3 (2n + 1)1 I 7r2 x2 · = .,._,00 lim 2 (2n + 3) ! 7r2n+l x n+l (2n + 3)(2n + 2) 1
=0<
1
for all x, so R = oo.
..
83
~ diverges
1
· X2n+ l
00
L: (1)"   for lxl < 1. In particular, for x
n=O
7r
~1 = 1 =/= 0, so its interval ofc~nvergence is ( n+
D
© 2012 Cenga;~e Learning. All RighiS Rese:n'\.'1.1. May not be scanned. copil.!d, or dupl icaled, or poslct.llo 3 publicly accessi ble wcbsih:, in whole or in part
I
84
0
CHAPTER 11
INFINITE SEQUENCES AND SERIES
9.
n
t<">(o)
f (nl (x)
0
2"'
1
1
2"'(ln 2)
ln2
2
2"'(1n 2) 2
(In 2) 2
3
2"'(1n 2) 3
(In2?
4
2"' (In 2) 4
(In 2) 4
11.
n
f (n)(x)
J (n) (0)
0
sinh x
0
1
cosh x
1
2
sinhx
0
3
cosh x
1
4
sinhx
0
n
J Cnl(x) 4 x  3x2 + 1
lim noo
I a,. I= I (n + = (In = < a ..+!
2 lim ) !xi noo n 11
J<~> (o)
0
so sinh x=
if n is odd
r
I II
r
an+
·I
=0<1
 1
12x2  6
6
3
24x
24
4
24
24
5
0
0
6
0
0
f(x)
for all x, so R
= 0 for n ;:: 5, so f
= x4 
2
6x
=
3x 2
 1·
0! (x 
=  1
+1= 1)
0
4
t<"l (1)
n=O
n!
 2
+ lT (x 
= In x =
0
In x
ln2
= 0! (x 2) 0 + 1! 21
1
1/x 1/x2
 1/2 2
ln 2
24/x5
 6/24
1)
1
1
= oo.
6
.
+ 2f (x 
(x 2)
+ 4J2d ~ (x 
2/2
5
1
1/2 3
 6/x
r
= 1.
1)
2
f: t
n=O
4
2
= x · n~ (2n + 3)(2n + 2)
A finite series converges for all x , so R = oo.
f(x)
2/x
I
24 (' + 3f x  1)3 + 24 (x 1)4 41 4 2(x 1) + 3(x  t? 1 4(x 1) 3 + (x 1)
J(n)(2)
3
x2n+l 00 I: ( _ ) • n=o 2n+ 1.1
2:::::   (x  1t
J(nl(x)
oo.
has a finite series expansion about a
n
3
(2n + 1)!
x2n+3
J(nl (l)
2
2
=
= (2n + 1)I,. then
"~ ~ = ,..:.,~ (2n + 3)! · x 2 n+l
4x3
15.
1 for all x , so R
0 if n is even
={ 1
I
x2n+l
1

0
Use the Ratio Test to find R. If an
f (n) (x)
13.
lim (ln2)n+lxn+l . n! n oo 1)! (In 2)"x"
= ln2 +
2) 4
f:(1)"+dn ~:)! n=l
n.2
1
+
1 2 2 )3 2! 22 (x 2) + 3! 23 (x 2
+~ (x 5!25
2)5
+ ·· ·
(x 2)''
24/2 5
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TAYLOR AND MACLAURIN SERIES
SECTION 11.10
.
r
lan+l l
J~ ~ = n~
lx 
=
l(lt+2(x~2)"+ (n + 1) 2n+l
2
2
1
1 ,._n;.,
n2" . (  1)"+ 1 (x 2)" 
lx 
1 < 1 for convergence, so
j(x)= e 2"'· =
17. n
t<">(x) 2
1 2 3 4
2e
2ae2x 24e2"'
8e6
f
e
= 0! (x  3)
r . (
n)
lx21
":~ n + 1  2
2.
n. ·
0
2e6
4e 6
+ l ! (x a? +. 2! (x 8e6
4e0
l( l)(x2)n,_ (n + 1)2 
J'"',(a) (x 3)"
6
6
2e "' 22e2"'
< 2 and R =
n= O
t<"l(3) e6
e2"'
0
21
li
0
+ 3! (x 
3) 3
3)
2
16e 6
+ 4! (x 3) 4 + ...
2n 6 +<x  3)n n= O n. oo
16e6
=
L
19.
n
J.'"'(x)
j (n ) (7r)
0
cosx
 1
1
 sin x
0
2
 cosx
1
3
sin x
0
4
COSX
 1
= 1 + (x  7r?2!
= Loo
(1)"+1
lim an+l _ lim
I a,. I
noo
lim ••oo
= 1r"+1,. IRn(x) l ::=; .
7rn+l 1 ( l )l lxl" + n+ .
+
(x  7r)G 6!
)2n
(2n)!
=
So n.lim R,.(x) +c:x:>
X 'lr
n=O
n oo
with a= 0 and M
{
(x7r)4 4!
[I x(2n +12)!
1r 2n+2 .
(2n ) 1. lx 7rl2 "
lx  7rl2 (2n + 2)(2n + 1)
]
=0 < 1
for all x, so R = oo.
17rx l" +l
= (n+ 1 ) 1. • Thus, IRn(x) l + Oas n + oo by Equation 10.
= 0 and, by Theorem 8, the series in Exercise 7 represents sin 1TX for all x . .
23. If f(x) = sinhx, then for all n, j
jJ
~ d, then l!(n+ll(x)l ~ cosh x ~ coshd, so by
= coshd; we have I.Rn.(x) ! ~ (~0:~!
1
lxln+ . 1tfollowsthat IR.,(x) l + 0 asn+ oofor
Ixi ::=; d (by Equation 10). But d was an arbitrary positive number. So by T heorem 8, the series represents sinh x for all x.
©
2012 Ccngage uaming. All Righi:< Resc" ·cd. Moy not be sconned, copied, or duplicated. or poSlcd lo • publid y ocecs.ible website, in whole or in pan.
85
86
0
CHAPTER 11
INFINITE SEQUENCES AND SERIES
f (1n/ 4) ·(x)"= 1 +t(x)+ H,2 ~)(x) 2 + H~),(D(x? + · .. 3.
25. {/1 x = [1 +(x)FI4 =
n=O
= 1 _ !x + ~ ( 1)" (1)" · [3 · 7 · · · · · (4n  5)]x" 4 n=2 4n · n! 1
4
· .. (4n I: 3 · 7 ..4n. n=2 n!
¢}
lxl < 1, so R = 1.
1 = 1  x
and 1xl < 1 27.
1 (2 + x) 3
=
00
1 [2(1 + x/2)] 3
= 81 ( 1 + ~2 ) 
5)
3
xn
f: (n3) (~2 ) ". The binomial coefficient is
1 = 8 n=O
I
 3) = ( 3)( 4)(  5) ..... (3  n + 1) = ·c3)(  4)( 5) ..... [ (n + 2)] ( n n! n! _ (  1)" · 2 · 3 · 4 · 5 · · .. · (n + l )(n + 2) _ (  l)"(n + l )(n + 2) 2 ·n! 2 Thus 1 = ! ~ ( l)"(n + 1)(n + 2) x" = ~ ( 1)"(n + 1)(n + 2)x" for ~ ~~ < 1 ' (2 + x) 3 8 n=O 2 2n n=O 2"+4 2 . X2n+ l
00
) ~ 2n + 1 1·
29. sin x= L::(1)" ( n=O
31. ex =
oo
x"
7r2n+ 1
lxl < 2, so R = 2.
) = L::(1)" ( ) x ~+ ,R = oo. 2n + 1 1· n = O 2n + 1 1·
2
f(x) =sin(1rx) = L::( 1)" ( n=O
~ (2x)" ~ 2" x"  = L.J => e2x = L.J  ,so J( x ) =ex + e2o: n=O n.1 n=O n1
EI n = O n.
oo
( 1fX) 2n+ 1
00
¢}
~ 1 n ~ 2" n = n=O L.J x + L.J 1 1 x n. n =O n.
=
1
~ 2" + 1
L.J n=O

n.1
 xn,
R =oo. 33. cosx =
00
x2n
n=O
(2 ·), n ·
I: (1)"
35. We must write the binomial in the form (1 +expression), so we' ll factor out a 4. X
X
J4 + x 2 = y'4(1 + x2 /4)
X
X ~(
2
2
=  +  ~
_
X
n= l

00
 2 + ..~ {1)
. 2
37. sm x
1
:i;
= 2 y'1 + x2 /4
X (
=2
1
•
X2)l/2
X
oo
= 2 ,~0
+4
(1)(X2)" 4 2
n
l )"' 1 · 3 · 5 · · · · · (2n 1) X 2n 2" . 4" . n ! n
1 · 3 · 5 · · · · · (2n 1)
n! 23,.+ 1
x
2n+l
and
X
2
4
<1
!xj ¢>
2
< 1
¢>
lxl <2, soR = 2.
= !  ! [  oo ( 1)" (2x?" ]  ! [   oo ( 1)"(2x?"] = oo (1)n+l22n l x2n (2n)!  2 1 1 ,~1 ( n)! n~l (2n)! ' 2 (1 cos2x) 2 1 ..~o 2
R = oo ® 2012 Ccngagc Learning. AURights Reserved. Muy not be scanned, copied, or tlupli cated. or posted to u publicly accessible website:, in whole or in part.
SECTION 11.10
TAYLOR AND MACLAURIN SERIES
D
87
1.5
 1.5 tr.'+~ .. :11.5
= 1 ~x''
+ fixs
7~oxl2
\ . T 8 = T9 = T 10 = T11
+ ...
\ 't
.
The series for cos x converges for all x , so the same is true of the series for
f(x), that is, R =
•
\
c;o· Notice that, as n increases, Tn(x) becomes a better
 1.5
approximation to f(x). oo
41 . e:r; (11) =
n
"" X , so e  x L I
n =O
n.
f (x)
oo ( )" "" L ....=::___1
=
n =O
n.
oo
·= ""{1)" Ln =O
n .:._,SO
n1
= xe "' = f: (1)" ~ x"+l · n.
n=O
The series for e"' converges for all x, so the same is true of the series for f(x); that is, R = oo. From the graphs off and the fi rst few Taylor polynomials, we see that T,. (x) provides a closer fit to 7T
43. 5° = 5°
(
1800
7r
)
=
36
f (x) near 0 a!; n
. oo x2" radians and cos x = ..~o ( 1t (2n) ! = 1 
increases. x2
2f +
7T (7r/ 36) 2 (7r/ 36)4 (7r/ 36)6 (7r/ 36) 2 cos  = 1    + + ... Now 1 36 2! 4! 6! . 21
x4 x6  6f + ·· ·,so 41 ~
.
0 99619 and addmg .
(7r/36)~ 4!
~ ~
24 .
does not affect the fifth decimal place, so cos 5° ~ 0.99619 by the Alternating Series Estimation Theorem.
1
(b)sin lx= l
dx=C + x +
V1 x 2
=x+
f; n= l
(16)
f: 1·3 · 5 ·· .. · (2n1)x n+l 2
(2n + 1)2n · n !
n= l
1 · 3 · 5 · · · · · {2n  1) x 2 n+l (2n + 1)2". nl
since 0 = s in 1 0 = C.
x 2n
00
47. COSX = "~ ( 1 )" {2n) !
0
I (16)
49. cosx
I
=
COS X 
X
oo
n
x2"
.~o ( 1) ( 2n)! 1
oo
oo
n
X 2n
2n
·(
X Gn+2
oo
3 )
dx
= C + .~o (x2"
=> cosx  1 = .~ (  1)" {2n)! 1
dx = C + 2:::{1) n =l
x cos(x
1)" (6n + 2 )(2n)! , with R = oo. .
cos x X
1
=
oo
2n  1
n=l
(2n) !
2: {1)" .::_____ =>
•
) , wtth R = oo. 2n 1.
@ 20 12 Ccngagc Lcunlin g. All Rights Ri:st:rvc:J. May not be scwmct.l. cupicd, or duplicated, or pO!'Itr.:d to a publicly IU:cc~ siblc website, in whole or in pan.
X
wG
88
0
INFINITE SEQUENCES AND SERIES
CHAPTER 11 •
oo
x2n+l
2: ( 1)" 2
51 . arctanx =
n+1
n=O
I
for lxl
5
(1 /2) 1.5 
(1/ 2r
~
x2n+ 4
n =O
2n+ 1
for Jxl < 1 and
x2n+5
oo
3
x arctanx dx
00
< 1, so x3 arctanx = 2: (1)"  
= C + ,..;;:o ( 1)"' (2n + 1)( 2n + 5)' Since ~ < 1, we have
9
11
(1/ 2) · · {1/ 2) +~ ~ 0.0059 and subtracting T i l ~ 6.3 x 10 6 does not affect the fourth decimal place,
so J0112 x 3 arctan x dx ~ 0.0059 by the Alternating Series Estimation Theorem.' 53. v'1 + x4 = (1 + .x 4 ) 112 =
00
(1/ 2) (x 4)", so n
2: n =O
I
}1 + x 4 dx = C + ,
L:
oo n =O
( 1/ 2) x4n+1  and hence, since 0.4 < 1, n 4n + .1
we have I=
1
0.4
o
00
} 1+x4 dx = E
n=O
(1/ 2) (0.4) 4n+l n 4n + 1
(o.4) 1 ~ (o.4) 5 H4) (o.4) 9 Ht)(~) (o.4) 13 H4)(~)(~) (o.4) 17 =( 1) o r + l !  5  +2,9+ 3! ~+ 4! rr +· ·· (0.4) 5
(0.4) 1 3
(0.4) 0
= 0.4 + w  ;:;2 + 208 Now
(O~~)o ~ 3.6 x 10
6
<5x
5(0.4) 17 2176
+ ...
10 6 , so by the Alternating Series
5
Esti~ation Theorem, I ~ 0.4 + (0~~) ~ 0.40102
(correct to five decimal places).
55. l im x0
x ln(1 + x) x (x  l x2 + l x 3  lx4 = lim 2 3 ., 4 x2 x .0 x· = m+0 lim (!2  l3 x
2 + lx 4
5 + lx fi
• • ·)
= lim
.,_,o
lx2 2

lx3 + l x 4 :l
4 X2

l x5 + · · · 5
ix3 + : .. ) = !2 5
since power series are continuous functions.
57. lim
xo
sinx x
+ l6 x 3
x5
since power series are continuous functions. .
_59. From Equation 1 I, we have ex 2
2
=1
~
I
1.
Equation 16. Therefore, e"' cosx = (1  x 2 2
degree < 4 ' we get e"' cos x
~
~
~
~
+   I + · · · and we know that cos x .= 1   + 41  · · · from 21. 21. 3. .
+ ~x4  · ·) (1 
2 4 = 1  lx + .l.x 2 24

x 2 + lx" 2
~x2 + i4 x 4

  • ).
Writing only the tenns with 
2 4 + lx + . · . = 1  !!.x + 24 ~x4 + ·. . 2 2
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SECTION 11 .10
61 _ x_ <~> · sin x 
x tx
TAYLOR AND MACLAURIN SERIES
0
x
3
+ 1~0 x5  · · · ·
1
+ l~o xs  .. . j
x * x3
I
2
7
4
+ 6x +aGo X + · · ·
x x  ~ x3 + rlo xs  .. . l x3 _ ....Lxs 6 1
3
oX 
From the long division above, ·
oo
4n
oo
I
120 1 5 36X
+ .. . + .. .
~ = 1 + ~x2 + 3 ~x4 + · · ·. SlllX
·
I
(  x~~) n
a
63. L; ( 1) " ~ = L;  1  = e x· , by (I I ). n=O n. n =O n. 65.
~(1)"J n=l
3 nn n5
= ~( 1)"_ 1 (3 /5 )" = ln(1 + ~) n
n= 1
5
[fromTable l ]
= ln~
5
n.
71. lfp is an n thdegree polynomial, then p<'l (x)
(i ) ( )
= 0 for i> n, so its Taylor series at a is p(x) = L; ~(x  a)i. t.
i =O
Put x a.= 1, so that x = a+ 1. Then p(a + 1) =
n.
p(i) (a)
i=O
t.
2:  .1  . n
This is true for any a, so replace a by x : p(x
p(i)(x)
+ 1) = iL: .1 =O t.
73. Assume that lf" (x)l ~ M , so /" (x) ~ M for a ~ x ~a + d. Now fa"' !'11 (t) dt ~ 1
f"(x)  J"(a) ~ M (x  a)
=>
f"(x) ~ J"(a) + M(x  a). T hus,
J'(x)  f '(a) ~ J"(a)(x  a)+ ~M(x a )2
J: M dt
=>
I: J"(t) dt ~ I: [!"(a)+ M(t a)] dt
=> J'(x) :::; /'(a) + J"(a)(x  a)+ ~M(x a) 2 =>
.r: f(t )dt ~ .r: [!' (a) + J"(a) (t  a)+ tM(t  a?] dt
=>
f(x)  f(a) :::; · f' (a)(x a) + tf"(a)(x  a? + iM(x a) 3 . So f(x)  f(a)  J'(a)(x a) ~f"(a)(x  a) 2 ~ ~ M(x  a)3 . But R 2(x) = f (x) T2(x) = f (x)  f(a)  J'(a)(x a) ~f"(a) (x a) 2 , so R2(x) :::; !M(x  a) 3 • A sim ilar argument using f"'(x ) ~ M shows that R2(x) ~  t M(x a) 3 . So IR2 (x2)l ::; tM lx  a j3 . Although we have assumed that x > a, a similar calculation shows that.this inequality is a lso true if x
< a.
© 2012 Ccngngc Lc:uning. All Righls Rcscr'\'cd. May nol be scanned, copied, or duplicated, or posted lo • publicly accessible wcbsilc. in whole or in pan.
=>
89
90
0
CHAPTER 11
75. (a) g(x)
= :L; 00
n~
(1 + x) g'(x)
INFINITE SEQUENCES AND SERIES
(k)n x"'
= (1 +
00 => g'(x) = :L;
n~
00 :L;
x)
n=1
(k)n nxn
1
(k) nx"n
= :L; 00
n=1
1
,
so
(k)n nx"' + :L; (k)n nxn 1
00
n= 1
. .
I
=
f: (n+k1) (n + 1)xn + f: (k)nxn n
n=O 00
_

n~o (n + 1) 00
= n~O
[
n=O
Replace n with n + 1 ] in the first series
k(k1)(k2) .. ·(kn+1)(kn) n (n + 1)! x +
00
(n + l )k(k  '1)(k  2) ... (k n + 1)
(n+ 1)!
= k n~o k(k 
1)(k
.
[
n~O (n)
k(k 1)(k  2)···(kn+ 1) ] ,, nl
n
((k n) _+n]X
2l!... (k n + 1) x" = k fo (~) x" = kg(x)
Thus, g'(x) = k g(x).
1 +x
(b) h(x)
= (1 + x)k g(x)
=>
h'(x) =  k(l + x)  k 1g{x) + (1 + x) k g'(x)
[Product Rule]
= k(l + x)i.: 1 g(x) + (1 + x)  k kg(x)
[from part (a)]
1 +x
= k(l + x)k 1 g(x) +
k{1 + x)"' 1 g(x)
(c) From part (b) we see that h(x) must be constant for x E (1 , 1), so h(x) Thus, h{x) = 1 = (1 + x)k g{x)
11.11
<::>
=0
= h(O) = 1 for x E ( 1, 1).
g(x) = {1 + x)k for x E ( 1, 1).
Applications of Taylor Polynomials
.
1. (a)
n
J
0
cosx
1
sinx
0
2
cosx
 1
3
sinx
0
4
cosx
1
5
sin x
0
6
cosx
1
t <"'>(o) 1
Tn(x)
T,=
r,
2
1 1 1 ~x2 1  ~x2 1 ~ x2 + f4x 4 1 ~x 2 + f4x4 1  ~x2 + f4x4  7~oxs
T6
2
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x
SECTION 11.11
APPLICATIONS OF TAYLOR POLYNOMIALS
(b)
To =T1
f
X
T2 = Ts
T4 =Ta
1
0.6916
0.7074
0.7071
2
0
1
 0.2337
0.0200
 0.0009
1T
 1
1
3.9348
0.1239
1.2114
4 7f
(c) As n increases, Tn (x ) is a good approximation to
3.
n
f(nl(x)
j(n)(2)
0
1/x  1/ x 2
2
 4
2/x3 4 3 ·  6/x
4 3  8
1
Te
0.7071
1r
1
1
1
2
3
f (x) on a larger and larger interval.
J (n)(2)
Ts(x) = 2:::::   (x 2)" ·n =O n.1 1
1
= 2  l
0!
1!
1
(x  2) + 4 (x 2? 2!
t H x 
2) + ~(x  2) 2
n
J<"l(x)
f(n)( 1T/ 2)
0
cosx
0
1
sinx
 1
2
cosx
0
3
sin x
1
=

3
j_
3!
(x 2?
fa(x 2) 3
5.
0 1~'1.;;j '•'fr _!!__
2
1.1
= 
(X  ¥) + i (X  ¥)3 2
7. f (n) (x)
J<"l(1)
0
ln x
0
1
1/x
n
2
 1/x
3
2fx3
2
1  1
1 = 0 + li(x  1) =
2
 1 + T!(x 
(x 1)  tCx 
©
1) 2
1
)2
2 3 + 3!(x1)
+ ~(x 1?
2012 Ccngnge Learning. All Rights Rcsc.:rvcd. May not be scanned, copied, or duplic.oled, or posted lou publicly accessible wcb:;ile, in whole or in pnrt.
D
91
I
92
0
INFINITE SEQUENCES AND SERIES
CHAPTER 11
9.
J
J<">(x)
n
xe
0
1 2
2
0
"'
(1 2x)e
2
4(x  1)e
2
4(3  2x)e
3
"'
1
"'
4
2
12
"'
4
11. You may be able to simply find the Taylor polynomials for
J(x) =cot x using your CAS. We will list the values of J<"J (rr / 4) for n = 0 to n = 5.
2
=
1  2 (X

4,. )
+ 2 ( X  4,. )2 · 38(X  4")3 + 310 (X

64 (X  4,. )5 4,.. )4  15
For n = 2 to n = 5, Tn (x) is the polynomial consisting of all the terms up to and inc luding the
(a) f(x)
13.
n
J
f {n)(4)
0
Vx
2
1
4
2
lx 1/2 2  l x  3/2
3
1x5/2
4'
1
r term.
1 2 = Vx ~ T2(x) = 2 + !(x4)  /~ (x 4? 4 2.
= 2 + i(x (b) IR2(x)l ~

4) 
i4(x 4) 2
M 3 3! Jx 4J , where lf'"(x) l ~ M. Now 4 ~ x
~ 4.2 =>
1
32
8
lx 41 ~ 0.2
=> Jx  41 3 ~ 0.008. Since f'"(x) is decreasing
on [4,4.2], we can take M
IR2(x)l (c)
(X  .i
= 1/"'(4) 1 =
i4 5 / 2 =
2 ~ 6 , so
~ 3/~56 (0.008) = 0~~~8 = 0.000015625.
0.00002 r      ,
From the graph of IR2 (x)l =
lvx T2(x)l, it seems that the
6
error is less than 1.52 x 10 on (4, 4.2].
0
2012 Cengnse U:.vnins. All Rights Rescn.:d. Moy not be: scanned. copied, oc duplicnted. or posted 10 o publicly accessible website, in whole or in·p:ut,
SECTION 11.11
15.
(a) f(x)
n
f (n) (x)
J (n){1)
0
x2/ 3
1
1x l /3
1 2 3
2
_1x 4/ 3
 g
3
.§..x 7/3
8
4
 ~ x 10 / 3
3
=. x 213 ~ Ta(x) = 1 + ~(x 1) 
IR3(x)l5
~ lx 
93
2/9 8/27 2!(x  1) 2 + 3!(x  1) 3
= 1 + ~(x 1) ·~(xI?+ 841(x (b)
0
4 11 , where Jj< 4 >(~)~5 M. Now 0.8
1) 3
5 x 5 1.2
=?
2
9
27
lx l l 5 0.2 ::} lx  114 5 0.0016. Since jJ <4>(x) l is decreasing
27
on [0.8, 1.2], we can take M
81
~(0
IR3(x) l 5 81 ' 0.00006
(c)
APPLICATIONS OF TAYLOR POLYNOMIALS
8)10/3 24
=I
j <4 >(o.s)l
= ~(0.8)  1013 , so
(0.0016) ~ 0.000096 97.
From the graph of IRa(x) l = Jx 2/ 3

Ta(x) I• it seems that the
error is less than 0.000 053 3 on [0.8, 1.2).
17.
(a) f(x)
n
f(n) (x)
J (n){O)
0
secx
1
1
secx tanx
0
2
secx (2sec2 x 1)
1
= secx ~ T2(x) = 1 + ~x
2
2
sec x tan x (6 sec x  1)
3
(b) IR2(x) l 5
~ lx l3, where It <3>(x) l 5 M . Now  0.2 5 x 5 ~.2
=:>
lx l 5 0.2
=?
lxl 3
5 (0.2) 3.
f(3 ) (x) is an odd function and it is increasing on (0, 0.2) since secx and tanx are i"ncreasing on [0, 0. 2) ,
I
so J C3>(x) l 5 J <3>(o.2)
3
~ 1.085 158892. Thus, IR2(x) l 5 J< >i~· 2) (0.2) 3 ~ 0.001447.
0.0004
(c)
· From the graph ofiR2(x)l
= isec x 
T2(x) l, it seems that the
error is less than 0.000 339 on [0.2, 0.2). 0
19.
J
n· 0
e"'
2
2
1
e"' (2x)
2
e"' (2
3 4
2
+ 4x2 ) 2 ex (12x + 8x 3) 2 e"' (12 + 48x2 + 16x4 )
f (nl(o) 1 0 2 0
(b) IR3(x)l $
~ lxl 4 , where Jt <4 >(x) J 5 4
x 4 5 (0.1) , and letting x
M. Now 0
5 x $ 0.1
= 0.1 gives
IR3(x)l $ eo.Dl (12 + ~!8 + 0.0016) (0.1)4
~ 0.00006.
® 2012 Ccngngc Lc:uning. AU Ri~tsRc.scn·cd. M.D.)' not be .scrutncd. coried, orduplicattd. or posted to a publicly acc<."SSiblc wcbsilc, in whole or in part.
=?
94
D . CHAPTER11 (c)
INFINITE SEQUENCES AND SERIES
0.00008
~~
From the graph of jR3(x) l =
2
le~  T3(x) l, it appears that the
y = IR,(x)l
error is less than 0.000 051 on [0, 0.1].
21.
n
J
t<" l(o)
0
xsinx
0
1
sinx+xcosx
0
2
2cosx xsinx
2
3
 3sin x xcosx
0
4
 4cosx + xsin x
 4
5
5sinx + x cosx
(b) jR4(x) j ~ jxj
~~ jxj 5 , where I f(")(x)l ~ M. Now  1 ~ x ~ 1 =>
~ 1, and a graph of j15 l(x) show; that I j <5 l(x) j ~ 5 for  1 ~ x ~ 1.
Thus, we can take M
5 = 5 and get IJ4(x) l ~ 5T · 1r; =
1 24
= 0.0416.
0 .009
(c)
From the graph of jJ4(x) j = jxsinx T.1(x) j, it seems that the I
error is less than 0.0082 on [ 1, 1].
23. From Exercise 5, cosx =  (xIJ< 4 l(x) l IR 3 (
Ts
4 ;' )
~) +
Hx 
~)3 +
80°
= (90° 
= jcosxj ~ M = 1. Now x = I~
Rs(x), where jRs(x)l 10°)
~ ~ !x  ~ l4 with
= (~fa) = ~11' radians, so the error i.s
f.;_ (fa)4 ~ 0.000 039, which means our estimate would not be accurate to five decimal places. However,
= T4, so we can use !Rot'(;) I ~ 1~0 (fat
25. All derivatives of e"' are e"' , so IRn (x)l
~
~ 0.000 001. Therefore, to five decimal places,
(n ~ I)! lxl"+ 1 , where 0 < x < 0.1. Letting x
= 0.1,
.R,,(O.l)
~
Rs(O. l )
< 0.0000046. Thus, by adding the four terrns of the Maclaurin series for e"' corresponding ton = 0, 1, 2, and 3,
( eO.l ) ' (0.1)"+1
n+I .
< 0.00001, and by trial and error we find that n = 3 satisfies this inequality since
we can estimate e0 · 1 to within 0.00001. (In fact, this sum is 1.10516 and e0 · 1 ~ 1.10517.)
@) 2012 Ccn~agc Lcamins;. All Rights Reserved. May not be scanned, cop ic~ or dupliC4ted, or posted to a publicly accessible website. in whole or in p:u1.
SECTION 11.11
1 3 +Six 1 s  · · ·. Bth · senes · 27. sinx = x 3!x y e A Iternatmg
APPLICATIONS OF TAYLOR POLYNOMIALS
~x3 is less than I ;, x 5 1 < 0.01
lx 5 l
< 120(0.01)
¢:}
lxl
r~~~~~
y=sinx+ o.oV / /
< (1.2) 115 ~ 1.037. The curves
3
+
< 0.35
¢:}
_ . __ _ _ _ _ __ __ . /
1.2
esti~ate
only for x
> 0. Thus, the desired range of
< x < 1.037. x5
7
x 5 7
+ · · · . By the Alternating Series
Estimation Theorem, the error is less than l ~ x 7 1 jx 7 j
"L,__L,.:...__
0.8
and the given approximation are odd functions, we need to check the
x3
3
y=xkx
O.QJ
0.9
the graph confirms our estimate. Since both the sine function
29. arctan x = x 
/
/~ 'fooox
¢:}
y = x  ix3 andy= sin x 0.01 intersect at x ~ 1.043, so
. values for xis  1.037
95 ·
0.9
Estimation Theorem, the error in the approximation sin x = x 
D
< 0.05
y =arctan x + 0.05
¢:}
lxl < (0.35?1 7 ~ 0.8607. The curves
y = x tx 3 + kx 5 andy= arctan x + 0.05 intersect at
x
~
0.9245, so the graph confirms our estimate. Since both the
arctangent function and the given approximation are odd functions, we need to check the estimate only for x range of values for xis  0.86 31 . Let
1
> 0. Thus, the desired
y = x  .!. x 3 + .!. xs
3
5
< x < 0.86.
s(t) be the position function of the car, and for convenience set s(O) = 0. The velocity of the car is v(t) = s' (t) and th~
acceleration is a(t ) = s" (t), so the second degree Taylor Polynomial is T2(t) = s(O) + v(O)t +
a~O) t 2 =
20t + t 2. We
estimate the distance traveled during the next second to be s(1) ~ T2(1) = 20 + 1 =21m. The function T 2 (t) would not be accurate over a full minute, since the car could not possibly maintain an acceleration of 2 m/s 2 for that long (if it did, its final speed would be 140 m/ s ~ 313 mi/ hl).
33. E =
~2  (D!d)2 = ~2 D2(1:d/ D)2 = ; 2 [1 
(1+
~r2l
We use the Binomial Series to expand (1 + d/ D )  2 : .
when D is much larger than d; that is, when P is far away fro m the dipole. 35. (a) If the water is deep, then 27rd/ L is large, and we know that tanh x tanh(27rd/ L) ~ 1, andsov2 ~ f!L/ (27r)
©
¢:}
+
1 as x
+
oo. So we can approximate
v ~ ..jgL j (21r).
20 12 Cengage Learning. All Rights Reserved. May not be SC41lncd, copied, or duplicaled, or poSied 10 a publicly accessible website. in whole o r in part.
96
D
CHAPTER 11
INFINITE SEQUENCES AND SERIES
(b) From the table, the first term in the Maclaurin series of
f (nl(o)
j (n) (x)
n
tanhx is x, so ifthe water is shallow, we can approximate
0
tanhx
1
sech2 x
2
2sech xtanhx
0
3
2sech2 x (3tanh 2 x 1)
 2
0 1
2
(c) Since tanh x is an odd function, its Maclaurin series is alternating, so the error in the approximation 3
2trd 2trd. . . IJ"'(O)I (2trd) 1 (2trd) tanh L ~ L IS less than the first neglected term, wh1ch IS  ,  L = 3 3 L If L
than
> lOd, then ~ ( gL 7r 2
tr3 •
375
~
2
~d)
3
< ~ ( 2tr ·
1 ) 10
3
3
.
3
=; .
5
, so the error in the approximation v
2
= gd is less .
0.0132gL.
37. (a) Lis the length of the arc subtended by the angle B, soL
B=L/R.NowsecB=(R+C)/R. =>
= RB =>
RsecB=R+C =>
C = RsecB R = Rsec(L/R) R. (b) First we'll find a Taylor polynomial n(x) for f(x)
n
/(nl(x)
/(n) (0)
0
secx
1
1
secx tanx
0
2
secx(2tan x
+ 1) + 5) 4 2 secx(24tan x + 28tan x + 5)
1
secxtanx(6tan 2 x
0
2
3 4
Thus, f(x) = secx ~ T4(x) = 1 + ~(x
(c) Taking L
C
= secx at X = 0.
5
o? + fi(x W = 1 + ~x + f4x 4 • By part (a), 2
= 100 km and R = 6370 km, the formula in part (a) says that
= Rsec(L/ R) R
= 6370 sec(100/6370) 6370 ~ 0.785 009 96544 km.
£2 . . The formula m part (b) says that C ~ R 2
+
5£4 R 24 3
1002
5. 1004
= 2 . 6370 + 24 . 63703
.
~ 0.785 009 957 36 km.
The difference between these two results is only 0.000 000 008 08 km, or 0.000 008 08 m! 39. Using f(x)
= Tn(x) + Rn(x) with n =
1 and x
polynomial off at a. Because a= Xn, J(r) and we have 0 =
f (Xn) + J' (Xn) (r

Xn)
= r, we have J(r)
= T1(r)
+ R 1(r) , where T1
is the firstdegree Taylor ,
= f(xn) + f'(x n)(r x,.. ) + R1(r). But r is a root off, so f(r)
+ R1 (r). Taking the first two terms to the left side gives us
© 2012 Ccngagc Learning. AU RighlS Rcscr.·cd. May not be scanned, copied. or duplicated, or poslci.lto u publicly accessible website, in whole or in pan.
= 0
CHAPTER 11 REVIEW
J'(xn)(xn  r)  f (xn) = R 1 (r). Dividing by f'(xn), we get Xn  r 
;,~:~)
IR 1 (r)l ::=;
I J'~~r)llr  xnl2 • Combinin~ this inequality with the facts lf"(x) l ::=; M
I·
Tay lor's Inequality gives us
and
IJ'(x)l ~/(gives us
lxn+l  rl ::=; 2M/( \xn  r \2 ·
11
Review
CONCEPT CHECK 1. (a) See Definition 11.1. 1.
(b) See Definition 11.2.2. (c) The terms ofthe sequence {an} approach 3 as n becomes large. (d) By adding s ufficiently many terms of the series, we can make the partia l sums as close to 3 as we like. 2. (a) Sec the definition on page 721 [ET page 697].
(b) A sequence is monotonic if it is either increasing or decreasing. (c) By Theorem 11.1.1 2, every bounded, monotonic sequence is convergent. 3. (a) See (4 ) in Section 11.2. (b) The
4. If
pse~ies
2: a ,. =
f: J.. is converge~! ifp > 1.
n=l 77,1'
3, then lim an = 0 and lim n  oo
Sn
n t ex>
= 3.
5. (a) Test for Divergence: If lim a,, does not exist or if lim an n+oo
n + oo
=I 0, then the series 2:"':.. an is divergent. n 1
(b) Imegral Test: Suppose f is a continuous, positive, decreasing fun ction on
:2:;::'=1 an (i) If
f1
00
(ii) If f 1
00
is convergent if and only if the improper integral
f1
00
[1, oo) and let an= f(n). Then the series
f(x) dx is convergent. In other words:
f(x) dx is convergent, then 2:;::'=1 a, is convergent. f( x) dx is divergent, then
I::;"=1 a,.
(c) Comparison Test: Suppose that E an and
is divergent.
2: b,. are series with positive terms.
(i) If E b,. is convergent and an ::=; bn for all n, then (ii) If E br. is divergent and an ~ bn for a ll n, then
E a,. is also convergent.
2: an is a lso divergent.
(d) Limit Comparison Test: Suppose that E a,. and E bn are series with positive tenns. If lim (a ,/bn) n. oo
= c, where c is a
finite number and c > 0, then either both series converge o r both diverge. (e) Alternating Series Test: Ifthe alternating series :2:;::'= 1 (  1)n 1 bn
= b1
 b2
+ b:3 
b4
+ bo 
ba
+ ···
[b,.
satisfies (i) bn+l :::; bn for all n and (ii) lim b,. = 0, then the series is convergent. n
0
97
:(~::).By the formula for Newton's
=
I ;(~:})
method, the left side of the preceding equation is Xn+I  r, so lxn+l  r l =
0
20 12 Cengoge Learning. All Rights Resem:d. May not be seo.nned. copied, or duplicoll:d. or posted to n publicly oecessiblc website, in whole or in p!ttt.
>
OJ
98
0
CHAPTER 11
INFINITE SEQUENCES AND SERIES
(f) Ratio Test: (i) If lim noo
Ian+l ) I
(.ll") If 1"un an+1 n oo
= L
an
an
< 1, then the series
f
no=l
an is absolutely convergent (and therefore convergent).
I= L > 1 or 1" IO.n+l I= oo, then the senes . ~ a n .IS d"Ivergent. 1m
Ln=l

n+oo
an
I an I
(iii) If lim an+l = 1, the Ratio Test is inconclusive; that is, no n+co
c~nclusion can be drawn about the convergence or
divergence of 2:: an. (g) Root Test:
< 1, then the series 2::;:'= 1
a,. is absolutely convergent (and therefore ~nvergent).
(i) If lim
\li(i:j =
(ii) If lim
\li(i:j = L > 1 ~r n+oo lim ~ = oo, then the series 2::;:'= 1 an is divergent.
(iii) If lim
~
n co
n.oo
11.+ 00
L
= 1, the Root Test is inconclusive.
6. (a) A series L an is calledab~olutely convergent if the series of absolute values
2:: Iani is convergent.
(b) If a series L an is absolutely convergent, then it is convergent. (c) A series L:al' is called conditionally convergent if it is convergent but not absolutely convergent. 7. (a) Use (3) in Section 11.3. (b) See Example 5 in Section 11.4. (c) By adding terms until you reach the desired accuracy given by the Alternating Series Estimation Theorem. 00
8. (a)
2:: en(x 
a)"
n=O 00
(b) Given the power series
L
en(x a)", the radius of convergence is:
n=O
(i) 0 if the series converges only when x
=a
(ii) oo if the series converges (or all x, or (iii) a positive number R such that the series converges if lx 
al < Rand diverges if lx al > R.
(c) The interval of convergence of a power series is the interval that consists of all values of x for which the series converges. Corresponding to the cases in part (b), the interval of convergence is: (i) the single point {a}, (ii) all real numbers, that is, the reaL number line ( oo, oo ), or (iii) an interval with endpoints a  R and a + R which can contain neither, either, or both of the endpoints. In this case, we must test the series for ·convergence at each endpoint to determine the interval of convergence. 9. (a), (b) See Theorem 11 .9.2. 10. (a) Tn(x)
(b)
n
f(i)(a)
i =O
~.
= L  .1 
oo j (n){ ) I: __a_
(x  a)'
(x  a)"
nl
n=O
®
2012 Cengnge Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted ton publicly ncccssiblc website, in wbole or in part.
CHAPTER 11 REVIEW
(c)
oo f (n) (O)
L  1 n.
n=O
xn
0
99
[a= 0 in part (b)]
(d) See Theorem 11.10.8. (e) See Tay.lor's Inequality ( 11.10.9).
11 . (a)(f) See Table I on page 786 [ ET 762]. 12. See the binomial series ( 11 .10.17) for the expansion. The radius of convergence for the binomial series is 1.
TRUEFALSE QUIZ 1. False.
See Note 2 after Theorem I 1.2.6.
3. True.
If lim an
5. False.
For example, take en = (  1)" j (n6n) .
n  oo
= L, then as n 7+ 00, 2n + 1 + oo, so a2n + l
. . I_+_ an I= lim . I(n +1 1) a,. 1
7. False, smce hm
noo
n  oo
3
.· · n3 ~ 1 = lun 1
n+oo
+
L.
I(n + 1) . · 1j1/n31  = n n3
3
3
. hm n oo
1 (1 + 1/ n) 3
= 1.
See the note after Example 2 in Section 11.4.
9. False. 11. True.
See (9) in Section II . I .
13. True.
By Theorem 11.10.5 the coefficient of x 3 is ~
111 /
(0)
= 31 => !'" (0) = 2.
Or: Use Theorem 11 .9.2 to differentiate f three times. For example, let a..
15. False.
= bn = (1r.
[2:: (1)" a,.
17. True by Theorem 11.6.3.
+ 0.9(0.1? + 0.9(0.1) 3 + ·· · =
f: (0.9)(0.1)nl = __2.:Q_ = 1 by the formula 1  0.1
n=l
for the sum of a geometric series [S = a 1 / (1 ; r)] with ratio r satisfying lrl 21 . True.
= 1, so {a,. b.. } is convergent.
is absolutely convergent and hence convergent.]
0.99999 . . . = 0.9 + 0.9(0.1) 1
19. True.
Then {an} and {bn} are divergent, but a,.b,.
< 1.
A finite number of terms doesn't affect convergence or divergence of a series.
EXERCISES 2 + n3
1. { 1 + 2n 3
}
.
n +00
3
3. lim a,. n+OC
.
converges smce lim
= n+ lim ~ = lim oo 1 + n n+ oo
2 + n3
1+
2n 3
/ ~
1 n
·
.
= lim n+00
+1 = 11n 3 + 2
2/ n 3
1
 . 2
+ 1 = oo, so .the sequence diverges.
© 2012 C
100
0
CHAPTER 11 INFINITE SEQUENCES AND SERIES
nsinn l::; ~ n 1 · <  ,so Jan !> 0 as n> oo. Thus, 1 1 + n + n l
5. !an i = 


n2
7.
.
lim a,.= 0. The sequence .{an} is convergent. n +oo
•
{(1 + 3)4"}is convergent. Let y (1 +;3)4:t. Then ~
~
lim lny = lim 4xln(1 + 3/x) = lim ln( 1 + 3/x) :r:oo :z:+oo x + oo 1/ (4x}
(1+ n3)""
limy= lim
.:coo
n.oo
~ lim ~ ( ~)  1/ (4x2)
. IliD
x+oo
an
< 2. Note first that 1 < a2
fo r n = 2. Now assume that ak1 < a k < 2. Then ak = ~(ak 1
+ 4)
= t (1 + 4) = 1 <
< ~(a~.:+ 4) < ~(2
and the induction is complete. To find the limit of the sequence, we note that L =
lim a n 'ft.oo
= HL + 4) n
n
11. g < 3 n +1 n
1 + 3/x
= 12,so
=e12 .
9. We use induction, hypothesizing that an 1 <
L
12
x  oo
2, so the hypothesis holds
+ 4) =
2. Soak < ak+l < 2,
= noo lim an+l =>
=> L = 2. 00 n ·· 1 g converges by the Comparison Test with the convergent pseries L; 2 [ p n=l n + 1 n=l n 00
= 21 , so L; n
= 2 > 1].
. 1 )3 · 1 =1 < 1, so LJ ~ na converges .by the Rat1.o .,.,,est. · 13. )1m   = 1'1m [ (n+1)3 + · 35n ] = l'rm (1 +Tl.___.OO an n  00 n noo n 5 5 n=l 511. 5n 1
lan+ll
~. Then f
=
15. Let f (x)
xv ln x
is continuous, positive, and decreasing on [2, oo), so the Integral Test appl ies.
00
1 2
f (x)dx= lim t +oo
= lim
t+oo
. LJ ~ so t I1e senes n=2
17. Jan! =
0
L; ( ~ )" [r =
Tt+ 00
1 ~ dx
XV In X
[u = ln x. du=_:
= too lim
!.In uL
1 2 /
ln 2
!n t du = t+oo lim (2Fu] In 2
(2 vlfrit  2v1ii2) = oo,
eliverges.
<
(l.~)n = (~J, so n~1 ianl converges by comparison with the convergent geometric
~ < 1] . lt fo llows that
n=l
19. lim
2
11 : (1~;)n I::; 1 + (~. 2 )" 00
series
1
~
nv lnn
1t
00
L; an converges (by Theorem 3 in Section
I 1.6).
n =l
Ian+l I= n+oo lim 1 . 3. 5 . . . . . (2n  1)(2n + 1) 5n+J (n + 1)! Qn
5n n ! 1 • 3 • 5 • • • • • (2n 1)
·lim 2n + 1 n+oo 5(n + 1 )
=
2
S < 1, so
th
converges by the Ratio Test.
21. bn
= Fn1 > 0, {bn} is decreasing, and lim bn = n+ n+oo
0, so the series
E ( 1)n1 nVn converges by the Alternating +1
n= l
Series Test.
@ 2012 O:ng•gc U:;uning. All RighlS Reserved. Moy not be seaMed, copied, or duplicoted, or posted to Dpublicly occcssiblc website, in whole or in p;u1.
.
e scnes
CHAPTER 11
En
23. Consider the series of absolute values:
3
I/ is a pseries with p
= t ~ 1 and is therefore divergent.
REVIEW
0
101
But if we apply the
1'1=1
Alternating Series Test, we see that bn
converges. Thus,
E(1)"
1
= yn 3~ > 0, {bn} is decreasing, and
lim bn nHxl
= 0, so the series n=l E( 1)" 1 n  1 / 3
n 113 is conditionally convergent.
tt = l
2 an+ll ~ (  1t+ 1 (n+2) 3n+l . 2 "+1 l  n+2 3 _ 1+(2/n) 3 3 25. ~ ·~ · ( 1)"(n + 1) 3" . n+ 1 · 4  1 + ( 1/ n) · 4 + 4 22n+3 l Test,
27·
oo
I: n=I
+
oo
I:
n=l
( 1)" (n 1)3" 2 "+ 1
2
( 3)n 1 23n
is absolutely convergent.
oo
( 3)nl (23)n
1
8
= n=l I:
.
< 1. asn+ oo, so by the Ratro
oo
= n=l L
( 3)"1 8"
1
( 3)n l 8" 1
oo
= 8 n= I:l
1
oo
= 8 n=l L
(
3 ) n l 1( · 1 )  8 = 8 1  (  3/8)
1
=s· u = u 00
29. :L[tan 1 (n+ 1)tan 1 n ) = lim n= l
Bn
noo
= lim [(tan 1 2 tan 1 1) + (tan 1 3 tan 1 2) noo
= lim
noo
[tan 1 (n + 1) tan 1 1) =
e2 e3 e4 oo e" 31. 1  e + 21.  31. + 41.  .. · = L (1)"1 n=O n.
1!. 2
oo ( e)n
1!. 4
=
+ · · · + (tan 1 (n + 1) tan 1 n)J
1!. 4
oo
x"
n=O
n.
= E 1 = e e since e"' = I: I n=O
n.
.
for all x .
1 · 1 ( oo x" oo (x)" ) 33. cosh x = (e"' +e"') =  I: 1 + L 12 2 n=O n. n=O n.
' 35·
oo
n~l
( 1) n+l n5
· 1
= 1
32
1
+ 243
1 1 1 1 1  1024 + 3125  7776 + 16,807 32,768 + ... ·
. 1 1 oo ( 1)"+1 Srnce ba = 85 = < 0.000031, I: 32' 768 n=l n5 37.
00
1
n=l
+5
I: 2
" ~
~
( 1)"+1 n=l n5 7
E
~
0.9721.
8 1 1 1 E  " ~ 0.18976224. To estimate the error, note that 2  · < ,so the remainder term is n=l 2 + 5 + 5" 5n .
00 00 1 1 Ra = n~o 2 + 5" < n~o S"
=
1/5 9 1 _ 115
= 6.4 ~ 10, 7
[geometric series with a=
b and r =
i ].
© 20 I2 Ceng~ge lc::tming. All Rights Resc:r•cd. May not be sc:umcd, copied, or duplicated. or posted io n publicly ncccs.siblc website, in ,whole or in part.
l
102
~HAPTER
0
39. Use the
11
INFINITE SEQUENCES AND SERIES
Limi~ Comparison Test.
lim
n+00
I(~an)a,, I =
Since E Ia,. I is convergent, so is I: \ ( n :
+
I
41. lim a,.+ll = lim [ lx 21"+1 . n  oo an !'oo (n + 1) 4" + 1
lx + 21
< 4
¢:?
+2 <
4 < x
4
1
lim n
n  oo
I,
) a,. by the Limit Comparison Test.
n4" ,.] = lim noo
lx + 21 ¢:?
1 + = n+oo Jim (1 + .!) = 1 > 0. n n
[__2!:___ lx +2[] n+1 4
 6 < x < 2. lf x
=
= [x + 21 < 1
4
oo E
6, then the series
.
•
n =l
(x
¢:}
lx +21 < 4, so R = 4.
+ 2)n becomes n 4"
f: (44r = f: ( nl )", the alternating harmonic series, which converges by the Alternating Series Test. When x = 2, the
n=l
71.
n=l
series becomes the harmonic series
f: .!,n which diverges. Thus, I= [ 6, 2).
n= l
1
2
3
43. lim I Un+l l = lim 1 n+ (x )"+1 · a,.
noo
soR = ~
I:
n=O y
../n + 4
noo
lx 31<
~= n +3
~
¢:?
I: n;/
2
alternating series, so I
,
which diverges
/("l(x)
/ (n)(:g)
0
sinx
2
1
cosx
.il
2
 sin x
 2
3
 cosx
sin x
¢:}
[P = ~
$
= 2 jx 31
lim noo
Jn + 3 = 2lx 31< 1
n+4
¢:}
jx 31 <
· · oo 2"{x  3}" ~ < x < ~ · For x =~. the series I: .;n+3 becomes n =l n +3
1),
but for X
= ~' we get I: ~ .which is a convergent n=O v n + 3
= [~ , ~).
n
4
~ < x 3 < ~ ·
n=3
45.
y'n+3 ,., 2"(x  3)
1
2
1
_ _il 2 1
2
•.
"C) {3) (7r) {4}C) 4 2 3 sinx = f(~)+f'(~)(x~)+!.____i_(x~) + / 6 (x~) +/ 6 (x ~) + ... 6 6 . 6 2! 6 3! 6 4! 6
1 47. 1 +X = 1 _
~X)
=
n~O ( x)" = n~O {1)" x" for lxl <
2
1 =>
00
~ =I: {1f'x"+2 withR= 1. 1+x
n=O
© 2012 Ccngnge Learning. All RighiS R=n.:d. May not be S<31lncd, copied, or duplicotcd, or posted too publicly accessible website, in wbolc or in port .
~.
CHAPTER 11
49./I
1  dx = ln{4 x) 4x
I
1 dx = 1 4 X . 4
REVIEW
D
+ C and
 1dx = 1 1  x/4 4
I L: (X)" 00
n=O 4
dx
/E oo x" oo xn+1 + C So = 1  dx = .41 n=O E 4 n=O 4" 4"(n + 1} ·
1 00 xn+1 00 xn+1 00 x" . ln{4  x )= E ,.( n +C.Puttmg x=O,wegetC = ln4. ) +C=  E ,.+1( + ) +C =  E 1 4 4 n=O 4 n+ 1 4 n=O n n=1 n . Thus, f(x) = ln{4 x) = ln4
x" " . The series converges for Jx/41 < 1 rt=1 n 4 oo
E
<:::}
JxJ < 4, so R = 4.
Another solution:
ln{4  x) = ln[4(1  x/4)] = In 4 + ln{1  x/4) = ln 4 + ln[1 + ( x/4))
f
4 = ln4 + 'f(1t+ 1 (x/ )n (fromTablel ) =ln 4+ f{  1) 2"+1 x "n = ln4 x" . n=1 n n=1 n4 n=1 n4" 51. sin x =
oo
( 1)"x2n+1
•
E ,__,.~,..,. n =O (2n + 1)!
oo
4
=* sm(x ) =
E n=O
{1)"(x4?n+l oo (2n + 1)1· = .n=O E
(
. 1 )"x8n+4 ) 1 for all x , so the rad1us of 1 · ,
(2n +
convergence is oo. 1
1
53. f(x)= ~
=
1
V'i6 (1 
{/16(1  x/16)
 1(1...!..x)1/ 4 ie x) l / 4  2 1e
~ [1 + (~) (~) + (t)(~) (~) 2 + ( t) (~) (~) (~)3 + .. ·] 2
4
16
2!
16
3!
16
.
.
1 oo 1 . 5 . 9 ..... (4n  3) 1 oo 1 · 5 · 9 · .. · · ( 4n  3) x" =  + 'E xn =  + 2:: ::::::':  !.. 2 n=l 2 . 4" . n ! . 16" 2 n=1 26n+1 n ! for
J 1~ J < 1 x"
00
55. e"' =
I
Jxl < 16, so R = 16.
#
e"'
E I• so X n=O n.
1 = 
ex
X
dx = C+ lnJxJ+ X
57. (a)
00
xn
E1 n=O n.
00
xn1
1 =  + X
00
xn1
2::  n=1 n.1·
and
x"
oo
n=l n·~
f(n)(x)
j (n)( 1)
0
x1/2
1
1
~x  1 12
2
 ±x3/2
3
l!x 5/2·
4
wx  7/2
©
xn1
E   = x 1 + n=1 E  n.1><=0 n 1.
E  ,.
n
8 15
00
=
Vx :=::::: T3(x) = 1 + 1/ 2 (x  1)  1/ 4 (x  1? + 3 / 8 (x  ·1)3 1!
2!
= 1 + Hx  1) Hx 
31
1?
+ fo(x 
1) 3
1
2 ..,.. 41 3 8 15
 Iii
2012 Cengoge Le=ing. All rughiS Reserved. Mlly not be sconned. copied, or duplicated, or posted to a poblidy accessible website, in "1101e or in prut.
103
104
0
CHAPTER 11
INFINITE SEQUENCES AND SERIES
1.5
(b)
(c) IR3 (x) l $
~~ ix 
4
11 , where J! (4 ) (x) J $ M with
j (4>(x) =  ~~x 7 1 2 . Now 0.9 ~ x $ 1.1
 0.1 ~
X
1 ~ 0.1
=>
(x 1) $ ·(0.1) 4 ,
. x = 0.9 gtves . M = . and Ietting
IRa(x) l $
2
0
=>
16 (0.~~712 41(0.1)
4
15 ( _ )7 , so 16 0 9 12 4 ::::::
0.000005648
:::::: 0.000 006 = 6 X lO G 5 X 10 6
(d)
From the graph of IR3(x)l = IJX  T3(x)l. it appears that the error is less than 5 x
x3 x6 x7 . x3 xs x7 ) 1 = x  +    +· ··,sosmx  x=  +    +··· and 1 + . 3! 5! 7! 3! 5! 7!
• x2n +l
oo
59.sin x = 2:( 1t ( 2n n=O
61. f(x ) =
w 6 on [0.9, 1.1].
00
2: Cn x"
00
00
=> f( x)= l: cn(xt = 2:( 1)"c.,x" n =O
n=O
n=O
(a) Iff is an odd function, then f(  x) =  f(x)
=>
00
00
u=O
n=O
2: ( 1)"cnx" = 2:  c,,x". The coefficients of any power series
are uniquely determined (by Theorem 11.10.5), so ( 1)" Cn =  en.
If n is even, then (  1)" Co
= C2 = C4 =
= 1, so en =
c,.
=> 2en
=0
· · · = 0. 00
(b) If f is even, then f (x) = f(x)
=>
2: {lt
If n is odd, then (  1 )" = 1, so en C3
= C5 =
···
= Cn =>
'
00
c,. xn
n=O
that is, Cl =
=> c,. = 0. Thus, all even coefficients are 0, that is,
= 2: Cn x"
=> {1)" Cn = Cn.
n=O
2c,..
=0
=>
Cn
= 0. Thus, all odd coefficients are 0,
= 0.
@) 2011 Ccngoge Lcorning. All Rigbls Reserved. Moy not be scanned. copied, or
D
PROBLEMS PLUS
1. It would be far too much work to compute 15 derivatives of f. The key idea is to remember that j <••l (0) occu~s in the .
coefficient ofx" in the Maclaurin series of f. We start with the Maclaurin series for sin: smx ·
Then sin(x3 )
/ (15) (O)
xl5
xg
= x 3  3f + 5f 
···,and so the coefficient of xlu is
1
5f· Therefore,
=
151
x5
x3
= x 3! +  ... . 5!
151 = 6 . 7 . 8 . 9 . 10 . 11 . 12 . 13 . 14 . 15 = 10,897,286,400. f ( 1"") (0) = 5f . Append.IX D , 3. (a) From Formu Ia 14a tn
2 cot 20 =
1  tan2 8
tan
e
"h x
Wit
=y=
2tanB., , so cot 28 = 1  tan28 1tan 8 2tan8
n u, we get tan 28 =
=cot 8. tan8. Replacing 8 by
tx. ~e get 2 cot
X=
cot ~X tan
=?
!x. or
tan ~x =cot ~x  2cotx. X
'
X
X
Sn
=
tan(x/ 2) 2
tan(x/ 4) 4
+
= [ cot~/2) _
+
tan(x/ 8) 8
for x
=
~
L ra.=l
1
X
tan  is 2n 2n
tan(x/2") 2n
1
2n
t anX = }"1m
2n
ntoo
x/2" . ( )
SIDX 12 n
X
"1 0. Therefore, if x "1 0 and x "1 k1r where k 0
so the n th partial sum of
,
_ cot(x/2n l) ] _ _ . .. _ cot x 2 1
cos(x/2n)
cos(x/2") 2" sin(x/ 2n)
n=l
1
cotx]+ [cot(:/4) _ cot(;/2) ] + [ cot~/8) _ cot~/4)] + ... + [ cot(x/2") n 2
Now cot(x/2n) 2"
+ .. •+
00
X
(b) From part (a) with  1 in place of x, tan  = cot   2 cot 2n. 2n 2tt 2n
1
> 
X
·
2") + cot(x/ ,. 2
[telescoping sum]
, 1
1 =  as n > oo since xf2n > 0 X
is any integer, then
s,. = Iim ( cotx+  1 cot X n+oo
2n
2n
)
= cot x+ 1 X
If x = 0, then all terms in the series are 0, so the sum is 0. 5. (a) At each stage, each side is replaced by four shorter sides, eacb of length
k of the side length at the preceding stage. Writing s 0 and Eo for the number of sides and the length of the side of the initial t1:iangle, we generate the table at right. In general, we have sn = 3 ·
e,. =
( ~) n' so the length of the perimeter at the nth stage of construction
is Pn
= Snfn = 3 ·. 4" · (~)" = 3 · (1)".
(b) P n =
4"
nl = 4 3
(4)nl 3 .
Since
so= 3 81
S2
S3
= 3·1 = 3 · 42 = 3 · 43
eo = 1
e1 = 1 / 3 e2
= 1/ 3 2
e3
= 1/ 3 3
1 > 1, Pn > oo as n> oo.
(c) The area of each of the small triangles added at a given stage is onen inth of the area of the triangle added at the preceding stage. Let a. be the area of the original triangle. Then the area an of each of the small triangles added at stage n is
e 2012 CcngUl_;C Lcamin& All Rig)tts RL"SCr\'cd. Ma)' nol be scanned. copied. or duplic31Cd. or pos1ld to a publicly accessible websile, in whole or in part.
105
106
CHAPTER 11 PROBLEMS PLUS
0
an = a · _.!:_ = .!::.. . Since a small triangle is added to each side at every stage, it follows that the total area An added to the 9n 9n figure at the ntl1 stage is An =
Sn1
4n1
9
·an = 3 · 4n 1 · ~, = a· 32 n_ 1 . Then the total area enclosed by the snowflake
.
2
3
curve is A = a + A1 + A2 + As + · · · = a + a · ~ + a · ; + a · : 5 + a · : 7 + · · · . After the first term, this is a 3 . senes . w1t . h common ratlo . !) 4 , so.A = a'+ a/3 8a But t I1e area of .the ongma . . I eqUI'IateraI _ .1 = a + a · 9 = S. geometric 1 9 3 5 .
. Ie w1t . h s1'd e 1 .IS a = '12 · 1 · sm . 7r = tnang
2 J3 J3 S h I db h wfl k . 8 J3 4 . o t e area enc ose y t e sno a e curve IS 5 · 4 = T .
3
7. (a) Let a= arctanx and b =arctan y. Then, from Formula 14b in Appendix D,
(b) tanatanb tan(arctan x)tan(arctan y) xy tan a = 1 +tan a tan b = 1 + tan(arctanx) tan(arctan y) = 1 + xy . Now arctan x  arctan y = a  b = arctan( tan( a  b))
xy
= arctan 1+xy
since
t < a 
b<
¥.
(b) From part (a) we have 120
arc t an
arc t an
120 119 
1 t 2311 .arc .an
1
Wi  239 = arc t an 120 1 1 + Wi . 239
28,561 28,441 28 _561 28,441
= arct an 1 =
47r
(c) Replacing y byyin the formula of part (a), we get arctanx +arctan y = arctan x + y . So 1 xy 4 arctan 51
=
2(arctan
= arctan
1
+1
1 1) 2 5 5  2 arc t an 5  arc t an 12 5 + arc t an 12 5 5 + arctan 5 = arctan 15. 1 12 51
..!!... +..!!... 12 12 5 5 1 12'12
Thus, from part (b), we have 4 arctan
= arctan g~
t  arctan 2i 9 = arctan ~ i~  arctan 2; 9 = "i.
(d) From Example 7 in Section 11.9 we have arctanx 11 arctan 5 = 5

x3
5 · x x7 x9 x 11 + 5  7+ 9 ~ +· ··,so
= x 3
1 1 3 · 53 + 5 · 55
1 1 7 · 57 + 9 · 59


1 11 · 511 + .. .
This is an alternating series and the size of the terms decreases to 0, so by the Alternating Series Estimation Theorem, the sum lies between ss and s6, that is, 0.197395560 <arctan (e) From the series in part (d) we get arctan
2 ~9 = 2~9 
3
.~
t
< 0.197395562.
393 +
5
.;
395
  ·. The third term is less than
2.6 x 10 13 , so by the Alternating Series Estimation Theorem, we have, to nine decimal places, arctan
2; 9
~ s2 ~ 0.00~184076. Thus, 0.004184075 < arctan 2 ~ 9 < 0.004184077.
(f) From part (c) we have 1r
= 16 arctan k 4arctan 2 ~ 9 , so from parts (d) and (e) we have
16(0.197395560)  4(0.004184077) < 3.141592652 <
1r
1f
< 16(0.197395562) 4(0.004184075)
=?
< 3.141592692. So, to 7 decimal places, 1r ~ 3.1415927.
© 2012 Ccngagc Lt;.uning. All Rights Rcscn·cd. Mny not be scanned. copied, or duplicntcd, or posted to a publicly accessible website, in whole or in pan.
CHAPTER 11 PROBLEMS PLUS
f;
9. We start with the geometric series
x" = 
n=l
X
_ ) = (  1 ) 2 for lx l < 1 = dd ( 1
)
1
X
n= O
l
 , lx l < 1, and differentiate: oo
00
oo nxn l = d }:
107
1 X
n. =O
d( }: xn
1
D
1
X
L
=>
X
X
oo
nxn
=XL
~
2
nxn l = ..,.~ n =l (1  x)2
n =1
for lx l < 1. Differentiate again: 00
}: n 2 x"
_
1
n= 1
x
(1  x?  x · 2(1  x)(  1) dx (1  x)2 (1  x) 4 d


f; n3xn  1 =

dx (1  x) 3
f; n 3 x" =
x+ 1 (1  x)3
=>
L n n =l
,..
x =
x2 + x (1  x)S
=>
!!:.._ x 2 + x = (1  x? (2x + 1)  (x2 + x)3(1  x)2 (  1) = x 2 + 4x + 1 ==>
n=l
n=l
=
3
x
t
4
(1  x)~
(1  x )6
)~ x , lx l < 1. The radius of convergence is 1 because that is the radius of convergence for the
2
x
1 X
geometric series we started with. If x = ± 1, the series is }:n3 (±1)n, which diverges by the Test For Divergence, so the interval of convergence is (1, 1). 11 . ln(1 :
2
)
= ln(n~~ 1 ) ,;, In (~+ ll~n  1)
= ·in[(n +1)(n 1)] ln n 2
= ln(n + 1) + ln (n  1)  2ln n = ln(n  1)  In n ln n + In(n + 1) n 1 n 1 n = In  n   [ln n  ln(n+ 1)] = In  n  In n+ . 1
Ek
Let Sk =
n= 2
Sk
In ( 1  21) = n
Ek
n=2
(
In n1   In  n  ) fork ~ 2. Then n n+ 1
3) + · · · + (In  1  In k+1 = . (In12 In +k1 ) = In 12 1 = hm
= ( ln 1  In 2) + ( In 2  In 2 3 3 4
1) = .
00 }: In ( 1   2 n= 2 n
lim
Sk
k oo
=
k k
)
ln
k
koo
13. (a)
k
1  In  k  so In2 k+ 1 '
·I
ln 1 In 2  In 1 =  In 2.
The xintercepts of the curve occur where sin x = 0
<=> x = mr,
nan integer. So using the formula for disks (and either .a CAS or l+l~~+''r7"""''...,.._t 40
sin2 x = ~ (1  cos 2x) and Formula 99 to evaluate the integral), the volume of the nth bead is
(  :r:/10 sm . x )2 dx = rr Jrmr :r:/ 5 sm . 2 x dx v.n = rr Jf"'' (,.l ) .. e (n 1 ) .. e = 2;g,rr (e (nl)lf/5 e mr/5)
 I
(b) The total volume is rr .{000 e :r:/S sin 2 x dx =
f n =l
V,,
= 2~it
f; [e (n1)"/
5 
en"/ 5 ] =
;g;
2
[telescoping sum].
n=l
Another method: If the volume in part (a) has been written as Vn =
2
~g; en"/ 5 (e"/5

1), then we recognize
·
as a geometric series with a= 2~g; (1  e"15 ) and r
©
= e"/ 5 .
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f; Vn n=l
108
D
CHAPTER 11 PROBLEMS PLUS
15. If Lis the length of a side of the equilateral triangle, then the area is A= ~L · 4L = 4£2 and so £ 2
=fa A.
Let r be the radius of one of the circles. When there are n rows of circles, the figure shows that L ( y'3 )' 2 n+ 31
L = v'3r+r+(n2)(2r)+r+v'3r=r(2n2+2../3),sor =
1 The number of circles is 1 + 2 + · · · + n = n(n + ) , and so the total area of the circles is 2
An =
=
n(n + 1)
1fT
2
2 n(n+ 1) 2
1f
=
n(n + 1)
· 7r
2
£2 2 4 (n+v'3 1)
4A/J3 n(n+1) 1rA 2 2 4(n+v'3  1) = {n+../31) 2../3
n(n + 1) 1r A = (n+../31) 2 2.;3 1 + 1/n =
=>
An
1r
1r
>   asn >oo
2
[1+(J31)/n) 2v'3
r
2../3
2r
17. As in Section 11.9 we have to integrate the function x"' by integrating series. Writing x"' = (e1"
Maclaurin series for e"', we have x"' integrate this series termbyterm:
= (eln "')"' =
1 ·1
0
f \"(lnx)"dx
=
}0
=
e"' 1" "' =
oo 2:
x"' dx =
11
n=O
withu = (lnx)",dv =x" dx,sodu=
0
n(ln x)" 1 · x
E (x Inn.x) 1
n=O
n
11 1 x"(lnxf dx. We integrate by parts n=O n. 0 oo 2:
I
x"+1
dxand v =   : . n+ 1
1
1
to+ t
= e"' 1""' and using the
n.
n=O
x"(ln xt dx = n.1
"')"'
= E x" (~ x) n . As with power series, we can
lim /. x"(ln xfdx= lim [x"+l (lnx)"] to+ n
.
t x"(Inxf
o _ n _
2r
~L~
1
n+ 1 } 0
+1
1
t
lim ( ___!!:__ x"(1nx f 1 dx
to+
it n + 1
dx
(where !'Hospital's Ru le was used to help evaluate the first limit). Further integration by parts gives
~ 1 n +1 0
1
x"(ln x)k dx =
( }0
1 1
1
x"(ln x)k 1 dx and, combining these steps, we get
1x"(ln x)"dx = (1)"n!11 x"dx= (1)"n! (n+1)"
0
"' d
1
, 0
X
X
=~ L...
n=O
0
..!_ 1 1 "(In ·)"' d n.I
0
X
.
X
X
=
19. By Table 1 in Section 11. 10, tan 1 x =
~ ..!_ ( 1t n! = ~ L...
n=O
n.I ( n
+ 1) n +1
have
6
= tan1
(
1 )
y'3
L... (
n=O
(1)" n
+ 1) " +1
= ~ ( 1)"1 n" 1
x 2n+l
00
n+ 1
2
r.;•
v3
·
we
1
00 (1/../3) "+ (1)" i 1 = n~O (  .1t 2n + 1 = n~O ( 1)" 3 y'3 2n + 1' SO 00
6 00 (1)" 00 (1)" ( 00 ' (1)" )  2 3  2 3 1  ../3 n~O (2n + 1)3"  y'3 n~O (2n + 1)3"  y'3 + f1 (2n + 1)3n
·rr  
L...
n=l
E ( l t 2'   f or JxJ < 1. In particular, for x =
n=O
7r
=?
(n+ 1)n+l
·
. =>
E 00
n=1
(1)" 7r =1. (2n + 1)3" 2 y'3
<[) 2012 Ccngogc Learning. All Rights Reserved. Moy not be sconncd. copio;cJ, or dUJ>licutcd, or posted to a publicly oeccssiblc website, in whole or in p:tr1.
CHAPTER 11
21. Let f (x ) denote the lefthand side of the equation l no solutions oftheequation. Note that / ( x
< 0 are given by x =
x
2
x
+ 2T +
x2 ! 4
x2
= 1  2! +
)
+ 6f + 8f + .. · = 0. lfx;:::: ~· then f(x ) ;::::
x4
4
x4
x3
·
PROBLEMS PLUS
xs
! 
6
!
xs
= 0 are x = 
(i  1rk)
k is a positive integer. 23. Call the series S. We group the terms according to the number of digits in their denominators:
Now in the group gn , since we have 9 choices for each of then digits in the denominator, there are 9n terms.
10 ~_ 1
[except for the first term in gl]. So 9n
00
Now
L: 9 ( 190 ) n  l
is a geometric series witl1 a
= 9 and r
=
< gn · 10~_ 1 = 9( {0 )"'
fa < 1. Therefore, by the Comparison Test,
n=l
S=
oo
oo
n= l
n=l
L: 9n < L:
9
9(w)
n 1
=
9 1 _ 9110
= 90.
Use the Ratio Test to show that the series for u, v, and w have positive radii of convergence (oo in each case), so Theorem 11'.9.2 applies, and hence, we may differentiate each of these series: du 3x 2 6x5  = +
dx
. .
dv
x6
x3
3!
6!
9x8
x2
x5
x8
++··· = 2! ++ 8! +· ·· = w 9! 5!
x9
dw dx
x4
x7
x 10
S imilarly, dx
= 1 + 3! + 6f + 9! + · · · = u, and
So u' = w, v'
= u, and w' = v. Now differentiate the le fthand side of the desired equation: d
dx (u 3
+ v3 + w 3 
3uvw)
2
3
13
3
+ wa 
+ 03 + 03 
4!
+ 7T +
lO!
+ · ·· = v.
= 3u2 u' + 3v2 v' + 3w 2 w'  3(u'vw + uv'w + uvw') = 3u w
u +v
=x+
+ 3v2 u + 3w 2 v 
3(vw 2 + u 2 w
+ u~ 2 ) = 0 =>
3uvw = C. To find the value of the constant C, we put x = 0 in the last equation and get 3(1 · 0 · 0)
= C =>
C
= 1, so u 3 + •v 3 + w3 
109
1 and t~ere are
+ 8f  · · · = cosx. The solutions of cosx =
i  1rk, where k is a positive integer. Thus, tl1e solutions off(x)
Furthermore, each term in 9n is less than
0
3uvw = 1.
® 2012 Ceog:age Lc.1ming. All Rig.hiS Resen·cd. Mny not be scanned, copied, or duplicated, or posted Lo a publicly occessiblc website, in whole or in p.'ltt.
1
•
0 for
2 ,
where
12 12.1
D VECTORS AND THE GEOMETRY OF SPACE ThreeDimensional Coordinate Systems
1. We start at the origin, which has coordinates (0, 0,. 0). First we move 4 units along the positive xaxis, affecting only the xcoordinate, bringing us to the point ( 4, 0, 0). We then move 3 units straight downward, in the negative zdirection. Thus only the zcoordinate is affected, and we arrive at {4, 0,  3). 3. The distance from a point to the yzplane is the absolute value of the xcoordinate of the point. G(2, 4, 6) has the x coordinate with the smallest absolute value, soC is the point closest to the ·y zplane. A(  4, 0, 1) must lie in the xzplane since the distance from A to the xzplane, given by the ycoordinate of A, is 0. 5. The equation x
+ y = 2 represents the set of all points in y=2  x
JR3 whose x andycoordinates have a sum of2, or equivalently where y = 2  x. This is the set
y=2 x,z= O
{(x, 2  x , z) I x E JR, z E R } which is a vertical plane that intersects the xyplane in the line y
=2
x, z
= 0.
X
7. We can find the lengths of the sides of the triangle by using the distance formu la between pairs of vertices:
IPQI = )(7 IQRI =
3)2 + [0  ( 2)]2 + [1  ( 3)] ~
j(l 7)2 + {2  0) 2 + {1  1) 2
IRPI = )(3 
= yf16 + 4 +
16 = 6
= vf36 + 4 + o = .J40 = 2 v'15
1)2 + ( 2  2)2 + ( 3  1) 2
The longest side is QR, but the Pythagorean Theorem is not satisfied:
= vf4+ 16+ 16 = 6
IPQI 2 + IRPI 2
=f:.
IQRI 2 . Thus PQR is not a right
triangle. PQR is isosceles, as two sides have the same length. 9. (a) First we find the distances between points:
IABI = ){3 2)2 + {7 4)2 + (2 IBGI = IACI
2)2
= v'26
){1 3) 2 + (3  7) 2 + [3 (  2)]2 =
= )(1 2)2 + (3  4) 2
v'45 = 3 v'5
+ (3 2) 2 = J3
In order for the points to lie on a straight line, the sum of the two shortest distances,must be equal to the longest distance. Sin<;:e
v'26 + J3 =f:. 3 v'5, the three points do not lie on a straight line. © 2012 Ccngagc Lcarnini;t. Al l Rights Reserved. Mny not be scanned, copied, or duplicaled, or pos1ct.11o .1 publicly acccssihle website. in whole or in pari.
111
112
0
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
(b) First we find the distances between points:
IDEl = J(l 0)2 + [ 2 ( 5)]2 + (4 IEFI = )(3  1) 2
+ [4 
5) 2
(  2)]2 + (2  4) 2 =
IDFI = )(3  0) 2 + [4 (5)]2
+ (2 
5) 2
:::::
VIT
v'44 = 2 VIT
= v'99 = 3Vll
Since IDEl + IEFI = ID F I, the three points lie on a straight line. 11. An equation of the sphere with center ( 3, 2, 5) and radius 4 is [x  (  3)]
(x + 3) 2
+ {y 
2) 2
+ (y 
2) 2
+ (z 
2 5) = 4 2 or
+ (z 5) 2 = 16. The intersection of this sphere with the yzplane is the set of points on the spher~
whose x coordinate is 0. Putting x
(y 2)2 + (z 5) 2
2
= 0 into the equation, we have 9 + (y 
2)
2
+ (z 5) 2 = 16, x = 0 or
= 7, x = 0, which represents a circle in the yz plane with center (0, 2, 5) and radius /7.
13. The radiusofthesphere is the distance between (4,3,  1) and (3,8, 1): r = )(3  4)2 Thus, an equation of the sphere is (x  3) 2
+ (y 
8)
+ y 2 + z2 
15. Completing squares in the equation x 2
2
+ (z 1)2 =
2x  4y + 8z
+ (8 
3)2
+ [1 
(1)]2
= V3o.
30.
= 15 gives
recognize as an equation of a sphere with center (1, 2,  4) and radius 6. 17. Completing squares in the equation 2x2
2(x2

4x + 4) + 2y2
(x  2)2 radius

8x + 2y2 + 2z 2
+ 24z = 1 gives
+ 2~z + 12z + 36) = 1 + 8 + 72 2
+ y 2 + (z + 6) 2 =
81 , 2
:::}
2(x  2) 2
+ 2y2 + 2(z + 6? = 81
:::}
which we recognize as an equation of a sphere with center (2, 0, 6) and
j¥ = 9/ v'2.
19. (a) lfthe midpointofthe line segment from P 1(x 1 ,y1,z1) to P2 (x 2 then the distances
IP1QI =
,y ,z 2
2)
1 1 2 is Q = (x ;x , Y
;v
2
,
1 2 z ; z ),
IPl Q l and IQ P21are equal, and each is half of IPt P21· We verify that this is the case:
v[~
(x1
+ X2) 
1 1 )2 = y / ( 2X2  2X1
= /(~) 2 ((x2 
X1]
2
+ a
1 1 )2 + ( 2Y22Y1 + ( 21 Z2 
x l)
2
+ [t(z1 + Z2 ) Z1]
Y1t
+ (y2  vd + (z2 
2
21 Z1) 2
z1)
2
]
= ~J(x2  xd + (y2 y1) 2 + (z2
 z1?
= ~ IPtP2 1
® 2012 Ccngogc Looming. All Rigllls Reserved. Mny not be scaMe~. copied, or duplicooed. or posted 10 • publicly =iblc wcbsioc, in whole or in port.
SECTION 12.1 THREEDIMENSIONAL COORDINATE SYSTEMS
!QP2\ =
J[x2  ~(X 1 + X2)] 2 + [y2 ~(Yl + Y2)] 2 + [z2 
=
(~x2 ~ x1) 2 + (h2 ~·yt) 2 + (tz2  ~z1) 2 =
.=
1
~J(x2 
~ (z1 + z2))
JG)
2
0
113
2
[(x2
xd + (y2 
yl) 2 + (z2  z1) 2)
xd + (y2 y1) 2 + (z2 zd = ~ \P1P2\
So Q is indeed the midpoint of P 1P2. (b) By part (a), the midpoints of sides AB, B C and C A are P1 ( ~, 1, 4), P2 (1, ~. 5) and P3 (~ , ~, 4). (Re~all that a median of a triangle is a line segment from a vertex to the midpoint of the opposite side.) Then the lengths of the media ns are:
IAP2\
=
\BP3\ = \CP1\ =
Jo + (~  2) + (5  3) = jf+4 = fj = ~ J(~ + + (~) + J¥ + *+ = J2j = ~JW J(~ + + J¥ + ~.J85 2
2
2)
2
2
4)
2
2
2
(4 5) =
(1 :__ 1}
2
1
(4 5? =
1=
21 . (a) Since the sphere touches the xyplan.e , its radius is the distance from its center, (2, 3, 6), to the xyplane, namely 6.
Therefore r
= 6 and an equation of the sphere is (x 2? + (y + 3? + (z  6? =
62
= 36.
(b) The radius of this sphere is the distance from its center (2,  3, 6) to the yzplane, which is 2. Therefore, an equation is
(~ 2) 2
+ (y + 3) 2 + (z 
6) 2
= 4.
(c) Here the radius is the distance from the center (2,  3, G) to the xzplane, which is 3. Therefore, an equation is
(x 2) 2
+ (y + 3) 2 + (z 6} 2 = 9.
23. The equation x = 5 represents a plane parallel to the yzplane and 5 units in front of it.
25. The inequality y
< 8 represents a halfspace consisting of all points to the left of the plane y = 8.
27. The inequality 0
~
29. Because z
=
z
~
6 represents all points on or between the horizontal planes z
1, all points in the region must lie in the horizontal plane ~
= 0 (the xyplane) and z = 6.
= 1. In addition, x 2 + y 2
= 4, so the region
consists of all points that lie on a circle with radius 2 and center on the zaxis that is contained in the plane z = 1.
+ z 2 ~ 3 is equivalent to jx 2 + y 2 + z2 ~ J3, so the region consists of those points whose distance from the origin is at most V3. This is the set of all points on or inside the sphere with radius ../3 and center (0, 0 , 0).
31. The inequality x 2 + y 2
33. Here x 2
+ z2
~ 9 or equiva lently Jx2
+ z2
~ 3 which describes the set of all points in JR3 whose distance from theyaxis is
at most 3. Thus, the inequality represents the region consisting of all points on or inside a circular cylinder of radius 3 with axis the yaxis. 35. Thi s describes all points whose xcoordinate is between 0 and 5, that is, 0
< x < 5.
37. This describes a region all of whose points have a distance to the origin which is greater than r, but smaller than R. So inequalities describing the region are r
< jx2 + y2 + z2 < R, or r 2 < x 2 1 y 2 + z 2 < R 2 .
@ 2012 Ccngugc l earnlnu. All R i~lus RescrvcU. May nut be scanned, copit."t.l. or duplicated, or poSicd ton public\)' :~cccssibtc website. in whoJc or in part.
114
0
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
39. (a) To find the x andycoordinates of the point P , we project it onto £2 I
and project the resulting point Q onto the x and yaxes. To find the zcoordinate, we project P onto either the xzplane or the yzplane (using our knowledge of its x  or ycoordinate) and then project the resulting point onto the zaxis. (Or, we could draw a line parallel to
QO from P to the zaxis.) The coordinates of Pare (2, 1, 4). (b) A is the intersection of £ 1 and £ 2, B is directly below the yintercept of L2, and 0 is directly above the xintercept of £2. 41. We need to find a set of points { P(x, y , z)
J(x + 1)2 + (y  5)2 + (z 3)2
(x + 1) 2 x
2
+ (y 5) + (z 
+ 2x + 1 + y 2 
lOy
3) 2
II API = IBPI}.
= J(x 
= (x 
+ 25 + z 2 
6)
6z
2
6)2 + (y  2)2 + (z
+ (y 
2) 2
+ 9 = x2 
+ 2)2
~
+ (z + 2) 2 =>
12x
+ 36 + y2 
4y + 4 + z
2
+ 4z + 4 '=>
14x  6y  lOz
= 9.
Thus the set of points is a plane perpendicular to the line segment joining A and B (since this plane must contain the perpendicular bisector of the line segment AB).
43. The sphere x 2 gives (x
2

+ y2 + z 2 =
4x + 4) + (y2
4 has center (0, 0 , 0) and radius 2. Completing squares in x 2 
4y + 4) + (z
2

4z + 4) =  11 + 4 + 4
+ 4 =>

4x + y 2
(x  2) 2
+ (y 

4y
2?
+z2 
4z = 11
+ (z 2) 2 = 1,
so this is the sphere with center (2, 2, 2) and radius 1. The (shortest) distance between the spheres is measured along the line segment connecti ng their centers. The distance between (0, 0, 0) and (2, 2, 2) is
.J(2  0)2 + (2 0)2 + (2  0)2 =
ffi = 2 J3, and subtracting the radius of each circle, the distance between the
spheres is 2 v'3  2  1 = 2 .J3  3.
12.2 Vectors 1. (a) The cost of a theater ticket is a scalar, because it has only magnitude. (b) The current in a river is a vector, because it has both magnitude (the speed of the current) arid direction at any given location. (c) If we assume that the initial path is linear, the initial fli ght path from Houston to Dallas is a vector, because it has both magnitude (distance) and direction. (d) T he population of the world is a scalar, because it has only magnitude. 3. Vectors are equal when they share the same length and direction (but not necessarily location). Using the symmetry,of the .
parallelog~am
®
+
+ +
as a guide, we see that AB = D C, DA
+ +
= CB, DE =
+
__.
EB, and EA
+
= CE.
2012 Ccngogc l.cnming. All Rights Reserved. May not be sc.mncd, copied, orlluplicatcd. or posted to a publicly accessible website. in whole or in part.
SECTION 12.2 VECTORS 5. (a)
un
(b)
(c)
w~
uNw
u
(d)
u
u v ~
(e)
u
0
v+w
(f)
Dw v+u+w
u
:Gv uwv
>
7. Because the taif of dis the midpoint of QR we have QR = 2d, and by .the Triangle Law, a + 2d = b c
=b
 d
=?
2d=ba
= b  ( ~ b  ~a)
=
d=!{ba) = th  ~a. AgainbytheTriangleLawwehavec + d = bso
=?
~ ~ + ~b.
9. a= (3  ( 1), 2  1}=(4, 1}
11.a =(2  (1), 2  3} = (3,1}
y
y
B(3, 2)
0
A( 1, 3)
X
13. a = ·(2  0, 3  .3,1  1} = (2, 0,2}
X
15. (1 , 4} + (6,  2} = (1 + 6, 4 + ( 2)} = (5, 2) y
A(0, 3, 1)
X
X
17. (:~, 0, 1)
+ (0, 8, 0) = (3 + 0, 0 + 8, 1 + 0) = (3, 8, 1)
(3, 0, 1)
X
©
2012 Ceng3ge Learni ng. All Rights Reserved. May not be scallll(((. copied. or duplicoltcd. or posted too publicly accessible wcbsilc. in who le or in part
115
116
0
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
19. a + b
= (5 + (3) ,12 + (6)) = (2,  18)
2a + 3b = (10, 24) + (9, 18) = (1, 42) lal
= y'52 + (  12) 2 = .jl69 = 13
Ia b l = 1(5 (3),  12 (6))1
= 1(8, 6) 1= y'82 + (  6)2 = v'100 = 10
21. a+ b = (i'+ 2j  3 k ) + ( 2 i  j + 5 k )
2a + 3b lal
= 2 (i +
=
i + j + 2k
2j  3 k ) + 3 (2 i  j + 5 k ) = 2 i + 4j  6 k  6 i  3j + 15 k
=
= y'12 + 22 + (3)2 = v'i4
Ia  b l = l(i+ 2j  3 k)  (2 i j +5 k )l = l3 i + 3 j  8 k l = y'32 +3 2 + (8)2 23. The vector  3 i . . .IS d1rect1on
+ 7j
1
r,:n
v58
25. The vector 8 i  j
has length l3 i
+ ~j I=
y'( 3)2 + 72
= v'82
= v'58, so by Equation 4 the unit vector with the same
(  3.1 + 7 J' ) =  r.;o 3 1. + r.;oJ 7 .·
v58
+ 4 k has length 18 i 
the same direction is ~ (8 i  j
v58
j + 4 k l = y'82 + ( 1) 2 + 4 2 =
+ 4k) = ~ i 
~j
J8I =
9, so by Equation 4 the unit vector with
+ ~ k.
.
y
27.
4 i + j + 9k
From the figure, we sec that t a n 0 =
.j3
 1 = J3 =>
B = 60°.
X
29. From the figure, we see that the xcomponent ofv is v1
= Ivi cos(1r/ 3) = 4 ·. ~ =
V2
= Ivi sin(1r / 3) = 4 · .,;; = 2 v'3. Thus
y
2 and theycomponent is v Vz
V
= (th, V2) = (2, 2 .j3 ). v,
X
31. The velocity vector v makes an angle of 40° with the horizontal and
has magnitude equal to the speed at which the footba ll was thrown. From the figure, we see that the horizontal component of v is
Ivi cos 40° =
60 cos 40°
is Ivi sin 40°
= 60 sin 40° ~ 38.57 fils.
~
45.96 ft/s and the vertical component
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SECTION 12.2 VECTORS
0
117
33. The given force vectors can be expressed in terms of their horizontal and vertical components as 300 i and
~00 cos 6~0 i +
200 sin 60°j
these two vectors: F
= 200 ( ~)
= (300 +
100) i
IFI ~ V(200) 2 + (100v'3) 2 =
i + 200 (
+ (0 +
~) j = 100 i + 100 v'3 j . The resultant force F =
100 v'3) j
.j70,000 = 100 V7
200 i
is the sum of
+ 100J3j. Then we have
~ 264.6 N. Let() be the angle F makes with the
· · xax1·s. Then tan () = lOO . I pomt . of F 11es · m . t he second qua drant, so positive _ J3 = . J3 an d the termma 2 200 () = tan
1
( 
~) + 180° ~ 40.9° f. 180° =
139.1° .
35. With respect to the water's surface, the woman's velocity is the vector sum of the velocity of the ship with respect to the water, and the woman's velocity with respect to the ship. If we let north be the positive ydirection, then
v=
(0, 22) + (3, 0) = (3, 22). The woman 's speed is
with the east, where()= tan
1
(
lvl = .jg + 484 ~ 22.2 mi/ h. The vector v makes an angle()
~;) ~ 98°. Therefore, the woman's direction is about N(98 90)
0
W
= N8°W.
37. Let T 1 and T 2 represent the tension vectors in each side of the clothesline as shown in the figure. T 1 and T 2 have equal vertical components and opposite horizontal components, so we can write
T1
=  a i + bj and T 2 =
a i + bj [a, b > 0]. By si milar triangles, !!_ = a
0 08 ·
=> a = 50b. The force due to gravity
4
=
acting on .Ule shirt has magnitude 0.8g ~ (0.8)(9.8) = 7.84 N, hence we have w of the tensile forces counterbalances w , so T 1 ( 50b i
+ bj ) +
(50b i + bj )
= 2bj
= 7.84j
+T2 =
=> (a i + bj ) + (a i + bj )
w
=> b =
7
·:
7.84j. The resultant T 1 + T 2
4
= 3.92 and a= 50b =
= 7.84 j
=>
196. Thus the tensions are
T 1 =  a i + bj =  196i + 3.92j and T 2 = ai + bj = 196i + 3.92j . Alternatively, we can find the value of() and proceed as in Example 7. y
39 .. (a) Set up coordinate axes so that the boatman is ·at the origin, the canal is bordered by the yaxis and the line x = 3, and the current flows in the negative ydirection. The boatman wants to reach the point (3, 2). Let() be the angle, measured from the positive yaxis, in the direction he should steer. (See the figure.)
X
In still water, the boat has velocity v b = (13 sjn B, 13 cos()) and the velocity of the current is V c (0, 3.5), so the true path of the boat is determined by the velocity vector v
= v b + V c = (13 sin B, 13 cos() 
3.5). Let t be the time (in hours)
after the boat departs; then the position of the boat at time t is given by tv and the boat crosses the canal when tv = (13sin0, 13cos0 3.5) t = (3, 2). Thus 13(sinO)t = 3
=>
3 . ()and (13 cosB 3.5) t = 2. 13 sm ·
t = 
<1) 2012 Cengnge Le:uni ng. All Rig)lts Reser\'cd. Mny not be scanned, copi<d, or duplleoted, or posted ton publicly ne<es.•ible website, in whole or in par1.
118
0
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
Substituting gives {13 cos (}  3.5)
13
:in(} = 2
=>
+ 110.25 = 676 sin2 () = 676 ( 1 
2
1521 cos 8  819 cos 8 2197 cos 2 (}

39 cos 8 10.5 = 26 sin(} (1). Squaring both sides, we have cos2
9)
819 cos(}  565.75 = 0
The quadratic formula gives
_ 819 ± v'C819)2 4{2197)( 565.75)
cos 8. 
·2{2197)
·= 819 ± ~5g~42 ' 572
:::::: 0. 72699 or  0.35421
The acute value for(} is approximately cos 1 (0.72699) :::::: 43.4°. Thus the boatman should steer in the direction that is 43.4° from the bank, toward upstream.
Alternate solution: We could solve ( 1) graphically by plotting y
= 39 cos 8 10.5 andy = 26 sin 8 on a graphing device
and finding the appoximate intersection point (0.757, 17.85). Thus 0:::::: 0.757 radians or equivalently 43.4°. (b) From part (a) we know the trip is completed when t
= ·13 sm ~ (} .· But 8 :::::: 43.4°, so the time required is approximately
3 . :::::: 0.336 hours or 20.2 minutes. 13sm 43 .40 41 . The slope of the tangent line to the graph ofy
= x 2 at ,the point {2, 4) is ·
and a parallel vector is i are ±~
I
dy dx x=2
+ 4 j which has length Ji + 4j J =
= 2xl .)1 2
= 4 x=2
+ 42 = ..JPi,so unit vectors parallel to the tangent line
(i + 4 j ). ~
_..
43. By the Triangle Law, AB + BC
+ +
t
,_..
= AC. Then AB +
+
+
BC + CA
+
+
._.
+
+
= AC + CA, but AC + CA = AC +
(
+) = 0 .
AC
+
SoAB+BC+CA = 0. (c) From the sketch, we estimate that s :::::: 1.3 and t:::::: 1.6.
45. (a), (b)
(d) c = sa + t b
#
7
= 3s +
Solving these equations gives s
2t and 1
= 2s 
= t and t =
t.
V.
47. Jr ro J is the distance between the points (x, y , z) and (xo, y 0 , z 0 ), so the set of points is a sphere with radius 1 and
center (xo, Yo, zo).
Alternate method: Jr  ro J = 1
(x  xo? + (y  yo) 2
+ (z 
#
J(x xo) 2
+ (y vo)2 + (z zo)2
= 1
#
zo? = 1, which is the equation of a sphere with radius 1 and center (xo, yo, zo).
© 2012 Ccngagc Learning. All Rights Reserved. Moy not be scanned. copied. orduplicotcd, or J>
SECTION 12.3 THE DOT PRODUCT
0
49. a+ (b + c) = (at, a2} + ((bt , b2} + (c1, c2}) = (at , a2} + (b1 + c1, ~ + c2) = (a 1 + b1 + c1, a2 + ~ + c2} = ((a1 + bt) + c1, (a2 + ~) + c2) = (a1 + b1 ,a2 + b2} + (ct, c2} = ((a1,a2} + (b1 , b2}) + (c1,c2} .= (a+ b } + c _____.
+
+
51 . Consider triangle ABC, where D and E are the midpoints of AB and BC. We know that AB. + BC = AC +
+
_.
I
.;.
+
+
+
(1)
+
..
and
t
DB +BE= DE (2). However, DB = ~AB, and BE = ~BC. Substituting these expressions for DB and BE into +
+
+
+ .
t
..
+
· (2) gives ~AB + ~BC = DE. Comparing this with (1) gives DE = ~AC. Therefore·AC and DE are parallel and
12.3 The Dot Product 1. (a) a · b is a scalar, and the dot product is defined only for vectors, so (a · b) · c has no meaning.
(b) (a . b) c is a scalar multiple of a vector, so it does have meaning. (c) Both [a[ and b · care scalars, so [al (b ·c) is an ordinary product of real numbers, and has meaning. (d) Both a and b +care vectors, so th~ dot product a · (b +c) has meaning. (e) a · b is a scalar, but cis a vector, and so the two quantities cannot be added and a · b 1:' c has no meaning. (f) Ia ! is a scalar, and the dot product is defined only for vectors, so lal · (b + c ) has no meaning. 3. a· b = (  2, ~) · ( 5,1 2} = (2)(5} + (~)(12) = 10 + 4 = 14
5. a · b = (4, 1, i ) · (6, 3,  8) = (4)(6) + (1)( 3) +
(i) ( 8) =
7. a· b = (2 i + j ) · (i  j + k ) = (2)(1) + (1) (  1) + (0)(1) '
19
=1
'  1\
'•
9. By Theorem 3, a · b = lal lb1 cos 0 = (6)(5) cos
2 ;
= 30 (  ~) = 15.
11. u , v , aJ)d ware all unit vectors, so the triangle is an equilateral triangle. Thus the angle between u and v is 60° and
u ·v
=
lullvl cos60° = (1)(1) (~)
=
~. lfw is moved so it has the same initial point as u , we can see that the angle
between them is 120° and we have u · w = lul lw l cos 120° = (1)(1) ( ~) =  ~. 13. (a) i · j = (1, 0, 0) · (0, 1, 0) = (1)(0) + (0)(1) + (0)(0) = 0. Similarly, j · k
= (0)(0) +
(1)(0) + (0) (1) ,;, 0 and
k . i = {0){1) + {0){0) + (1)(0) = 0. Another method: Because i, j , and k are mutually perpendicular, the cosine factor in each dot product (see Theorem 3) is cos~= 0. (b) By Pr~perty I of the dot product, i · i = lil = 1 2 = 1 since i is a unit vector. Similarly, j · j = lj l = 1 and 2
k .k
= ©
lkl
2
2
= 1.
2012 Ccngage Le:sming. All Ri!lhiS Reservod . May no1 be scanned. copied, or dupl.ical
119
120
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
D
15. ja j =
vW +
a· b
cos(}= jajjb/ =
5
5 . v'5 =
1 vg· So the angle between a
17. jaj = .,j32 + (1)2 + 52 = .;35, jbj = cos B =
v'5, and a· b
32 = 5, jbj = }22 + ( 1)2 =
l:l·l~l
= .;35 ~ v'29 =
.,j(
= {4){2) + {3)( 1) = 5. ~rom Corollary 6, we have
· . 1 ( 1 ) · and b IS(}= cosv'5 ~ 63°.
2) 2 + 42 + 32 =
v'1~ 15 and the angle between a and b is.B =
19. jaj = .J42 + ( 3)2 + 12 = .;26, jbj = .,j22 + 02 + ( 1)2 a· b · T hen cos (} = jajjbj =
v'29, and a · b
7
7
v'26 . v'5 = v'130 an
d
= v'5, and a. b
. _1 ( 8 = cos
7
v'130
)
~ {3){ 2) + (1)(4) + {5)(3)
cos
1
(
= 5. Then
v'1~15 ) ~ 81° .
= {4){2) + (  3)(0) + (1)( 1) = 7.
o ~ 52 .
21. Let p, q, and r be the angles at vertices P, Q, and R respectively. >
'+
Then p is the angle between vectors PQ and P R , q is the angle >
>
between vectors QP and QR, and r is the angle between vectors +
> .
RPaildRQ. >
>
.
2 4 2 ( • 3 ) · (1 • ) =  + 12 =___!Q_and =cos 1 (___!Q_) .,j( 2)2+32 .j12+ 42 y'l3ffi v'ffi p y'ffi
!?O!!nl
PQ·PR =
Thuscos = p +
r
~
y
+
QP . QR .
cosq =
~48° . Similarl,
· (3, 1) 6 3 3 IQ?IIQRI= v'4(2, 3) + 9 v'9 + 1 = v'13 v'f5 = v'13Q so
 1(
q =cos
3 )
v'l3Q ~ 75
d
o
an
180°  {48°+ 75°) = 57°.
.
.
Alternate solution: Apply the Law of Cosines three times as follows: cos p
=
,___.,2 ,___.,2 1,+12 ~~n ,__. , ,__.
2 PQ 2
2
_i?R( I?QI . IQJil. , and ,___.,, __. ,
cos q ·
23. (a) a . b
2 PQ
= (5){6) +
2
_
cosr 
j?Q(  lnl IQ'RI .
QR (3)(  8) + {7){2)
1
2
..
2 PR QR 1
PR
11
1
= 40 :/= 0, so a and bare not orthogonal. Also, since a
is not a scalar multiple
ofb, a and bare not parallel. (b) a · b = (4) ( 3) + (6)(2) = 0, so a and bare orthogonal (and not parallel). (c) a· b = (1){3) + (2){4) + {5)(1) = 0, soaand bare orthogonal (and not parallel). (d) Because a =  ~ b , a and b ~e parallel. +
+
+
+
+
+
25. QP = (1,  3, 2), QR = (4,  2,  1), and QP · QR = 4 + 62 = 0. Thus QP and QRare orthogonal, so the angle of the triangle at vertex Q is a right angle.
© 2012 Cengogc Learning. All RighiS Resen'
SECTION 12.3 THE DOT PRODUCT
0
121
+ a 2 j + a 3 k be a vector orthogonal to both i + j and i + k , Then a · (i + j ) = 0 <=> a1 + a2 = 0 and a . (i + k ) = 0 <=> a 1 + a3 = 0, so a 1 =  a2 ~ a3. Furthermore a is to be a unit vector, so 1 = ai + a~+ a§ = 3ai . . .  7a 1. 1 1 '+ 7a 1 J' + 73 1 k are two .sueh umt . vectors. 1mphes a 1 = ± 1 . Thus a= l 1 J l k .and a =  73
21. Let a = a 1 i
73
73
73
y ·= 2x  3 has slope 2, so a vector parallel to the line is a = (1, 2}. The line 3x + y = 7
29. The line 2x  y = 3 <=>
<=>
y =  3x + 7 has slope  3, so a vector parallel to the line is b = (1,  3). The angle between the lines is the same as the
angle 8 between the vectors. Here we have a · b = (1)(1) + (2)(  3) =  5, lal = v'P + 22 = [hi = J1 2
+ (3)2 = v'lQ, so cos 8 =
5 a ·b Ialibi = y'5. v'1Q =
 5 5
J2
v'5, and
1 v'2 = J2 or ""'2· Thus 8 = 135°, and the
acute angle between the lines is 180°  135° = 45°.
<=> x3
31. The curves y = x 2 andy= x3 meet when x 2 = x 3
x2 = 0 <=> x 2 (x  1) = 0 <=> x = 0, x = 1. We have

i:x2 = 2x and .!£x3 = 3x2 , so the tangent lines of both curv~s have slope 0 at x = 0. Thus the angle between the curves is dx
0° at
dx
th~ point (0, 0). For x =
1,
! x 1:c=l 2
= 2 and
d~ xt:=l = 3 so the tangent lines at the point (1, 1) have slopes 2 and
3. Vectors parallel to the tangent lines are (1, 2} and (1, 3}, and the angle (} between them is given by cos(}
~3. Since 1(2, 1, 2) 1 = v'4 + 1 + 4 =
V9 =
=
(1, 2} . (1, 3} 1(1, 2} 1 1(1,3}1
=
1+6 = v'5v'W
_ 7_ 5v'2
3, using Equations 8 and 9 we have cos a=~ . cos/3 =~. and cosy = ~· The
direction angles are given by a = cos 1 ( i) ~ 48°, /3 = cos 1 (~) ~ 71°, and 1 = cos 1 (~) = 48°. 35. Since I i  2 j  3kl = v'1 + 4 + 9 =
v'l4, Equations 8 and 9 give cos a =
7IT• cos/3 = * ' and cosy = 7&, while
a = cos 1 (~ ) ~74° , {3 = cos 1 (~) ~ 122°, and /=cos 1 ( vih) ~ 143°. 37. l(c, c, c) I = v'c2
39. IaI =
+ c2 + c2
J( 5)2 + 122 =
= v'3c [since c
> 0], so .cos a = cos/3 =
cos:y =
~
v3c
=
~ and
v3
r;an a ·b 5 · 4 + 12 · 6 v w:.r = 13. The scalar projection of b onto a is comp 11 b =  = 13 1a 1
vector projection ofb onto a is, proj 11 b = (
j~~)
: 1
1
= 4·
f3 (5, 12} =
41 . la l = v'9 + 36 + 4 = 7 so the scalar projection of b onto a is compab = . . of b onto a .IS,. prOJa . b = 7j;f 9 a proJeCtiOn
= 79 · 7l (3 , 6 , 
= 4 and the
( ~,  ~ ).
j~~
=
~ (3 + 12..:. 6) =
l
The vector
2} = 4o o (3 , 6 ,  2} = ( 49 27 54 18 ) • 49• 49 ·
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122
0
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
43. lal =
J4 + 1 + 16 = J2I so the scalar projection of b onto a is compa b =
0
a · bl =
a
1
. . fb . . b prOJeCtion 0 Onto a IS prOJa
1 a 1 j +4k = J2T jaj = J2T · 2iJ2T
(
•
•
1 = 2l 21 J
; ; 21
2
+ 4 k ) = 212 i 
1 =   while the vector
J2I
1 2T j
+ 2T4 k.
' a ·b a·b 2 45. ( ortha b ) · a = (b  p roja b ) · a = b · a  (proja b } · a = b · a   2 a· a = b · a   2 Ia I = b · a  a· b Ja l lal
= 0.
So they are orthogonal by (7). 47. compa b =
a·b
j;f =
~
2
a · b = 2lal = 2 v'IO. lfb = (b1, b:l, b3) , then we need 3bl
+ Ob2 1b3 = 2 v'IO.
One possible solution is obtained by taking b1 = 0, b2 = 0, bs =  2 v'IO. In general, b = ( s, t, 3s  2 v'10 ), s, t E llt
+ (20 8) k = 6 i + 2j + 12 k so, by Equation 12, the work done is 6j + 9 k ) · (6i + 2j + 12k} = 48  12 + 108 = 144joules.
49. The displacement vector is D
W = F · D = (8 i 
51. Here ID I = 80 ft, IF I
w
=
= (6 
0) i + (12  10} j
30 lb, and B = 40°. Thus
=F . D = IF IIDI cos B = (30)(80) cos 40° = 2400 cos 40° ~ 1839 ftlb.
53. First note that n = (a, b) is perpendicular to the line, because if Q 1 = (a 1, bt) and Q2 +
n · Q1 Q2 = aa2  aa1
+~ 
bb1
= 0, since aa2 + bb:l =
 c = aa1
+ bb1
=
(a2 , b2) lie on the line, then
from the equation of the line.
I
Let P2 = (x2, y2) lie on the line. Then the distance from P1 to the line is tJ:te absolute value of the scalar projection +p. (P t P. ) In · (x2 x 1, Y2  Yl ) l lax2 ax1 + by2 lnJ1 I lax1 + by1 + cl ofp1 2 onto n . compn 1 2 = I I = = n vr:?'7l:? a2 + b2 vr::?7'l:? a2 + b2
since ax2
+ by2 =c. The required distance is 1(3)(  2) + ( 4 )(3) + 51 = .
)32
+ ( 4)2
,I
13 . 5
55. For convenience, consider the unit cube positioned so that its back left corner is at the origin, and its edges lie along the coordinate axes. The diagonal of the cube that begins at the origin and ends at (1, 1, 1} has vector representation (1, 1, 1). The angle B between this vector and the vector of the edge which also begins at the ori~in and runs along the xliXis [that is,
. . b B (1, 1, 1) · (1, 0, 0) 1 (1,0,0)] 1sg1ven ycos = I( , , }I I( , 0, 0) I = J3 1 1 1 1 57. Consider the H CH combination consisting of the sole carbon atom and the two hydrogen atoms that are at (1, 0, 0) and (0, 1, 0) (or any H  C H combination, for that matter). Vector representations of the line segments emanating from the
carbon atom and extending to these two hydrogen atoms are ( 1 
(0 
t,0  t, 0 
t) =
(
t , t, t) and
t.1 t.0 t) ~ ( t, t,  t ). The bond angle, B, is therefore given by
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SECTION 12.4 THE CROSS PRODUCT , 0
123
59. Let a = (a1, a2, a3) and= (b1, b2, b3).
Property 2:
a· b = (a1, a2 , a3 ) · (b1, b2, bs) = = b1a1
Property 4:
+ ~a2 + b3a3 =
a1b1
+ a2b2 + asbs
(b1, ~. bs ) · (a1 , a2, as ) = b ·a
(ca) · b = (ca1, ca2, ca3) · (b1, b2, bs) = (ca1)b1 + (ca2)b2 + (ca3)b3
= c(a1b1 + a2b2 + asbs) = c (a· b) = a1(cb1) + a2(~) + a 3(cbs) = (a1 , a2, as)· (cbt, c~, cbs) = a· (c b )
Property 5:
0 ·a = (0, 0, 0) · (a1, a2, a3) = {O)(al) + (O)(a2) + (O)(a3) = 0
61 . Ia · b l = l lal lbl cos8l = lallbl lcos 81. Since lcos81 ~ 1, Ia · bl = lal lbl lcos 81 ~ lal lbl.
Note: We have equality in the case ofcos8 = · ± 1, so 8 = 0 or 8 =
1r,
thus equality when a and bare parallel.
t he Parallelogram Law states that the sum of the squares of the
63. (a)
lengths of the diagonals of a parallelogram equals the sum of the squares of its (four) sides.
2
2
2
2
(b) Ia + b l =(a + b) · (a + b )= lal + 2(a ·b)+ lbl and Ia  b l = (a b )· (a b)= lal
2

2(a · b )+ lb l2 •
2
Adding these two equations gives Ia + b l2 + Ia b l2 = 2 lal + 2 lbl 2 .
12.4
The Cross Product k
j
1. a x b
=
= I~
6 0 2 0 8 0
2 . 1 01

6  2 . J 0 01 1
+
6 oI k 0 8 1
= [0  ( 16)]i  (0  O)j + (48  O) k = 16 i + 48k Now (a x b)· a = (16, 0, 48) · (6, 0, 2) = 96 + 0  96 = 0 and (a x b)· b = (16, 0, 48) · (0, 8, 0) = 0 + 0 + 0 = 0, so a x b is orthogonal to both a and b . j
3. a x b =
k
1 3  2 1 0 5
= I ~ ~~I i  ~~ ~
·= (15  O)i  (5  2) j
+ [0 
lj l~~ ~I +
k
( 3))k = 15 i 3j +3 k
Since (a x b) · a= {15i  3j + 3k) · (i + 3j  2k) = 15  96 = O, a x bisorthogonal to a . Since (ax b)· b
®
= (15 i 
3j + 3 k) · (  i
+ 5 k) = 
15 + 0 + 15 = 0, a x b is orthogonal to b .
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124
0
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
k
j
5. ax b
=
1  1 1 1
1
1
2
= ,~
= {~ i 
=
t
e = (1 
j+ ~
1 1/t t
+ ~ k)
2
1
=
2
· (i  j  k)
= ~ + 1 ~
~ lk ~ 0 and
= ~  1 + ~ = 0, so a x
1/t
t2
t 2 ) k = (1  t) i
t2). (t, 1, 1/t) =

2
b is orthogonal to both a and b .
I I t21 t I·J + I t t2 I
+ (t3 
Since {a X b ). a = (1 t,O,t3
 11, J. + 111
1
2
1 1/t i
1t21
t) i  (t t) j
Since (a x b ) · b
11
k) · Gi + j + ~ k)
k
j
7. a x b
li
2
Now {ax b )· a= {~ i  j
(a x b ) · b
~
= ( 1  t, 0, t 3  t 2 )
• (
+ (t3 
t
t 2 , t 2 , 1) = t 2
1 k
t2) k
e +0 +t

2
t

t3 + 0 + t 3

= 0, a t2
X b is orthogonal to a.
= 0, a
x b is orthogonal to b .
9. According to the discussion precedin.g Theorem II , i x "j = k, so (i x j ) x k = k x k = 0 [by Example 2).
11. (j  k)
X
(k i) = (j  k)
X
k
+ (j 
= j X k + (  k)
X
k) .X (  i)
by Property 3 of Theorem I I
k + j X ( i) + ( k) X (  i)
by Property 4 of Theorem 11 2
= (j X k ) + (1){k X k) + ( 1)(j X i)+ (1) (k X i) = i + ( 1) 0 + (1)( k) + J = i + j + k
by Property 2 of Theorem I 1 by Example 2 and the discussion preceeding Theorem II
13. (a) Since b x cis a vector, the dot product a· (b x c) is meaningful and is a scalar. (b) b ·cis a scalar, so a x (b ·c) is meaningless, as the cross product is defined only for two vectors. (c) Since b x cis a vector, the cross 'product a x (b x c) is meaningful and results in another vector. (d) b · cis a scalar, so the dot product a · (b ·c) is meaningless, as the dot product is defined only for two vectors. (e) Since (a . b ) and (c . d) are both scalars, the cross product (a . b )
X (c
. d) is meaningless.
(f) a x band c x d arc both vectors, so the dot product (ax b )· (c x d ) is meaningful and is a scalar.
15. If we sketch u
an~
v starting from the same initial point, we see.that the
angle between them is 60°. Using Theorem 9, we have
lu x vi = lullvl sin 0 = {12)(16) sin 60° = 192 ·
v; =
96 v'3.
v
By the righthand rule, u x v is directed into the page. . © 2012 Ccng.age Learning. All Rights Reserved. May nol be sc.::umed. copit."ll, or duplicated, or polo1ctl too publicly accessible website. in whole or in part.
SECTION 12.4 THE CROSS PRODUCT
j k 17. axb =
2  1 3 = 4 2 1 j k
b xa=
4
2 1
2 1 3 Notice a x b
~~ :Ii  I~
: lj +
=1 ~ ~ li I: ~ lj +
I~
.0
1, k = ( 16) i  (2 12)j + [4 ( 4)] k =  7i+10j+8 k 2 .
21 k = [6 (  1)] . i {12  2)j + ( 4  4) k = 7 i  lOj 8 k 1: 1
=  b x a here, as we know is always true by Property I of Theorem II .
19. By Theorem 8, the cross product of two vectors is orthogonal to both vectors. So we calculate
3i
2j
k1 = 12
 1
1
0
(3,2,1)x( 1, 1, 0) =
1
11 i 
0
I13
11 · j + I 3
0
 1
21 k=  ij +5 k. 1
. I b h ± (1, 1,5) ±(1, 1,5) h . ( 1 1 5 ) So two umt vectors orthogona to ot are v' + + = ../3 , t at 1s,  M, M ~ 37:3 1 1 25 3
21 . Let a= (a1, a2, as). Then
O x a=
0
j
k
0
0
= Ia: :31i
a1 a2 as
j
k
axO = a1 a2 as =
 ~~
Ia~ I
as i _
0
0
0 lk= O,
:sl j + 1:1 a2
I a1 as, . + 0 0 J
I
a1 a21 0 O k = O.
0 0 0
25. ax (b+c) =a x (b1 +c1,b2 +c2,bs +c3)
= (a2(bs + cs) a3(b2 + c2), as(b1
= ((a2bs  asb2)
+ (a2cs 
+ c1) a1(b3 + cs) , a1(b2 + c2) a2(b1 + q))
asc2), (asb1 a1bs) + (asc1  a1c3), (a1b2  a2b1)
+ (a1c2 
= (a2bs asb2 , aab1  a1bs, a1b2 a2b1 ) + (a2cs  asc2, asc1 a1c3, a1c2  a2c1 ) = (a x b) + (a
X
125
c)
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a2c1))
126
0
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE y
27. By plotting the vertices, we can see that the parallelogram is determined by the +
vectors AB
+
= (2, 3) and AD= (4, 2}. We know that the area of the parallelogram
determ ined by two vectors is equal to the length of the cross product of these vectors. .
+
In order to compute the cross product, we consider the vector AB as the three+
dimensional vector (2, 3 , 0) (and similarly for AD), and then the area of parallelogram ABCD is j
IA13 x ml=
2 3 4  2
k
o =l(o) i (o) j+ (4  12)kl=l16kl = 16 0 +
+
29. (a) Because the plane through P, Q, and R contains the vectors PQ and P R, a .vector orthogonal to both of these vectors +
(such as their cross product) is also orthogonal to the plane. Here PQ +
+
= ( 3, 1, 2} and P R = (3, 2, 4}, so
+
PQ x PR = ((1)(4) (2)(2), (2)(3) (3)(4) , (3)(2) (1)(3)) = (0, 18,  9) Therefore, (0, 18,  9) (or any nonzero scalar multiple thereof, such as (0, 2,  1)) is orthogonal to the plane through P , Q, andR. (b) Note that the area of the tr~angle determined by P, Q, and R ls equal to half of the area of the parallelogram determined by the three points. From part (a), the area of the parallelogram is
I.PQ x t
31 . (a) PQ
PR I = I(0, 18, 9) I = y'O + 324 + 81 = v'4o5 = 9v'5, so the area of the triangle is ~ · 9v'5 = ~ .J5.
= (4, 3, 
+
+
2) and P R
= (5, 5, 1}, so a vector orthogonal to the plane through P, Q, and R is
+
PQ x P R = ((3)(1)  ( 2)(5), ( 2)(5)  (4)(1), (4)(5)  (3) (5)) = (13,  14, 5) [or any scalar mutiple thereof]. +
+
(b) The area of the parallelogram determined by PQ and PR is
IPQ x P'RI = 1(13,14, 5)1 = yl132 + ( 14)2 +52 = v'395,sotheareaoftriangle PQR is
~v'39Q.
33. By Equation 14, the volume of the parallelepiped detennined by a , b , and cis the magnitude of their scalar triple product, '
which is a · (b x c) .
=
1 2 3
1 1 2
= 1 111
21 4
 2
1 1 2
24 1 + 3
112 111 = 1(42)2( 44) + 3(12) = 9.
2 1 4 Thus the volume of the parallelepiped is 9 cubic units. t
+
+
35. a= PQ = (4, 2, 2}, b = PR = (3, 3, 1), and c = P S = (5,5, 1).
so the volume of the parallelepiped is 16 cubic units. ® 201 2 Cengage LC1tning. All Richts Rescrnd. M.:ty not be scnnncd. copied, or duplicntcd, or posccd to u publicly accessible website, in whole or in p:lrt.
SECTION 12.4 THE CROSS PRODUCT
1
D
127
5 2
1 0 3 0 3 0 =1 1 · 1 5 1 1+( 2) 1 ll = 4 + 60  64 = 0,whichsaysthatthevolume 9 4 5 4 5 9 . . 9  4
37. u · (v x w)= 31
5
of the parallelepiped determined by u , v and w is 0, and jhus these three vectors are coplanar.
39. The magnitude of the torque is Jr J = Jr x FJ = JrJJF JsinO = (0.18 m)(60 N) sin(70 +lOt = 10.8 sin80° r::::: 10.6 N·m.
41 . Using the notation of the text, r = {0, 0.3, 0) and F has direction (0, 3, 4). The angle(} between them can be determined by
cos 9 =
(0, 0.3, 0) . {0, 3, 4) . J(o, o.3, o)JI{o,3, 4)1
100 = 0.3IFisin53.1°
9
=> cos =
0.9 (0.3)(5)
=> cosO = 0.6 =>
(J
r::::: 53.1°. Then Jr l = lri iFI sin(} =>
=> 1F i r:::::417N.
43. From Theorem 9 we have Ia x bl = la l lb l sin 9, where 0 is the angle between a and b , and from Th,eorem 12.3.3 we have
Ja l lbl = a · be . Substituting the second equation into the first gives Ia x b l = a · b sin B, so cos cos (J
=>
a· b = lal lb l cosO
Ia X b l = tan9. Here Ia . a· b
X
b l = 1(1, 2, 2)1 = V1 + 4 + 4 = 3, so tan() = Ia x bbl = ~ = a· v3
J3 => () =
60° .
The distance between a point and a line is the length of the perpendicular
45. (a)
from the point to th~ line, here
d=
In I= IQP' sin
(J
In I=d. But referring to triangle PQS,
= lbl sin 9. But 8 is the angle between
QP =
+ Ia x bl and QR = ~·Thus ~y Theorem 9, sinO = lai Jb l
a
d an so d +
= lb1
.
'I sm 9
= Jbi Ja x bl = Ia x b J Jai Jbl laJ
+
(b) a= QR = ( 1,  2,  1) and b = QP = (1,  5,  7}. Then a x b = ((2)( 7)  ( 1){7 5), ( 1)(1) (  1)(  7) , ( 1)( 5) (  2)(1)} = (9,  8, 7). Thus the distance is d = Ia [:Ibl =
~ JB l + 64 + 49 =
ji¥ = j¥.
47. From Theorem 9 we have Ia x b l = [al[b[ sin 9 so
Ja x b] 2 = [a[ 2 [b[ 2 sin2 () = [a[ 2 lb[2 (1  cos2 B) 2
= fa i Jb[
2

2
2
2
([ai Jb [cos 9) = [a[ lb[

(a · b)
2
by Theorem 12.3.3.
©
201 2 Cengage l c'3ming. All Rights Reserved. May not be scanned, cor lcd, or duplicated, or j>ost<'
b
128
D
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
49. (a  b ) x (a + b) = (a  b) x a + (a  b ) x b
.
by Property 3 of Theorem 11
=ax a+ (b)x a +ax b+( b)x b
by Property 4 ofTheorem I I
=(a
by Property 2 ofTheorem II (with c
X
a) (b x a)+ (a
X
b ) (b x b)
= 0  (b x a ) + (a x b ) 0
by Example2
=(ax b)+ (a x b)
by Property I of Theorem ll
=: 1)
= 2(a x b ) 51. a
X
(b
X
c) + b
X
(c
X
a) + c
X
(a X b )
= [(a · c)b (a · b)c] + [(b · a)c (b · c)a] + [(c · b)a  (c: a )b]
by Exercise 50
= (a · c)b  (a· b )c+ (a · b)c (b · c)a+ (b · c)a (a · c)b = 0 53. (a) No. If a · b = a · c, then ·a · (b  c) = 0, so a is perpendicular to b  c, which can happen if b :f. c. For example, let a = (1, 1, 1), b = (1, 0, 0) and c = (0, 1, 0). (b) No. If a x b =ax c then a x (b c) = 0, which implies that a is para llel to b c; which of course can happen ifb :f. c. (c) Yes. Since a· c = a· b , a is perpendicular to b  c, by part (a). From part (b), a is also parallel to b  c. Thus since a :f. 0 but is both parallel and perpendicular to b  c, we have b  c = 0, so b = c.
12.5 Equations of Lines and Planes 1. (a) True; each of the first two lines has a direction vector parallel to the direction vector of the third line, so these vectors are each scalar multipl~s of the third direction vector. Then the first two direction vectors are also scalar multiples of each other, so these vectors, and hence the two lines, are parallel. (b) False; for example, the x and y axes are both perpendicular to the zaxis, yet the x andyaxes are not parallel. (c) True; each of the first two planes has a normal vector parallel to ~he normal vector of the third plane, so these two normal vectors are parallel to each other and the planes are parallel. (d) False; for example, the xy and yzplanes are not parallel, yet they are both perpendicular to the xzplane. (e) False; the x and yaxes are not paralle l, yet they are both parallel to the plane z
= 1.
(f) True; if each line is perpendicular to a plane, then the lines' direction vectors are both parallel to a normal vector for the plane. Thus, the direction vectors arc parallel to each other and the lines are paralle l. (g) False; the planes y
= 1 and z = 1 are not parallel, yet they are both parallel to the xaxis.
(h) True; if each plane is perpendicular to a line; then any normal vector for each plane is parallel to a direction vector for the line. Thus, the normal vectors are parallel to each other and the planes are parallel. (i) True; see Figure 9 and the accompanying discussion.
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SECTION 12.5 EQUATIONS OF LINES AND PlANES
0
129
(j) False; they can be skew, as in Example 3. (k) True. Consider any normal vector for the plane and any direction vector for the line. If the normal vector is perpendicular to the direction vector, the line and plane are parallel. Otherwise, the vectors meet at an angle B, 0° :S B < goo., and the line will intersect the plane at an angle goo B.
+ 2.4j + 3.5 k and v = 3 i + 2 j  k, so a vector equation is r = ro + t v = (2 i + 2.4 j + 3.5 k ) + t(3 i + 2j  k) = (2 + 3t) i + {2.4 + 2t) j + (3.5  t) k and parametric equations are X = 2 + 3t, y = 2.4 + 2t, Z = 3.5  t.
3. For this line, we have r 0 = 2 i
5. A Line perpendicular to the given plane has the same direction as a normal vector to the plane, such as n
= (1, 3, 1). So r 0 = i + 6 k , and we can take v
r = {i
2
2
y
Then a vector equation is
= (2 
0, 1  ~ .  3 1)
'
= 3t, z
= 6 + t.
= ( 2, ~ . 4) is parallel to the line. Letting Po = (2, 1, 3), parametric equations
. = 2 + 2t, y = 1 + 21 t , z = 3  4.t, wh'1le symmetnc· equat1ons are X  2 2
are x
9. v
+ 3j + k.
+ 6 k) + t(i + 3j + k ) = (1 + t) i + 3tj + (6 + t ) k , and parametric equations are x = 1 + t, y
7. The vector v
X 
= i
.
= y 1
112
Z +3 = =4 or
= 2y _ 2 = Z + 3.  4
= (3 ( 8) , 
=1
2 1,4 4)
= (11,  3,·0), and letting Po= ( 8, 1, 4), parametric equations are x
= 8 + llt,
. x + 8 :;: y _  1 , z = 4. Nottcc . here t hat t he d'trectton . number = 4 + 0t = 4, wI11·1e symmetnc. equations are u3
3t, z
z~
c = 0, so rather than writing 11. The line has direction v
4
in the symmetric equation we must write the equation z
= 4 separately.
= (1, 2, 1). Letting Po = (1,  1, 1), parametric equations are x = 1 + t, y = 1 + 2t, z = 1 + t
. equatiOns . an d symmetriC are x  1 =
y +2 1
= z  1.
13. Direction vectors of the lines are v 1 = ( 2  (  4), 0  (  6),  3  1) v 2 = (5  10,3  18, 14  4)
=
(2, 6,  4) and
= (5,  15, 10), and since v 2 = :~ v1 , the direction vectors and thus the lines are parallel.
15. (a) The li ne passes through the point (1,  5, 6) and a direction vector for the line is (  1, 2,  3), so symmetric equations for
.
the line are
x 1
y+5
=1 = 
2
z 6
= =3.
(b) The line intersects the xyplane when z = 0, so we need
Y
~ 5 = 2 =>
we need x
=0
y
=>
x 1 06 x 1  = =3 or =1' = =1' = y+5 2
= 1. Thus the point of intersection with the xyplane is (1
the x zplane, we need y
= y ~ 5 = z ~36 =0
=>
=> y
=
3, z
2
=>
x
=
1,
1,  1, 0). S imilarly for the y zplane,
= 3. Thus the line intersects the y zplane at {0,  3, 3). For
=>
x
=  ~, z =  ~ . So the line intersects the xzplane
([) 2012 Ccngagc Lc.1ming. All Rights Rcscr\'cd. MU)' not be scanned, copied, or duplicated, or posted to 3 publicly ncccss iblc website, in whole or in p,art.
130
D
\~~From
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE Equation 4, the line segment. from ro
~
= 4i+
2 i  j + .4 k to r 1
6 j + k is
r (t) = (1  t)ro +tr1 = (1  t)(2 i  j +4 k ) +t(4 i + 6j + k ) = (2 i  j +4k) +t(2 i + 7j 3k), 0 ::=; t ::=; 1.
l 19. Since the direction vectors (2, 1, 3} and (4,  2, 5} are not scalar multiples of each other, the Lfnes aren't parallel. For the '
\
\ 
'

lines to intersect, we must be able to find one value oft and one value of s that produce the same point from the respective
· 
..______:
~·"'
parametric equations. Thus we need to satisfy tlie following three equations: 3 + 2t = 1 + 48, 4 
...
t = 3  28,
1 + 3t = 4 + 5s. Solving the last two equations we get t = 1, 8 = 0 and checking, we see that these values don't satisfy the first equation. Thus the lines aren't parallel and don't intersect, so they must be skew lines.
r
·• 21. Since the direction vectors (1,  2,  3} and (1, 3, 7) aren't scalar multiples of each other, the lines aren't parallel. Parametric equations of the lines are L1: x = 2 + t, y = 3 2t, z = 1  3t and L2: x ·= 3 + 8, y lines to intersect, the three equations 2 + t
= 3 + 8, 3 
Solving the first two ~quations gives t = 2, 8
= 4 + 38, z =
2 78. Thus, for the
2t = 4 + 3s, and 1 3t = 2 78 must be satisfied simultaneously.
= 1 and checking,_we see that these values do satisfy the third equation, so the
lines intersect when t = 2 and 8 = 1, that is, at the point (4, 1, 5). 23. Since the plane is perpendicular to the vector (1, 2, 5), we can take (1,  2, 5} as a normal vector to the plane. (0, 0, 0) is a point on the plane, so setting a= 1, b = 2, c
= 5 and xo =
0, yo = 0, z 0 = 0 in Equation 7 gives
1(x  0) + (2)(y 0) + 5(z  0) = 0 or x 2y + 5z = 0 as an equation of the plane. 25. i + 4j + k xo
=
= (1, 4, 1} is a normal vector to the plane and ( 
1, yo = ~, zo
= 3 in Equation 7 gives 1 (x 
1, ~, 3) is a point on the plane, so setting a= 1, b = 4, c = 1,
( 1 )] + 4
~) + 1(z  3) = 0 or x + 4y + z = 4 as an equation of
(11 
the plane. 27. Since the two planes are parallel, they will have the same normal vectors. So we can taken the plane is5(x1) 1 (y ( 1)]1[z (1)]
= (5, 1, 
1), and an equation of
= Oor5x 11 z = 7.
29. Since the two planes are parallel, they will have the same normal vectors. So we can taken = (1, 1, 1), and an equation of the plane is 1(x 1) + 1 (y

~)
+ 1 (z 
~) = 0 or x
+y +z =
Jt or 6x + 6y + 6z = 11.
31. Here the vectors a= (1 0, 0  1, 1 1 1) = (1, 1, 0) and b = (1  0, 1 1, 0 1) = (1, 0,  1) lie in the plane, so ax b is a normal vector to the plane. Thus, we can taken= ax b
= (1 
(0, 1, 1), an equation of the plane is 1(x 0) + 1(y  1) + 1(z  1) 33. Here the vectors a = (8  3, 2  (  1), 4 2} = (5, 3, 2) and b
= (1, 1, 1}. If P o is the point
= 0 or x + y + z = 2.
= ( 1 
3;  2  ( 1) , 3 2} = (4,  i ,  5) lie in
= ( 15 + 2, 8 + 25,  5 + 12) = 2) = 0 or 13x + 17y + 7z = 42.
the plane, so a normal vector to the plane is n = a x b the plane is  13(x  3) + 17[y (1)) + 7(z 
0, 0 + 1, 0 + 1)
(13, 17, 7) and an equation of
35. If we first find two nonparallel vectors in the plane, their cross product wi ll be a normal vector to the plane. Since the given line lies in the plane, its direction vector a= (2, 5, 4) is one vector in the plane. We can verify that the given point (6, 0, 2)
©
2012 <:engage Learning. All Rights Rcscm:d. May not be scanned, copied, or duplicated, or posted on • publicly accessible website, in whole or in p.'lrt.
SECTION 12.5 EQUATIONS OF LINES AND PLANES
0
131
does not lie on this line, so to find another nonparallel vector b which lies in the plane, we can pick any point on the line and find a vector connecting the points. If we putt = 0, we see that ( 4 1 3 1 7) is ·on the line, so b = (6  4, 0 3, 2 7) = (2, 3, 9) and n = a x b = (45 + 12, 818,6 10) = ( 33,  10,  4). Thus, an
equation of the plane is 33{x  6) 10(y 0) 4[z  ( 2)] = 0 or 33x +lOy + 4z = 190. 37. A direction vector for the line of intersection is a= n1 x n 2 = {1, 1,  1) x (2, 1, 3) = (2 1  51  3), and a is parallel to the desired plane. Another vector parallel to the plane is the vector .connecting any point on the line of intersection to the given point (  1 1 2 1 1) in the plane. Setting x = 0, the equations of the planes reduce toy z = 2 and y+ 3z = 1 with simultaneous solution y = ~ and z = ~. So a point on the line is (0 1 ~ 1 ~) and another vector parallel to the plane is (  1 1 ~ , ~) .Then a normal vector to the plane is n = (2,  5 1 3) x ( 11 ~, ~ ) = ( 2,4,  8) and an equation of the plane is 2{x + 1) + 4(y 2)  B(z  1) = 0 or x  2y + 4z = :1. 39. If a plane is perpendicular to two other planes, its nonnal vector is perpendicular to the normal vectors of the other two planes. Thus (2, 1, 2) x (1, 0, 3) = (3 0, 2 6, 0  1) = (3, 8,  1) is a normal vector to the desired plane. The point (1, 5, 1) lies on the plane, so an equation is 3(x  1)  B(y  5) (z 1) = 0 or 3x  By z =  38. 41. To find the xintercept we set y and obtain 2x = 10 =>

=z =
0 in the equation 2x + 5y + z = 10
x = 5 so the xintercept is (5, 0, 0). When
x = z = 0 we get 5y = 10 => y = 2, so theyintercept is (0, 2, 0). Setting x = y = 0 gives z = 10, so the zintercept is (0, 0, 10) and we graph the portion of the plane that lies in the first octant.
X
43. Setting y = z = 0 in the equation 6x  3y + 4z = 6 gives 6x = 6 x = 1, when x = z = 0 we have 3y = 6
implies 4z = 6
=> z =
=>
=> y =  2, and x = y = 0
~.so the intercepts are (1, 0, 0), {0, 2, 0), and
(0 1 '0, ~). The figure shows the portion of the plane cut off by the coordinate planes. X I
@
ubstitute the parametric equations ofthe line into the equation of the plane: (3 t)  (2 + t) + 2{5t) = 9 =>
Bt = 8 => t = 1. Therefore, the point of intersection of the line and the plane is given by x = 3  1 = 2, y = 2 + 1 = 3, and z = 5(1) =5 1 that is, the point (2, 3, 5). 47. Parametric equationS for the line are x = t, y = 1 + t, z = ~t and substituting into the equation of the plane gives 4{t){1+t)+3(~t)=8
=> ~t=9 => t=2.Thusx = 2,y=1+2 .= 3,z=H2)=1andthepointof
intersection is (2, 3 1 1). ®
20 12 CenCJlgc Leaminc. All Rights Reserved. May no1 be scanned, wpl<
132
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
D
49. Setting x = 0, we see that (0, 1, 0) satisfies the equatio~s of both planes, so that they do in fact have a line of intersection. v
=
o 1
x
= (1, 1, 1} x
02
(1, 0 , 1)
= (1, 0, 1} is the direction of this line. Therefore, direction numbers of the intersecting
line are 1, 0,  1. 51. Normal vectors for the planes are 01
=
(1, 4, 3} and 0 2 .
But 01
· 0 2 ,;,
= (3, 6, 7), so the normals (and thus the planes) aren't parallel. .
3 + 24  21 = 0, so the normals (and thus the planes) are perpendicular.
= (1, 1, 1) and 02 = (1, 1, 1). The normals are not parallel, so neither are tl1e planes. 1 + 1 = 1 =F 0, so the planes aren't perpendicular. The angle between them is given by·
53. Normal vectors for the planes are 01 Furthe'rmore, o1 cos8
·
o2 = 1 
= lolllo21 O t . 0 2 = __ 1_ = ~ .j3.j3 3 =>
55. The normals are 0
1
O=cos
1(1)
3 ::::::70.5o.
= (1, 4, 2) and 0 2 = (2, 8, 4). Since 02 = 2o1 , the normals (and tlms the planes) are parallel.
57. (a) To find a point on the line of intersection, set one of the variables equal to a constant, say z
= 0. (This will fail if the line .o f
intersection does not cross the xyplane; in that case, try setting x or y equal to 0.) The equations of the two planes reduce to x
+ y = 1 and x + 2y = 1. Solving these two equations gives x = 1, y = 0. Thus a point on ilie line is (1, 0,·0).
A vector v in the direction of this intersecting line is perpendicular to the normal vectors of both planes, so we can take v = o1 x o2 line are .'1:
= (1, 1, 1)
x (1, 2, 2)
= (2 2, 1 
2, 2 1) = (0,  1, 1}. By Equations 2, parametric equations for tlle
= 1, y =  t, z = t.
. (b) The angle between the planes satisfies cos 0
= I0] II. 0 2 I = 1 +r;;2 +If\2 = 5r;;. T here fiore 0 = cos 1 ( 0] 02 v 3 v 9. 3v3
5r;; ) 3v3
::::::
15 . 80.
= 0, tlle equations of the two planes become 5x 2y = 1 and 4x + y = 6. Solving tllese two equations gives x = 1, y = 2 so a point on the line of intersection is (1, 2, 0). A vector v in the direction of this intersecting line is perpendicular to the normal vectors ofbotll planes. So we can use v = o 1 x 0 2 = (5,  2, 2) x (4, 1, 1} = (0,  13, 13} or
59. Setting z
equivalently we can take v
=
(0, 1, 1}, and symmetric equations for the line are x
61. The distance from a point (x , y , z) to (1, 0,  2) is d 1 (3 ,4 , 0) is d2 = ..j(x 3) 2
= 1, y _=12 =
I
or x
=
1, y
2= z.
= ..j(x 1)2 + y 2 + (z + 2)2 and the distance from (x, y, z) to
+ (y 4) 2 + z 2 . The plane consists of all points (x, y, z) where d1 =
+ (z + 2) 2 = (x  3) 2 + (y  4) 2 + z 2 <* x 2  2x + y 2 + z 2 + 4z + 5 = x 2 : 6x + y 2  By + z 2 + 25 4x + By + 4z = 20 or equivalently x + 2y + z = 5.
d2
=> d{ = di
<*
(x 1)2 + y 2
<*
4:1;
+ By + 4z = 20 so an equation for the plane is
Alternatively, you can argue that the segment joining points (1, 0, 2) and (3, 4, 0) is perpendicular to tlle plane and the plane includes the midpoint of the segment. 63. The plane contains tlle points (a, 0, 0), (0, b, 0) and (0, 0, c) . Thus the vectors a = (a, b, 0} and b
=
( a, 0, c) lie in tlle
plane, and o = a x b = (be 0, 0 + ac, 0 + ab) = (be, ae, ab) is a normal vector to the plane. The equation of the plane is (!) 2012 Ccngoge Leoming. All Ril:)>IS Reserved. Moy not be sconned, copkd. or duplicated, or posted to • publicly accessible website, in whole or in port.
SECTION 12.5 EQUATIONS OF LINES AND PLANES
there fore bcx + aetJ + abz = abc + 0 + 0 or bcx + acy + abz
0
133
= abc. Notice that if a ol 0, b ol 0 and c ol 0 then we can
' . . to remem ber.I rewnte the equat1on as X + '1j + Z = 1. Th'ts 1.s a goo d equat1on a b c
65. Two vectors which are perpendicular to the·required line are the normal of the given plane, (1, 1, 1), and a direction vector for the given line, (1,  1, 2). So a direction vector for the required line is (1, 1, 1} x (1, 1, 2} = (3,  1, 2}. Thus L is given by (x, y, z} = (0, 1, 2} + t(3, 1 ,  2}, or in parametric form, x
67. Let Pi have normal vector D i. Then n1
= (3, 6, 3}, n 2 =
= 3t, y = 1 t , z = 2 2t.
(4, 12, 8), n a
=
=
(3, 9, 6), ll4
(1, 2,  1}. Now n 1
= 3D4,
so n 1 and n4 are parallel, and hence P1 and P4 are parallel; similarly P 2 and P3 are parallel because n 2 = ! n 3. However, n 1 and n 2 are not parallel (so not all four planes are parallel). Notice that the point (2, 0, 0) lies on both P 1 and P 4 , so these two planes are identical. The point ( ~ , 0, 0) lies on P2 but not on P3, so these are different planes.
= (2, 1, 1),. points on the line corresponding to t = 0 and t = 1. .Let
69. Let Q = (1, 3, 4) and R
bl = /al
d = Ia x
.
tl
+
>
= QR = (1, 
P = (4, 1,  2). Then a
2, 3), b = QP = (3,  2,  6) . The distance is
1(1, 2,3} x (3,2,6}/ = 1(6,  3,4)1 = J62 + (3)2 + 42 = 1(1, 2, 3}/ 1(1,  2,  3}1 JP+( 2)2 + (3)2
71. BE y quat1on 9, 1e
d'ta . D iax1+by1 +cz1 +d/ IS nee IS = = /3(1) +2(  2)+6(4)5/ Ja2+b2+ c2 . J 32 + 22 + 62
=
J6I = {61
v'I4 /181 r.n v49
V14' = 18 . 7
73. Put y = z = 0 in the equation of the first plane to get the point (2, 0, 0) on the plane. Because the planes are parallel, the distance D between them is the distance from (2, 0, 0) to the second plane. By Equation 9,
D = /4(2)  6(0) + 2(0)  31 = _ 5_ = _5_ or 5 VT4. J42 + (  6)2 + (2)2 V56 2 VI4 28 75. The distance between two parallel planes is the same as the distance between a point on on~ ofilie planes and the oilier plane. Let Po = (xo, yo, zo) be a point on the plane given by ax+ by + cz + dt distance between Po and ilie plane given by ax + by + cz + d2 D = /ax o + byo
+ czo +
d2/
=
/dl + d2 / Ja2 + b2+ c2
Ja2 +b2+c2 77. £ 1: x
x= y
=y=z =
=?
x
=y
(1).
2. However, when x
£2: x + 1
= 2, x
lines do not intersect. For L1. v1
= z
,
byo + czo + d 1
= 0 and the
= 0 is, from Equation 9,
/d1  d2 /
Ja2+ b2+c2
= y/2 = =?
= 0. Then axo +
z
z/3
=
=?
x + 1
= y/2
2, but x + 1 = z/3
(2). =?
The solution of (l ) and (2) is
z =  3, a contradiction. Hence the
= (1, 1, 1), and for L2, v2 = (1, 2, 3}, so the lines are not parallel. Thus ilie lines are skew
lines. If two lines are skew, they can be viewed as ly ing in two parallel planes and so ilie distance between the skew lines would be the same as the distance between these parallel planes. The common normal vector to the planes must be perpendicular to boili (1, 1, 1) and (1, 2, 3), the direction vectors of the two lines. So set ®
2~ 12 Ccogogc
Lc;uning.. All Rights R.esen·cd. Mny not be SC4Med, copied, or duplicated, or posted to a publicly accessible website, in whole or in pan.
134
D
n
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
= {1, 1, 1) x
{1, 2, 3) = (3 2,  3 + 1, 2 1) = {1,  2, 1). From above, we know that (  2, 2,  2) and ( 2,  2, 3)
are points of £1 and~ respectively. So in the notation of Equation 8, 1{ 2) 2(  2) + 1( 2) + d 1 = 0
1( 2)  2( 2) + 1(3) + d2 = 0
~
~
d 1 = 0 and
d2 = 1.
By Exerci~e 75, the distance between these two skew lines is D =
.
IO  II
v 1 +4 + 1
Alternate solution (without reference to planes): A vector which is perpendicular to both of the lines is
n
= (1, 1,1) x
(1, 2, 3)
= (1,  2, 1). Pick any point on each of the lines, say (  2, 2, 2) and ( 2,  2,  3), and form the
vector b = (0, 0, 1) connecting the two points. The distance between the two skew lines is the absolute value of the 'sca,lar
. .
..
proJection of b along n , that IS, D .
= ln·bl = 11 · 02·0+1·11 =  
v 1 +4+1
n1
1
1 u;· v6
79. A direction vector for £1 is v1 = (2, 0, 1) and a direction vector for £ 2 is v 2 = (3, 2, 2). These vectors are not parallel so neither are the lines. Parametric equations for the lines are £1 : x = 2t, y = 0, z
=
t, and £2: x
= 1 + 3s, y =  1 + 2s,
z = 1 + 2s. No values oft and s satisfy ·these equations si multaneously, so the lines don't intersect and hence are skew. .we can view the lines as lying in two parallel planes; a common normal vector to the planes is n
= v 1 X v2 =
(2,  7, 4). Line
£ 1 passes through the origin, so (0, 0, 0) lies on one of the planes, and.(1,  1, 1) is a point on £2 and therefore on the other
plane. Equations of the planes then are 2x 7y between the two skew lines is D
=
+ 4z = 0 and 2x 7y + 4z 13 =
I0(13)1 v4+49 +16
=
0, and by Exercise 75, the distance
13 rcn. v69
Alternate solution (without reference to planes): Direction vectors of the two lines are v1
Then n
=v1 x v2 =
= (2, 0 , 1) and v 2 = (3, 2, 2).
(2,  7, 4} is perpendicular to both lines. Pick any point on each of the lines, say (0, 0, 0) and (1, 1, 1),
and form the vector b = (1, 1, 1) connecting the two points. Then the distance between the two skew lines is the absolute value of the scalar projection of b a long n , that is, D
81 . If a =F 0, then ax + by+ cz + d = 0 =?
3 = ln · bl = 12 + 7 + 41 = ~ . 1D 1 V4 + 49+16 YO::!
69
a(x + dj a) + b(y  0) + c(z  0) = 0 which by (7) is the scalar equation of the
plane through the point ( d/ a, 0, 0) with normal vector (a , b, c). Similarly, if b =F 0 (or if c =F 0) the equation of the plane can be rewritten as a(x  0) + b(y + djb) + c(z  0)
= 0 [or as a(x 0) + b(y 0) + c(z +
djc) = 0] which by (7) is the
scalar equation of a plane through the point (0,  d/b, 0) [or the point (0, 0, d/c)] with normal vector (a, b, c).
SECTION 12.6
12.6
CYLINDERS AND QUADRIC SURFACES
0
Cylinders and Quadric Surfaces
1. (a) In JR 2 , the equation y = x 2 represents a parabola.
X
(b) In R 3 , the equation y
z
= x 2 doesn't involve z, so any
horizontal plane with equation z = k intersects the graph in a curve with equation y
= x 2 • Thus, th~ surface is a
parabolic cylinder, made up of infinitely many shifted X
copies of the same parabola. The rulings are parallel to the zaxis.
(c) In R 3 , the equation z
= y 2 also represents a parabolic
z
cylinder. Since x doesn't appear, the graph is formed by moving the parabola z
= y 2 in the direction of the xaxis.
Thus, the rulings of the cylinder are parallel to the xax.is.
3. Since y is missing from the equation, the vertical traces
5. Since x is missing, each vertical trace z = 1  y 2 ,
= k, is a copy of the same parabola in the plane = k. Thus the surface z = 1  y 2 is a parabolic
x2
+ z 2 = 1, y = k, are copies of the same circle in the plane y = k. Thus the surface x 2 + z 2 = 1 is a
x x
circular cylinder with rulings parallel to theyaxis.
cylinder with rulings parallel to the xaxis.
.1'
© 2012 Cengage Le.1ming. /\II Rights Reserved. Mo.y not be scanned. copied, or duplicated, or pOSted too publicly n.cccssiblc website, in whole or in pa{t.
135
136
D
CHA~TER 12
VECTORS AND THE GEOMETRY OF SPACE
7. Since z is missing, each horizontal trace xy = 1,
= k, is a copy of the same hyperbola in the plane z = k. Thus the surface xy = 1 is a hype~bolic z
cylinder with rulings parallel to the zaxis.
9. (a) The traces ofx2
+ y2 
= k are y 2 
z 2 = 1 in x
oriented differently for 1
< k < 1 than for k < 
family of hyperbolas. The traces in z = k are x 2 xyplane, the circle is of radius 1. As
z 2 = 1  k 2 , a family of hyperbolas. (Note that the hyp!!rbolas are 1 or k
>
1.) The traces in y = k are x 2

+ y 2 = 1 + k 2 , a family of circles. Fork =
z 2 = 1  k 2 , a similar 0, the trace in the
Iki increases, so does the radius of the circle. This behavior, combined with the
hyperbolic vertical traces, gives the graph of the hyperboloid of one sheet in Table 1.
(b) The shape of the surface is unchanged, but the hyperboloid is
rotated so that its axis is the yaxis. Traces in y
= k are circles,
while traces in x =. k and z = k are hyperbolas.
(c) Completing the square in y gives x 2
+ (y + 1)2 
z2
= 1. The
surface is a hyperboloid identical to the one in part (a) but shifted one unit in the negative ydirection.
11. For x = y 2 + 4z2 , the traces in x = k are y 2 + 4z 2 = k. When k >. 0 we have a fam ily of ellipses. When k the trace is empty for k
= 0 we have just a point at the origin, and
< 0. The traces in y = k are x = 4z 2 + k 2 , a
family of parabolas opening in the positive x direction. Similarly, the traces in z
= k are x = y 2 + 4k2 , a family of parabolas opening in the positive
xdirection. We recognize the graph as an elliptic parabofoid with axis the xaxis and vertex the origin.
© 2012 Ccngngc Learning. Al l Rights Reserved. May not be scanned. copied, or duplicated. or posted ton publicly accessible website, in whole or in part.
SECTION 12.6 13. x 2 = y 2
+ 4z2 . The traces in x
traces in y = k are x 2 intersecting lines if k x2
= k are the ellipses y 2
+ 4z2
CYLINDERS AND QUADRIC SURFACES
0
137
= k 2 . The
4z2 = k 2 , hyperbolas for k =/; 0 and two

= 0. Similarly, the traces in 'z =
k are
y 2 = 4k2 , hyperbolas for k =/; 0 and two intersecting lines if k = 0.

We recognize the graph as an elliptic cone with axis the xaxis and vertex the origin.
+ 4y2 
15.  x 2 4y 2

z2
z2
= 4. The traces in x =
k are the hyperbolas
= 4 + k2 . The traces in y = k are x 2 + z 2 = 4k2 
circles for
lkl > 1, and the traces in z ·= k are 4y 2 
4, a family of
x 2 = 4 + k 2 , a family
of hyperbolas. Thus the surface is a hyperboloid of two sheets with axis the y axis.
17. 36x 2
+ y 2 + 36z2 = 36. The traces in x = k are y 2 + 36z2 = 36(1 
a family of ellipses for and are empty for 36x 2
lkl < 1. (The traces are a single point for lkl = 1
lkl > 1.) The traces in y = k'are the circles
+ 36z2 = 36 k 2
traces in z
x2
{"}
+ z2 = 1 
tok 2 ,
= k are the ellipses 36x 2 + y 2 = 36(1 
lkl < 6, and the
k 2 ), 1~1
graph is an ellipsoid centered at the origin with intercepts x
z
k 2 ),
< 1. The
= ± 1, y = ± 6,
= ±1.
19. y = z 2  x 2 • The traces in x
the traces in y
= k are the parabolas y
= k are k = z 2 
are oriented differently fork the parabolas y
= k 2 .
= z2
:
k2 ;
x 2 , which are hyperbolas (note the hyperbolas
> 0 than fork < 0); and the traces in z = k are
x 2 • Thus,
y
l =
21. This is the equation of an ellipsoid: x
2
z2
12
x2
 ]2 is a hyperbolic paraboloid. X
2
2
+ 4y2 + Dz2 = x 2 + ~ + ___!___, = (1/2)
(1/3t
1, with xintercepts ±1, yintercepts ±i 0

and zintercepts ±~. So the major axis is the xaxis and the only possible graph is Vll. 23. This is the equation of a hyperboloid of one sheet, with a
= b = c = 1.
Since the coefficient of y 2 is negative, the axis of the
hyperboloid is the yaxis, hence the correct graph is II.
® 20 12 Ccogage Learning. All Rights R~ned. May not be scanned, copied, or duplicated, or pos!cd to o publicly 3Cccssihlc website. in whole: or in p:u1.
138
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
D
25. There are no real values of x and z that satisfy this equation for y
< 0, so this surface does not extend to the left of the .
xzplane. The surface intersects the plane_y
.
= k > 0 in an ellipse. Notice that y occurs to the first power whereas x and z
occur to the second power. So the surface is an elliptic paraboloid with axis theyaxis. Its graph is VI.
27. This surface is a cylinder because the variable y is missing from the equation.' The intersection of the surface and the xzplane is an ellipse. So the graph is VIII .
= x 2 + ~z
29. y 2
2
or y 2 = x 2
. z2
+g
.
•
represents an elliptic
cone with vertex (0, 0, 0) and axis the yaxis.
31. x 2
+ 2y 
2z 2
= 0 or 2y = 2z2 
x 2 or y
2
= z2  ~
represents a hyperbolic paraboloid with center (0, 0, 0).
z
35. Completing squares in all three variables gives
33. Completing squares in y and z gives
4x
2
+ (y 
2)
2
x z + (y ~ ) 2
2
+ 4(z + (z 
(x  2)2  (y + 1? + (z 1)2 = 0 or
2
3) = 4 or
3) 2 = 1, an ellipsoid with
(y + 1? = (x  2) 2
+ (z 1?, a circular cone with
center (2, 1, 1) and axis the horizontal line x
center (0, 2, 3).
z =l. z
y
37. Solving the equation for z we get z. = ±)1 + 4x2
z
+ y 2 , so we plot separately z = ) 1 + 4x2 + y 2 and
=  ) 1 + 4x2 + y2 .
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= 2,
CYLINDERS AND QUADRIC SURFACES
SECTION 12.6
0
139
To restrict the zrange as in the second graph, we can use the option view=  4 . . 4 in Maple's p l"ot3d command, or Plo.t Range
 > { 4 , 4} in Mathematica's Pl ot 3D command.
39. Solving the equation for z we get z
z
41.
= ±y'4x2 + y 2 , so we plot separate ly z = y'4x2 + y2 and z = 
z=2
y'4x2
+ y2.
43. The surface is a paraboloid of revolution (circular paraboloid) with vertex at the origin, axis the yaxis and opens to the right. Thus the trace in the yzplane is a lso a parabola: y = z 2, x
= 0. The equation is y =
x2
+ z 2.
The parabola y=x2
X
= (x, y, z) be an arbitrary point equidist ant fro m ( 1, 0, 0) and the plane x = 1. Then the distance from P to
45. Let P
( 1, 0, 0) is y'(x + 1) 2 + y 2 + z 2 and the distance from P to the plane x = 1 is lx 11/Vf'i (by Equation 12.5.9). So lx 11 = y'(x + 1)2 + y 2 + z 2
x 2  2x + 1 = x 2
+ 2x + 1 +. y 2 + z 2
¢:>
 4x
¢::>
= lx 
(x 1) 2 = (x + 1) 2 + v. 2 + z 2
11
¢:>
= y 2 + z 2 • Thus the collection of all such points P
is a circular
paraboloid with vertex at the origin, axis the x axis, which opens in the negative direction.
47. (a) An equation for an ellipsoid centered at the origin with intercepts x = ±a, y
= ±b, and z =
2
±cis x 2 a
2
2
+ Y2 + .=.._2 = b
c
Here the poles of the model intersect the zaxis at z = ± 6356.523 and the 'e quator intersects the x andyaxes at
x
= ±6378.137, y = ±6378. 137, so an equation is x2 (6378.137) 2 •
•
(b) Traces to z
2
x +y
2
®
=k
~
~ I
are the Circles (6378.137)2
= (6378.137)2
y2
z2
.
+ (6378. 137) 2 + (6356.523)2 = 1
+ (6378.137)2 = 1 
~
(6356.523)2
2
(6378.137) k2 ~6.~ .
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1.
1
140
D
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
= m x into the equation of the ellipsoid:
(c) To identify the traces in y = mx we substitute y 2
x (6378.137) 2
. z2
(mx?
+ (6378.137)2 + (6356.523) 2 = 1 (1 + m 2 )x2 z2 1 (6378.137) 2 + (6356.523)2 = x2
z2
(6378.137)2/(1 + m2)
+
(6356.523)2 =
1
As expected, this is a family of ellipses. 49. If (a, b, c) satisfies z
= y2 
x 2 , then c
£2: x =a+ t, y = b t, z = c 2(b
= b2 
a2 . L 1 : x
= a + t, y = b +
.
c
+ 2(b  a)t = (b + t) 2
L 1 lies on z
= y2 

(a+ t) 2 = b2


(a+ t?
= b2

= y2 
a 2 + 2(b a)t => c = b2
2 x . Performing similar operations with £2 gives: z
c 2(b + a)t = (b t) 2
= c + 2( b 
a )t,
+ a)t. Substitute the parametric equations of £1 into the equation
of the hyperbolic paraboloid in order to find the points of intersection: z I
t, z
a2

2(b + a)t

=>
a2 . As this is true for·all values of t,
= y2 
=> c = b2
x2
'
x
2
=>
a 2 . This tells us that all of £2 also lies on

The curve of intersection looks like a bent ellipse. The projection
51.
of this curve onto the xyplane is the set of points (x, y, 0) which satisfy x x2
12
+
2
+ y2 = 1 y2
{1 //2)2
y
2
<=>
x
2
+ 2y 2 = 1
<=>
= 1. This is an equation of an ellipse.
Review
CONCEPTCH ECK 1. A scalar is a real number, while a vector is a quantity that has both a realvalued magnitude and a direction. 2. To add two vectors geometrically, we can use either.the Triangle Law or the Parallelogram Law, as illustrated in Figures 3 and 4 in ·section 12.2. Algebraically, we add the corresponding components of the vectors. 3. For c
> 0, c a is a vector with. the same direction as a and length c times the length of a. If c < 0, ca points in the opposite
direction as a and has length JcJ times the length of a. (See Figures 7 and IS in Section 12.2.) Algebraically, to find c a we multiply each component of a by c. 4. See (I) in Section 12.2. 5. See Theorem 12.3.3 and Definition 12.3.1. © 2012 Ccn~a.gc Lcnrnin~. All Rights Rcscr'\'ctl. Muy not be scamtcd: copied , or duplicated, or Jl(r.itcd ton publicly i1Ccess ihlc website.!, in whole or In p.1rt.
CHAPTER 12 REVIEW
0
141
6. The dot product can be used to find the angle between two vectors and the scalar projection of one vector onto another. In particular, the dot product can determine if two·vectors are orthogonal. A lso, the dot product can be used to determine the work done moving an object given the force and di~placement vectors.
7. See the boxed equations as well as Figures 4 and 5 and the accompanying discussion on page 828 [ET 804].
B. See Theorem 12.4.9 and the preceding discussion; use either (4) or (7) in Section 12.4. 9. The cross product can be used to create a vector orthogonal to two given vectors as well as to determine if two vectors are parallel. The cross product can also be used to find the area of a parallelogram determined by two vectors. In addition, the cross product can be used to determine torque if the force and position vectors are known.
•
10. (a) The area of the parallelogram determined by a and b is the length of the cross product:
Ia x
bl.
(b) The volume of the parallelepiped determined by a, b, and cis the magnitude of their sca lar triple product: 11. If an equation of the plane is known, it can be written as ax
+by + cz + d =
Ia . (b
x c )l.
0. A normal vector, which is' perpendicular to the
plane, is (a , b, c) (or any scalar multiple of (a , b, c)). If an equation is not known, we can use points on the plane to find two nonparallel vectors which lie in the plane. The cross,product of these vectors is a vector perpendicular to the plane. 12. The angle between two intersecting planes is defined as the acute angle between their normal vectors. We can find this angle
using Corollary 12.3.6. 13. See (1), (2), and (3) in Section 12.5. 14. See (5), (6), and (7) in Section 12.5. 15. (a) Two (nonzero) vectors are parallel if and only if one is a scalar multiple of the other. In addition, two nonzero vectors are
parallel if and only if their cross product is 0. ( b) Two vectors are perpendicular if and only if their dot product is 0. (c) Two planes are parallel if and only if their normal vectors are parallel. +
+
16. (a) Determine the vectors PQ = (a1 , a2 , a3 ) and P R = (b1, ~ . b3) . If there is a.scalar t such that . (a 1 , a 2, a 3) = t (b1, ~ . b3 ), then the vectors are parallel and the points must all lie on the same line. ~
+
+
+
Alternatively, if PQ x PR = 0, then PQ and PR are paralle l, ~o P, Q, and Rare collinear. Thirdly, an algebraic method is to determine an equation of the line joining two o.f the points, and then check whether or not the third point satisfies this equation. +
+
+
(b) Find the vectors PQ =a, P R = b , PS = c . a x b is normal to the plane formed by P, Q and R, and so S lies on this plane if a x band care orthogonal, that is, if(a x b)· c = 0. (Or use the reasoning in Example 5 in Section 12.4.) A lternatively, find an equation for the plane determined by three of the points and check whether or not the fourth point satisfies this equation.
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142
D
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
17. (a) See Exercise 12.4.45.
(b) See Example 8 in Section 12.5. (c) See Example 10 in Section 12.5. 18. The traces of a surface are the curves of intersection of the surface with planes parallel to the coordinate planes. We can find the trace in the plane x = k (parallel to the yzplane) by setting x = k and determining the curve represented by the resulting equation. Traces in the planes y = k (parallel to the xzplane) and z = k {parallel to the xyplane) are found similarly.
19. See Table I in Section 12.6.
TRUEFALSE QUIZ 1. This is false, as the dot product of two vectors is a scalar, not a vector. 3. False. For example, ifu = i and ·v = j then iu · v i= 101 = 0 but iu l lvl = 1 · 1 = 1. In fact, by Theorem 12.3.3, lu · v i = liu llvl cos 91 . 5. True, by Theorem 12.3.2, property 2. 7. True. If 9 is the angl~ between u and v , then by Theorem 12.4.9,lu x vi = iu l lv l sin 9 = lvllul sin 9 = 1':' x ui. (Or, by Theorem 12.4.1l,lu x v i = 1v x ul = 1 lllv x ul = lv x ui.) 9. Theorem 12.4.11 , property 2 tells us that this is true. 11. This is true by Theorem 12.4.11, property 5. 13. This is true because u x vis orthogonal to u (see Theorem 12.4.8), and the dot product of two orthogonal vectors is 0. 15. This is false. A normal vector to the plane is n = (6,  2, 4) . Because (3,  1, 2) = ~n, the vector is parallel ton and hence perpendicular to the plane. 17. This is fa lse. In IR2 , x 2
+ y2
= 1 represents a circle, but { (x, y, z) I x 2 + y 2 = 1} represents a threedimensional surface,
namely, a circular cylinder with axis the z axis.
19. False. For example, i · j = 0 but i
i 0 and j =f. 0 .
21. This is true. Ifu and v are both nonzero, then by (7) in s~ c tion 12.3, u · v = 0 implies that u and v are orthogonal. But u x v = 0 implies that u and v are parallel (see Corollary 12.4.10). Two nonzero vectors can't be both parallel and orthogonal, so at least one of u, v must be 0 .
®
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CHAPTER 12 REVIEW
0
1~
EXERCISES 1. (a) The radius of the sphere is the distance between the points (  1, 2, 1) and {6, 2, 3), namely, J [6  ( 1)]2 + (  2  2) 2 + (3  1) 2 =
0f!}.
By the formula for an equation of a sphere (see page 813 [ET 789]),
an equation of the sphere with center ( 1, 2, 1) and radius
J69 is (x + 1? +
(y  2) 2
+ (z 1) 2
= 69.
(b) The intersection of this sphere with the yz plane is the set of points on the sphere whose xcoord inate is 0. Putting x = 0 into the equation, we have (y  2) 2 + (z  1) 2 = 68, x = 0 which represents a circle in the yzplane with center (0 , 2, 1) and radius
.J68.
(c) Completing squares gives (x  4) 2 + (y + 1) 2 + (z +
3? =  1 + 16 + 1 +9 =
25. Thus the sphere is centered at
(4, 1, 3) and has radius 5. 3. u . v = Ju J.JvJ c~s45° = (2)(3)
'{! =
3 .;2.
Ju x vJ = JuJJv Jsin45° =
(2)(3)~ = 3 .;2.
By the righthand rule, u x vis directed out of the page.
5. Forthetwo vectors to be orthogonal, we need (3, 2,x} · (2x, 4,x} = 0 ~ + 6x + 8 = 0
<=?
(x + 2)(x + 4) = 0
<=?
#
(3)(2x) + (2){4) + (x )(x) = 0
#
x =  2 or x = 4.
w) = 2
7. (a) (u x v ) · w = u · (v X
(b) u · (w x v ) = u · [ (v
X
w )] =  u · (v
X
w) =  2
(c) v · (u x w) = (v x u )· w = (u x v) · w =  2 (d) (u x v ) · v = u · (v
X
v)
=u ·0 =0
9. For simplicity, consider a unit cube positioned with its back left corner at the origin. Vector representations of the diagonals
joining the points (0, 0, 0) to (1, 1, 1) and (1, 0, 0) to (0, 1, 1) are (1 ,' 1, 1} and (1, 1, 1). Let 8 be the angle between these twovectors.
8 = cos 1 +
=> cosO = ~ =>
(1, 1, 1)·{ 1, 1, 1} = 1+1 + 1 = 1=\(1, 1, 1) \ \{ 1,1,1)\ cos8=3cos8
G) ~ 71°. +
11 . AB = (1, 0, 1}, AC = (0, 4, 3), so +
+
(a) a vector perpendicular to the plane is AB x AC = (0 + 4, (3 + 0) , 4  0} = {4 ,  3, 4}. (b)
t lAB X AC I= h116 + 9 + 16 = 4J..
13. Let H be the magnitude of the force directed 20° away from the direction of shore, and let H be the magnitude qfthe other force. Separating these forces into components parallel to the direction of the resultant force and perpendicular to it gives H cos 20° + Fz cos 30° = 255 (1), and F1 sin 20° into (1) gives F2 (sin30° cot20° + cos 30°) = 255
F2 sin 30° = 0
=> H
~
=> A = F2 s~n
°
3 0 (2). Substituting (2) s1n 2 0 °
114 N. Substituting this into (2 ~ gives F 1 ~ 166 N. I
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144
0
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
15. The line has direction v = ( 3, 2, 3). Letting Po X
=4 
3t, y
= 1 +
2t,
Z
= 2+
= (4,  1, 2), parametric equation~ are
3t.
17. A direction vector for the line is a normal vector for the plane, n = (2, 1, 5), and parametric equations for the line are .
X
= 2 + 2t, y = 2 
t,
Z
= 4 + 5t.
19. Here the vectors a= (4 3,0 (1) ,2  1) so n
=a
= (1,1,1} and b = (6 
'
3,3 (1), 1  1) = {3,4, 0) lie in the plane,
x b = ( 4, 3, 1) is a normal vector to the plane and an equation of the plane is
 4(x 3) +3.(y  ( 1)) + 1(z 1) = 0 or  4x+3y +z
=
14.
21 . Substitution of the parametric equations into the equation of the plane gives 2x y .
 t
+ 3 = 2 =>
. t = 1. When t = 1, the parametric equations give x = 2 
+z =
2(2 t)  (1 + 3t) + 4t
1 = 1, y = 1 + 3
= 4 and z =
=2
=?
4. Therefore,
the point of intersection is (1,4,4). 23. Since the direction vectors (2, 3, 4) and (6, ;1, 2) aren't parallel, neither are the lines. For the lines to intersect, the three equations 1
+ 2t =
equations gives t =
1 + 6s, 2 + 3t = 3  s, 3 + 4t = 5 + 2s must be satisfied simultaneously. Solving the first two
k. s =
~ and checking we see these values don't satisfy the third equation. Thus the lines aren't parallel
and they don't intersect, so they must be skew. 25. n 1 = (1, 0, 1) and n2 = (0, 1, 2). Setting z = 0, it is easy to see that (1, 3, 0) is a point on ,the line of intersection of
= 1 andy+ 2z = 3. The direction of this line is v 1 = n 1 x n 2 = (1, 2, 1). A second vector parallel to the desired plane is v 2 = (1,1, 2), since it is perpendicular to x + y 2z = 1. Therefore, the normal of the plane in question is
x z
n = v 1 x v 2 = (4  1, 1 + 2, 1 + 2) = 3 (1, 1, 1) . Taking (x0 , y 0 , z 0 ) = (1, 3, 0), the equation we are looking for is
(x 1) + (y 3) + z = 0 . 27. By Exercise 12.5.75, D 
29. The equation x
¢*
x
+y +z =
4.
l2 (24)1  22 .  ~· + I2 + (4)2 v26
y'32
= z represents a plane perpendicular to
the xzplane and intersecting the xzplane in the line
31. The equation x 2 = y 2 + 4z2 represents a (right elliptical) cone with vertex at the origin and axis the x axis.
x = z,y = 0.
X X
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CHAPTER 12 REVIEW .
.
33. An equivalent equation
.
 x2
IS
y2 +4 ; z 2 = 1, a
0
145
35. Completing the squate in y gives
+ 4(y 
1) 2
+ z 2 = 4 or x 2 + (y 
hyperboloid of two sheets with axis the yax.is. For
4x 2
IYI > 2, traces parallel to the xzplane are circles.
an ellipsoid centered at (0, 1, 0).
1) 2
z2
+4 =.1,
z (0, 1,2)
(O, l , 2)
. x2
<=>
y2
4 + 16
2
= l. The equation of the ellipsoid is
~
+
i
2
6
2
+ :2
= l, since the horizontal trace in the
plane z = 0 must be the original ellipse. The traces of the ellipsoid in the yz plane must be circles since the surface is obtained 2
by rotation about the xaxis. Therefore, c2 4x2
+ y 2 + z2 =
= 16 and the equation of the ellipsoid is ~ +
i
2
6
2
+ ~ =1
6
<=>
16.
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D
PROBLEMS PLUS
1. Since threedimensional situations are often difficult to visualize and work with, let
us first try to find an analogous problem in two dimensions. The analogue of a cube is a square and the analogue of a sphere is a circle. Thus a similar problem in two dimensions is the following: if five circles with the same radius rare contained in a square of side 1 m so that the circles touch each other and four of the circles touch two sides ofthe square, find r. The diagonal of the square is .J2. The diagonal is a lso 4r + 2x . But x is the diagonal of a smaller square of side r . Therefore
~
x = .J2r
.J2=4r+2x =4r+2.J2r= (4+2.J2)r
~ r=
4
/{,/2.
Let's use these ideas to solve the original threedimensional problem. The diagonal of the cube is
\/12
+ 12 + 12 = v'a.
The diagonal of the cube is also 4r + 2x where x is the diagonal of a smaller cube with edger. Therefore
x = Jr 2 + 1·~ + r 2 = v'3 r
~
The radius of each ball is ( v'3

r;;
r;;
v'a = 4r+2x .= 4r+2v3r = (4+2v3)r.Thus r=
v'3
r;;= 2 v'a  3
4 + 2v3
2
~) m.
3. (a) We find the line of intersection L as in Example 12.5. 7(b). Observe that the point (  1, c, c) lies on both planes. Now since
L lies in both planes, it is perpendicular to both of the normal vectors n1 and n 2, and thus parallel to their cross product
n1 x n2 =
c
j
k
1
1 = (2c, c2 + 1, c2  1). So symmetric equations of L can be written as
1  c
c
x +1 y c z c   = ~ = ~ , provided that c I 0,  2c (..  1 c +1
±1.
If c = 0, then the two planes are given by y + z = 0 and x =  1, so symmetric·equations of L are x =  1, y =  z. If
c =  1, then the two planes are given by x + y + z =  1 and x + y + z = 1, and they intersect in the line x = 0, y =  z  1. If c = 1, then the two planes are given by x + y + z = 1 and x  y + z = 1, and they intersect in the line y = 0,
X =
1 Z.
(b) If we set z = t in the
y  c=
(t c)(c2 c2 +1

symmet~ic equations and solve for x andy separately, we get x 1)
 2ct + (c
2
~
x=
c2 +
1

1)
2
, Y=
(c

+ 1 = (t  2c)(  2c) c +1
1)t + 2c El. . . fr th . . unmatmg c om ese equattons, we
&+ 1
have x 2 + y 2 = t 2 + 1. So the curve traced out by L in the plane z = tis a circle with center~~ (0, 0 , t) and radius J fl + 1.
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147
148
0
CHAPTER 12 PROBLEMS PLUS
+ 1), so V = J; A(z )dz = 1r ( ~ z3 + z] ~ =
(c) The area of a horizontal crosssection of the solid is A (z) = 1r(z2
53
53
54
50
4 ,.. 3
5" 2
jv5 1 = 24 . 32 jv 1j = 23 . 32 . Similarly, jv5j = 24 . , jv.71 :;= 25 . 34 , and in general, jv ,. j = n_ 2 . n_ 3 = 3(ir 33 2 3
2
.
Thus
f
Jv ,.J
n= l
f
= lv1 l + lv 2l + = 5+
00
E
n= l
3(i)"
n=3
2
=2 +3+
f
n=l
3(~)"
ll
1
.2.(!~)"= 5 + ___L_ 2 1 ~
G
[sum of a geometric series]
= 5 + 15 = 20
7. (a) When 8 = 8., the block is not moving, so the sum of the forces on the block I
+ F + W = 0 . This re lationship is illustrated
must be 0, thus N
geometrically in the figure. Since the vectors form a right triangle, we have
 J!l_ 11.•nn
tan(O.)  INI 


fl s·
(b) We place the block at the origin and sketch the force vectors acting on the block, including the additional horizontal force
H , with initial points at the origin. We then rotate this system so that Flies along the positive x axis and the inclined plane is parallel to the xaxis. (See the follo~ing figure.)
N F
w
IF! is maximal, so jFj = 11. n for f) > B_,. Then the vectors, in terms of components parallel and perpendicular to the inclined plane, are
N =n j W
= ( mg sin 0) i + (mg cos fJ) j
F
= (11.n) i H = (hmin cos B) i
+ (
hmin sin B) j
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l CHAPTER 12 PROBLEMS PLUS
0
149
Equ!lting components, we have JJ,n mg sin(}+ hm1n cosO= 0
,=?
hmin cos(}+
n mg cos() hrnin s infJ = 0
=?
hmin
(1)
JL•• n = mgsin(}
sin(}+ mgcos(} = n
(2)
(c) Since (2) is solved for n, we substitute into (1): cos() + JJ~ (hmin sin(}+ mg cos 9) = mg sin 9
hmin
hmin
cos fJ +
hmin/1s
=?
'*
sin fJ = mg sin fJ  mgJJ. cos(}
 JJ.• cos'(} ) 1tmin = mg (sin8 cos (}+ 1t. sin(}
.
= mg
( tau8  /1., ) 1 + JJ. tau(}
h ( tan8  tanfJ8 ) d . . . . . . b +tan (}, tau(} an usmg a tngonomctnc tdentlty, From part () a we know JJ~ = tan (} ., so th IS ecomes min = mg 1
this is mg tau(8 8. ) as desired. Note for fJ = 8., hmin = mgtau 0 = 0, which makes sense since the block is at rest for 8. , thus no additional force H is necessary to prevent it from moving. As(} increases, the factor tau(8 8.), and hence the value of hmin, increases slowly for small values ofB 8. but much more rapidly as 8 steeper the inclined
~lane,
the less the
hori zont~l
e. becomes significant. This seems reasonable, as the
components of the various forces affect the movement of the block, so we
would need a much larger magnitude of horizontal force to keep the block motionless. If we allow f)
+
90°, corresponding
to the inclined plane being placed vertically, the value of hmin is quite large; this is to be expected, as it takes a great amount of horizontal force to keep an object from moving vertically. In fact, without friction (so (} 8 = 0), we would have (} +
90°
:::?
hmin +
oo, and it would be impossible to keep the block from slipping.
(d) Since hmo.x is the largest value of h that keeps the block from s lipping, the force of friction is keeping the block from · moving up the inclined plane; thus, F is directed down the plane. Our system of forces is similar to that in part {b), then, except that we have F = (JJ.n) i. (Note that IFI is again maximal.) Following our procedure in parts (b) and (c), we equate components:  JJ,n mgsin(} + hmux cos(} = 0
:::?
hme.x cos(} /)8 n
n  mg cos (}  hm:l.X sin (} = 0
'*
hmax
= mgsin8
e
sin (} + mg cos = n
Then substituting, hmax
cos 8  /1s ( hmax sin 8 + mg cos B) = mg sin 8 hmax
'*
cos(}  hmn.x/1s sin(} = mg sin(} + mgJJ. cos
e '*
® 20 12 Ccngage Lc:uning. Ali Righls R=n·cd. Mll)' nol be scanned, copied. or duplicated, or posled 10 a publicly accessible wcb:iilc, in whole or in pan.
150
0
CHAPTER 12 PROBLEMS PLUS
· ( tan8 +J.£.) _ mg (sin0+J.£.cos8) _ mg h m axcos 0  J.£8 sin B 1  J.£8 tan B tanB +tan B. )
== mg ( 1 tan B. tanB
= mgtan(B +B.•)
We would expect hmax to increase as 8 increases, with similar behavior as we establ ished for hmin, but with hmax values al~ays
hmin
larger than hmin· We can see that this is. the case if we graph hmo.x as a function of 8, as the curve is the graph of
translated 20. to the left, so the equation does seem reasonable. Notice that the equation predicts hmo.x
+
oo as
. 8+ (90°  0., ). In fact, as hmnx increases, the nonnal force increases as well. When (90°  Bs) :50 :S: 90°, the horizontal force is completely counteracted by the sum of the f!Orlnal and frictional forces, so no part of the horizontal force contributes to moving the block up the plane no matter how large its magnitude:
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VECTOR FUNCTIONS
D
13
13.1 Vector Functions and Space Curves 1. The component functions v'4 t 2 , e 3 t, and ln(t + 1) are a ll defined wher4  t 2
2: 0 =>  2 ~ t
~ 2 and
t + 1 > 0 => t >  1, so the domain of r is ( 1, 2].
3. lim e  31 = e 0 = 1, t+O
1 lim~= lim  sin t . t.o sin t
t+o
2
2
=
~
1 1 _ I_ _ 1 = . t)2  !2  ' lim sin2 t ( tlim~ t+0 t2 +0 t
and lim cos 2t = cos 0 = 1. Thus t+0
lim
t+O
(est i + 4 j +cos 2t k) = [lim e t] i + [urn 4] j + [rim cos 2t] k = i + j + k. sm t sm t 3
t+O
t+O
t +O
. 1 + t2 (1/ t 2 ) + 1 0+ 1 _ t ,. . 1  e2 t 1 . . . ] 1 tan 1 = 2 • t~ 1 1 1 1 t 2 = t~ (1/t2) 1 = 0  1 =  1't2.~ t = t2.~ t S. ~~
e
1+ 1  e2 lim ( ,tan 1 t, t 2 t+oo 1 t
1
 te_2_t
= 0  0 = 0 · Thus
t) =(l,f,O). .
7. The corresponding parametric equations for this curve are x = sin t, y = t. We can make a table of values, or we can eliminate the parameter: t = y
x
=>
= sin y, with y E JR. By comparing different values oft, we find tl)e direction in
which t increases as indicated in the graph. X
9. The corresponding parametric equations are x = t, y = 2  t, z = 2t, which are parametric equations of a line through the point (0, 2, 0) and with direction vector
(1, 1, 2).
®
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151
152
D
CHAPTER 13 VECTOR FUNCTIONS
= 1, y =
11. The corresponding parametric equations are x
cos t , z
= 2 sin t.
2
Eliminating the parameter in y and z gives ~ 2 + (z/ 2) = cos2 t + sin2 t or y2 + z 2 / 4 the plane x
= 1. Since x = 1, the curve is an ellipse centered at (1, 0, 0) in
= 1.
X
13. The parametric equations are
for
=1
X
=
e. y = t
4
•
z
= t 6 . These are positive
I I
t =I 0 and 0 when t = 0. So the curve lies entirely in the first octant.
I I I
I
The projection of the graph onto the xyplane is y = x
2
,
y
I
> 0, a half parabola.
I
I I
Onto the xzplane z
= x 3 , z > 0, a half cubic, and the yzplane, y 3 = z 2 •
,I
,,,
,'
y= xl
,,' X
15. Tbe projection of the curve onto the xyplane is given by r (t) = (t , sin t, 0} [we use 0 for the zcomponent] whose graph
= 0. Similarly, the projection onto the xzplane is r (t) = (t, 0, 2 cost), whose graph is the cosine wave z = 2 cos x, y = 0, and the projection onto the yzplane is r (t) = (0, sin t, 2 cost) whose graph is the ellipse is the curve y =sin x, z
2
y
 I
2
xyplane
xzplane
yzplane
From the projection onto the yzplane we see that the curve lies on an elliptical cylinder with axis the x axis . The other two projections show that the curve oscillates both vertically and horizontally as we move in the xdirection, suggesting that the curve is an elliptical helix that spirals along the cylinder.
= (2, 0, 0) and r1 = (6, 2, 2), we have from Equation 12.5.4 r (t) = {1  t)ro +t r1 = {1 t) {2,0, 0) +t{6, 2,2), 0 ~ t ~ 1 or
17. Taking ro
Parametric equations are x
= 2+
4t, y
= 2t,
z
= 2t,
0
~
t
~
r {t)
= (2 +
4t,2t,  2t);O ~ t ~ 1.
1.
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SECTION 13.1 VECTOR FUNCTIONS AND SPACE CURVES
19. Taking r o = (0,  1, 1) and r 1 = (~ . ~. r (t)
=
.
(1 t) ro +t r1
=. (1 
Parametric equations are x 21 . x
= tcost, y = t, z = t ~in t,
y
= 1+
t ;:::: 0.
153
i), we have
t) (0,  1, 1) + tO,~.~), 0 ~ t ~ 1 or r (t) = (~t , 1
= ~t,
0
~t, z
=1
+ 1t,1  tt), 0
~ t ~ 1.
tt. 0 ~ t ~ 1.
At any point (x, y, z) on the curve, x 2 + z 2
= t 2 cos2 t +
t 2 sin2 t
= t 2 = y 2 so the
curve Hes on the circular cone x 2 + z 2 = y 2 with axis theyaxis. Also notice that y ;:::: 0; the graph is II. 23. x
= t,
y = 1/ (1 + t 2 )., z
= t2•
At any point on the curve we have z
= x 2 , so the curve lies on a parabolic cylinder parallel
to theyaxis. Notice that 0 < y ~ 1 and z ;:::: 0. Also the curve passes through (0, 1, 0) when t
= 0 and y + 0, z
+
oo as
t + ±oo, so the graph must be V. 25. x = cos Bt, y
= sin Bt, z =
e0 ·8 L, t ;:::: 0. x 2 + y 2 = cos2 Bt + sin2 Bt
= 1, so the curve lie~ on a circular cylinder with
axis the zaxis. A point (x, y, z) on the curve lies directly above the point (x, y, 0), which moves counterclockwise around the unit circle in the xyplane as t increases. The curve starts at (1, 0, 1), when t = 0, and z
+
oo (at an increasing rate) as
t + oo, so the graph is IV.
= t , then x 2 + y 2 = t 2 cos 2 t + t 2 sin 2 t = t2 = z2 , so the curve lies on the cone z 2 = x 2 + y 2 . S ince z = t, the curve is a spiral on
27. Ifx = tcost, y = tsint, z
this cone.
29. Parametric equations for the curve are x = t, y gives 2t t 2
= t 2 =>
2t
= 2t2 =>
t
= 0,
z = 2t  t 2 • Substituting into the equation of the paraboloid
= 0, 1. Since r(O) = 0 and r (1) = i + k , the points of intersection
are(O, O, O)and (1, 0, 1). 31. r (t) =(cost sin 2t,sint sin2t,cos 2t).
We include both a regu lar plot and a plot showing a tube of radius 0.08 around the · ·curve.
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154
0
CHAPTER 13 VECTOR FUNCTIONS
33. r(t) = (t, t sin t, t cost)
35. r (t)
= (cos2t,cos3t ,cos4t)
X
37.
:li
= (1 + .cos 16t) cost, y = (1 +cos 16t) sin t, z = 1 +cos 16t. At any
point on the graph,
x2 + y2
= (1 +cos 16t)2 cos2 t +
= (1 + cos 16t )2 =
(1 +cos 16t) 2 sin2 t
z 2 , so the graph lies on the cone x 2 + y 2
= z2 •
From the graph at left, we see that this curve looks like the projection of a X
leaved twodimensional curve onto a cone. 39. Ift =  1, then x = 1, y = 4, z
= 0, so the curve passes through the point (1,4, 0). 1ft= 3, then x =
9, y =  8, z
= 28,
so the curve passes through the point {9,  8, 28). For the point (4, 7,  6) to be on the curve, we require y = 1  3t = 7 ==>
t
= 2. But then z =
1 + (  2? = 7 # 6, so (4, 7, 6) is not on the curve.
41. Both equations are solved for z, s~ we ~an substitute to eliminate z : x 2 = 1 + 2y ==> parameter x
J x2 + y2 =
1+ y
==>
x 2 + y 2 = 1 + 2y + y 2
==>
y = ~(x 2  1). We can form parametric equations for the curve C of intersection by choosing a
= t , then y =
~(t 2  1) and
z = 1 + y = 1 + ~(t 2  1) = ~(t2 + 1). Thus a vector function representing C
is r (t) = t i + 1 (t2  1) j + ~(t 2 + 1) k.
"·
43. The projection of the curve C of intersection onto the xyplane is the circle x 2 + y 2 = 1, z = 0, so we can write x y = sin t, 0 $ t $ 27!'. Since C also lies on the surface z = x 2 Thus parametric equations for Care x = cost, y = sin is r (t) = cost i + s in t j +cos 2t k , 0 $ t

y 2 , we have z = x 2

= cost, ·
y 2 = cos2 t  sin2 tor cos 2t.
t, ; = cos 2t, 0 $ t $ 27l', and the corresponding vector function
$ 27!'. The projection of the curve C of intersection onto the
45.
xyplane is the circle x 2 + y 2 = 4, z
= 0. Then we can write
x = 2 cost, y = 2 sin t, 0 $ t $ 27l'. Since C also lies on z
the surface z X
= x 2 , we have z = x 2 =
(2 cost)~ = 4 cos2 t.
Then parametric equations for C are x = 2 cost, y = 2 sin t,
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SECTION 13.1
47. For the particles to collide, we require r1 (t)
gives t2 = 4t 3, 7t  12 =
= r 2(t)
¢:>
(
t 2, 7t  12, t 2)
VECTOR FUNCTIONS AND SPACE CURVES
= .(4t 
3, t2, 5t 
t2, and t 2 = 5t 6. From the first equation, t 2  4t + 3 = 0
D
155
6). Equating components (t 3)(t 1) = 0 sot = 1
¢:>
or t = 3. t = 1 does not satisfy the other two equations, butt = 3 does. The piuticles collide when t = 3, at the point (9, 9, 9).
49. Let u (t) = (u1(t), u2(t ), u3(t)) and v (t) = (v1(t), V2(t), va(t)). In each part of this problem the basic procedure is to use Equation l and then analyze the individual component functions using the limit properties we have already developed for realvalued functions. (a) lim u (t ) + lim v(t) = t+a.
t+a
I\ tlim u1(t), lim u2(t), lim ua(t)) +a t+a t+a
+ I Jim V1(t ), lim v2(t), lim v3(t) ) and the limits of these ta. \t+a
t+a
component functions must each exist since the vector functions both possess limits as.t
+
a. Then adding the two vectors
and using the addition property oflimits for realvalued functions, we have that lim u(t)
t ..o.
+ tlim v (t) = I Jim u 1(t) + lim v1(t) , lim u2(t) + + a \t+
lim v2(t), lim ua(t)
t  (1.
t+u
= lim (u1(t) +VI (t), U2(t) + V2(t ), 'U3(t) + V3(t)) t+a
.
+ lim v (t)) t~>a 3
[using (I) backward]
= lim (u(t ) + v(t)] ta
(b) lim cu(t) = lim (cu1 (t), cu2(t), C'U3(t)) t +a
t + a
=
= \I tlim cu1(t) , lim cu2(t) , lim cua(t) )  a t + a t +a
I\ c tlim u 1(t), c lim u~(t), c lim u3 (t) ) = c I lim u 1(t), lim u2 (t), lim ua(t)) a t + a t + a \t__,.a. t+a t+a
= c lim (u1 (t), u2(t) , u3(t)) = c lim u (t) ta
t+a
(c) lim u(t) · lim v(t) t+a
t+a
= \I tlim u 1(t), lim u2(t), lim u3(t) ) ·I lim v1(t), lim v 2 (t) , lim v3(t)) a t + a ta \t.n. · t+a t  a. = [lim u1(t)] t + a
[u~ v1(t)] t + a
+ [lim u2(t)] [lim v2(t)] _;_ .[lim us (t)] t + a t + a t + a
[urn v3(t)] t +a
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156
CHAPTER 13 VECTOR FUNCTIONS
0
(d) lim u (t) x ·lim v (t) = ( lim u1 {t), l.i m u2(t), lim u3(t)) x ( lim v1(t), lim v2(t), lim va(t)) ta
ta
t a
=
t+a
t.a
t+a
t+o.
tta.
([lim u2(t)] [lim v3(t)]  [lim us(t)] [lim v2(t)] , [lim ua(t)] [lim v1(t)]  [lim u 1(t)] [tim vs(t)], t+a
t+a
t +a.
t+n
t+a
t+a
t +a
t+a
[p~u1(t)) [~~ v2(t))  [l~ u2(t)] [E~ vl(t)]) =
(J~ [u2(t )v3(t) u3(t)v2(t)] , E~ [ua(t)v1(t)  u1(t)va(t)] ,
~~ [u1(t)v2(t) u2(t)v1(t)]) =
lim (·u2(t)va(t) ua(t)v2(t), t+a
U3
(t) v1 (t)  u1(t)v3(t), u1(t)v2(t) tt2(t)v1 (t)) ·
= t<> lim (u (t) x v (t)] 51 . Let r (t) = (! (t), g (t), h (t)) and b = {b1, b2, ba) . If lim r (t) = b, then lim r (t) exists, so by (1), ta.
b = lim r (t) t+a
t+a
~ ( ta lim f(t), lim g(t), lim h(t)). By the definition of equal vectors we have lim f(t) ·= b1, lim g(t) = b2 ta t + a ta t +a
and lim h(t) = b3 • But these are lim its of realvalued functions, so by the definition of limits, for every c tHl
> 0 there exists
<·o3
then lh(t)  ba l < c/3. Letting o = minimumof{o1,02,6a }, then ifO < It  a l < 8we have
!f(t)  ~1 l + lg(t) ~2 l
+ lh(t) bal
< c/3 + c/3 + c/3 =c. But
lr(t)  bl = l(f(t)  b1, g(t)  b2 , h(t)  ba) l = J(f(t) b1) 2
::; J[J(t) b1)2 Thus for every c
>
0 there exists
+ J[g(t) b2]2 + J[h(t) b3)2 =
o > 0 such that if 0 <
It  a l <
+ (g(t)
 b2 )2
a!< othen !r (t)  bl <
c
¢:>
J [f(t)  b!) 2 + [g(t) b2j2 + [h(t)  baj2 < c
othen
l(f(t) b1 ,g(t)  b2,h(t)  ba) l < c ¢:>
ba) 2
lf (t) bll + lg(t) b2 l + lh(t) b3l
lr.(t) b l ::; !f(t)  b1 l + lg(t)  b2 l + lh(t)  b3 l
+ (h(t) 
> 0, there exists o> 0 such *>
(J(t)  b1 ] 2 + [g(t) b2] 2 + [h(t)  baf < c 2. But each term
on the left side of the last inequality is positive, so if 0 < It  a l <
o, then [! (t) 
b1]2 < c 2, [g( t)  b2]2 < c 2 and
[h(t)  b3] 2 < c 2 or, taking the square root of both sides in each ofthe above, lf(t)  b1 l < c, lg(t)  b2 l < . c and
lh(t)  b3l
lim h(t)
t +a.
= b3. But by (1), t+a lim r (t) = I lim f (t) , lim g(t ), lim h(t)), so lim \ ta t  •a t a
t+c&
®
~0 1 2
t a
r (t) = (b1, ~. ba) = b.
Ccng;tgc Learning. All Rights Reserved. May not be scanned. copied, or duplicated, or posted too publicly ac:ccssibh: \\'cbsitc, in whole or in
p;ltl .
SECTION 13.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS
13.2
0
157
Derivatives and Integrals of Vector Functions
1. (a) r (4.5)  r(4) Q
r(4.2) r (4)
)(.
4
(b) r(4 ·5) r ( ) = 2[r(4.5} r(4}], so we draw a vector in the same
0.5
direction but with twice the length of the vector r(4.5} r(4).
r (4·2)  r (4) = 5[r(4.2)  r (4)], so we draw a vector in the same 0.2
direction but with 5 times the (ength of the vector r( 4.2)  r( 4).
(c) By Definition I, r'(4) =
J..i~ r(4 + ~  r (4).
T (4)
= ~~~!~I·
(d) T(4) is a unit vector .in the same direction as r' (4), that is, parallel to the tangent line to the curve at r(4) with length 1.
3. Since (x
y
+ 2) 2 = t2 = y 
1
=>
(b) r'(t) = (1, 2t},
(a), (c)
= (x + 2? + 1, the curve is a
r'(  1) = (1,·2)
parabola.
)(.
= sin t, y = 2 cos t so x + (y/ 2) 2 = 1 and the curve is
5. x
(a), (c)
(b) r'(t) ~ cost i  2sintj,
2
an e llipse.
©
r
'(71")v'2. 4
'2 .
21 y~ J
2012 Ccngage Lcnming. All Rights Rcsen"Cd. May 001 be scanned. copied. or duplicoicd, or poSiod lu a publidy accessible websile, in ..t.olc or in port
158
0
CHAPTER 13 VECTOR FUNCTIONS
· 7. Since x = e2 t = (et) 2 = y 2 , the
y
(a), (c)
(b) r '(t) = 2e2 t i
curve is part of a parabola. Note that here x
r '(O)
+ et j ,
= 2i +j
> 0, y > 0. X
9. r ' (t) =
\!
[t sin t],
:t [e) ·, !
[t cos 2t]) = (t
= (t cost+ sin t , 2t, cos 2t 
c~s t +sin t, 2t, t( sin 2t) · 2 +cos 2t)
2t sin 2t)
=> r'(t)=1 i + Oj +2(~t 1 1 2 )k = i+ ~ k
11. r(t) =t i+j + 2Vt k 2
13. r (t) = et i  j + ln(1 + 3t) k
3
2
=> r'(t) = 2tet i +   k 1 +3t
15. r ' (t) = 0 + b + 2t c = b + 2t c by Formulas 1 and 3 of Theorem 3. 2 2 r '(O) = (1, 2, . 2). So lr '(O) I = .JP + .2 + 2 = .J§ = 3 and
19. r '(t) =  sin ti + 3j + 4cos 2t k
T(~) = 1::~~~~ = 21. r(t) = (t, t 2 , t 3 )
T (1 )
=
r ' ( 1) ir' ( 1 )i
+~2 +42 (3 j +4k) = t(3j +4k) = ~j + ~ k.
.J02
=> r' (t)
1 = V'i4
=> r '(O) = 3j + 4k. Thus
(
= ( 1, 2t, 3t 2 ). Then r ' (1)
= (1, 2, 3) and lr' (1)1 =
) 1 1 2 a ) "( ) 1,2,3 = \V'i4'V'i4 ' V'i4 . r t
j
r '(t) x r"(t)
=
=
(
\,/1 2 + 22 + 3 2 = v'l4, so
) 0,2,6t ,so
k
1 2t 3t2 = 12t 3t21 i 2 6t 0 2 6t
I
11 3t2 1· + 11 2t k 0 6t J 0 2
= {12f  6t~) i  (6t O) j + (2  0) k 23. The vector equation for the curve is r(t) = ( 1 + 2 Jt, t 3

= ( 6f , 6t, 2)
t, t 3 + t ), so r '(t) = ( 1/ Vt, 3f  1, 3t2 + 1). The point
(3, 0, 2) corresponds tot= 1, so the tangent vector there is r ' (1)
= (1, 2, 4). Thus, the tangent line goes through the point
(3, 0, 2) and is parallel to the vector (1, 2, 4). Parametric equations are x = 3 + t, y = 2t, z ,;,. 2 + 4t. 25. The vector equation for the curve is r (t) = ( e t cost, e t s in t, et), so
r' (t)
= (et( sin t) +(cost)( et), et cost+ (sin t)( et), (e t)) = (  et(cos t +sin t), e t(cos t  sin t), e')
The point (1, 0, 1) corresponds to t = 0, so the tangent vector there is r '(O) = ( e0 (cos 0 +sin 0), e 0 (cos0 sin 0), e0 ) = (1, 1, 1}. Thus, the tangent line is parallel to the.vector ( 1, 1,  1} and parametric equations are x = 1 + ( 1)t = 1  t, y ®
=0+
1 · t = t, z
=1+
(  1)t = 1  t.
2012 Ccngogelcaming. All Rights Reserved. Mlly not be: scanned, copied. or duplicated, or posted to a publicly accessible website, )n whole or In pan.
SECTION 13.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS
0
159
27. First we parametrize the curve C of intersection. The projection of C onto the xyplane is contained in the circle·
x2
+y2 =
25, z
= 0, so we can write x = 5 cost, y =
near the point (3, 4, 2), so we can write z = )20 r (t) =
(5c~st,5sint, )20 25sin2 t)
y2
5 sin t. C also lies on the cylinder y 2 • + z 2 = 20, and z ~ 0
= V20 
25 sin2 t. A vector equation then for C is
r '(t) = ( 5 sint,5cost,
=}
~(20 25sin2 t) 1 12 (50sintcost) ).
The point (3, 4, 2) corresponds tot = cos 1 (~),so the tangent vector there is
The tangent line is parallel to this vector and passes through (3, 4, 2), so a vector equation for the line is r (t) = (3  4t)i + (4 + 3t)j + (2  6t)k. 29. r(t) =(t,e t , 2t  e )
=* r '(t)=(1,e t,2  2t). At(0,1,0),
t = 0 and r '(O) = {1,  1, 2). Thus, parametric equations of the tangent .line are x = t, y = 1  t, z = 2t.
31 . r(t) = (tcost,t,t sint)
At ( 1r, 1r, 0), t =
1r
=* r'(t) = (costtsint, 1,tcost+sint).
and r ' (1r) = (1, 1, 1r). Thus, parametric equations
of the tangent line are x
= 1r  t, y =
1r
+ t, z =
1rt.
33. The angle of intersection of the two curves is the angle between the two tangent vectors to the curves at the point of
intersection. Since r; (t) = (1, 2t, 3t2 ) and t = 0 at (0, 0, 0), r} (0) = (1, 0, 0) is a tangent vector to r 1 at (0, 0, 0). Similarly, r~(t)
= (cost, 2cos2t, 1) and since r 2(0) = (0, 0, 0), r~ (0) = (1, 2, 1) is a tangent vectortor2 at (0, 0,0). IfO is the angle
between these two tangent vectors, then cos 0 =
35.
J; (t i 
t3j
+ 3t5 k ) dt = =
7f7r, (1, 0, 0~ · (1, 2, 1) = js and 9 =
(J; t dt). i  (J; t
3
(~t 2 ]~
+ [~t0 ] ~ k
i 
[tt4 ]~ j
dt) j
+ (!02
col:l 1 (
js) ~ 66°.
3t5 dt) k
= H4 O)i  t{16  O)j + ~(64 O) k = .2i 4j +32 k
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160 . 0
CHAPTER 13 VECTOR FUNCTIONS
37. f~"' 12 (3sin2 t cost i + 3 sin t cos2 t j + 2 sin t cost k ) dt
=
(I; 12 3 sin
2
= [sin3 t) ~
12
(.ro" 12 3 sin t cos
t cost dt) i +
i + [ cos t] ~ 3
2
12
39. f (sec2 t i + t(e + 1) 3 j + t ln t k ) dt =
2
t dt) j +
(I; 12 2 sin t cost dt) k
12 j+ [sin2 t) ~ k = ( 1  0) i + (0 + 1) j + ( 1  0) k = i + j + k
(f sec2 t dt) i + (f t(t,2 +
1) 3 dt) j +
(f t 2 ln t dt) k
=tan t i + t(t2 + 1)4 j + (tt3 ln t ~t 3 )k + C , where C is a vector constant of integration. [For the zcornponent, integrate by pa~s with u = ln t, dv = t 2 dt.] 41. r'(t) = 2t i + 3t2 j + Vt k
=}
r (t) = t2.i +
e j + tt 12k + c , where cis a constant vector. 3
But i + j = r (1) = i + j + i k + C. Thus C = tk and r (t) = ·t 2 i + t 3 j + ( tt 312

~) k.
Fo.r Exercises 4346, 1et u (t) = (u1 (t), u 2 (t), u 3 (t)) and v (t) = (v 1 ( t) , v2(t), v 3 ( t)). In each of these exercises, the procedure is to apply Theorem 2 so that the corresponding properties of derivatives of realvalued funciions can be used.
d t
d t
43. d [u (t) + v (t)] = d (u1(t) +v1(t),u2(t) +v2(t),ua(t) +v3 (t))
.
= ( ! [u1(t) +v1 (t)],! [u2(t) +v2(t)],
;t
[ua(t) + va(t)])
= M(t) + vW), u~(t) + vHt), u3(t) +vW)) = M (t), u~ (t ), u3(t)) + M(t), v~(t), v3(t)) = u ' (t) + v' (t)
d
d
45. dt [u(t) x v (t)] = dt (u2(t)va(t)  ua(t)v2(t),ua(t)v1(t)  u1(t)v3(t),u1(t)v2(t)  u2(t)v1(t))
= (u~va(t) + u2(t)v3(t)  u3(t)v2(t)  u3(t)v2(t), u3(t)vl(t) + ua(t)v~ (t) uW)va(t) u 1(t)vW), u~ (t)vz(t) .+ U1(t)vW)  u2(t)vl (t)  uz(tM (t))
= (uW)va(t)  u3(t)v2 (t), u3(t)vl (t) u~(t)va(t), ui(t)'!!2(t) u2(t)vl(t)) + (uz(t)vW)  ua(tM(t), u3(t)vi (t) ul(t)v3(t), u1 (t)vW) : uz(t)v~ (t)) = u ' (t) x v (t) + u (t) x v' (t)
· Alternate solution: Let r(t) = u(t) x v(t). Then
r(t +h) r (t)
= [u (t +h) X v(t +h)] 
(u(t) x v(t)]
= [u(t + h) x v(t + h)]  [u(t) x v(t)] + [u (t + h) x v(t)]  [u (t = u (t + h) x [v (t +h)  v(t)]
+ [u (t +h) 
u (t)]
X
t h)
X
v (t)]
v(t)
(Be careful of the order of the cross product.) Dividing through by hand taking the limit as h+ 0 we have r'(t) = lim u (t +h) x [v~ +h)  v (t)] + lim [u(t +h)  u (t)] h o . h o h
X
v(t) = u (t) x v'(t) + u'(t)
X
by Exercise 13.1.49(a) and Definition 1.
®
2012 Cengogc Lcwning. All RighlS Rescn•!!d. May not be scnnncd. copied. or duplicalcd, or posted to u publicly accessible website, in whol e or in parl.
v (t)
SECTION 13.3 ARC LENGTH AND CURVATURE
47.
:t
[u (t) · v(t)] = u'(t) · v (t) + u (t) ·.v'(t)
= (cost, 
D
[by Formula 4 of Theorem 3]
sin t , 1) · (t, cost, sin t) + (sin t, cost, t) · (1, sin t, cost)
= t cos t  cos t sin t
+ sin t + sin t 
cos t sin t
+ t cos t
= 2tcost + 2sint  2cost sint 49. By Formula 4 of Theorem 3, f' (t) = u ' (t) · v (t)
+ u (t) · v' (t), and v' (t)
= (1, 2t, 3t2 ), so
j'(2) = u'(2) · v{2) + u {2) · v'{2) = (3, 0, 4) · (2,4,8) + (1, 2,  1) · (1, 4, 12) = 6 + 0 + 32 + 1 + 8 12 = 35.
d
51 .  [r(t) x r ' (t)] dt
.
= r ' (t) ~ r' (t) + r (t) x r" (t) by Formula 5 of Theorem 3. But r' (t) x r' (t) = 0
Section 12.4). Thus,! [r (t) x r '(t)]
53.
~ Jr (t)l dt
= dd [r(t) · r(t)]
11 2
r "(t) .
(1 )I r(t) · r '(t) 1r t
~ [r(t) · r (t)] 1 1 2 [2r (t) · r '(t)] =
=
t·
= r(t) x
(by Example 2 in

55. Since u (t) = r (t) · [r '(t) x r" (t)),
=r
u' (t)
1
(t) · [r' (t)
r" (t)] + r (t) · ! [r ' {t)
X
= 0 + r(t). · (r"(t) = r(t) · (r' (t) X r
x r "(t) + r'(t)
111
X
X
r 11 (t))
r 111 (t)]
[since r' (t) ..l r ' (t) x r " (t)]
(t)]
[since r"(t) x r "(t) = 0]
13.3 Arc Length and Curvature 1. r (t)
=
(t,3cos t,3sin t)
Jr '(t)l = ) 12
=?
r' (t) = (1,3 sin t,3cost)
+ ( 3 sint) 2 + (3cost) 2 =
Then using Formula3, we have L 3. r(t)
= v'2t i + etj + e tk
J(v'2/ + (et)
lr'{t)l = Then L
= }~
1
5. r(t) = i + t 2 j Then L = 7. r(t) L
+t3 k
=
=?
r '(t) = v'2 i + etj  etk
r'(t) = 2t j
J; tJ4 + 9t
2
+ 3t2 k
dt
2
=?
= fa·
r 1 (t) = (2t, 3t2 , 4t3 )
J; J4t
+ cos2 t) = v'IO.
ts v'Wdt = .fi0t] ~ 5 ~ 10 .fiO.
+ (e t)2 =:= J2 + e2 t + e
=?
J01 lr '(t) i dt =
lr'(t)l dt =
®
lr'(t)l dt
=?
u = .J(et + e t)2
lr ' (t)i dt = J~(et + e t) dt = [et  e t] ~
= (t 2 , t 3 , t 4 )
= J~
2
=?
= f~ 5
) 1 + 9(sin2 t
=?
= et + e t [since et + e t > OJ.
= e e 1 .
ir'(t)i
= J4t2 + 9t4 =
t(4 + 9t2 ) 31 2
t=
1 27
(13 31 2
t J4 + 9t2 . [since t 2': OJ. 
431 2 )
= 2\
(133 / 2

8).
=> lr ' (t) i = V(2t? + (3t2 )2 + (4t3 ) 2 = J4t2 + 9t4 + 1fit6, so
+ 9t4 + 16t 6 dt ~ 18.6833.
2012 CcnJ::age lc~rning. All Rights Rcscr\o'f..d. Muy not be scannOO, copied, or duplicated. or posh:.d to a public ly accessible website, in whole or in pan .
161
162
CHAPTER 13 VECTOR FUNCTIONS
0
9. r (t) = (sin t,cost,tant)
lr '(t)l
=}
r '(t)
= (cost, 
sin t,sec2 t)
(sin t)2 + (sec2 t)2 = v'l + sec4 t and L
= ..jcos2 t +
=}
= J0,.14
lr '(t)l dt
= J~"' 14 v'1 +sec4 t dt ~ 1.2780.
11. The projection of the curve C onto the xyplane is the c6rve x 2 = 2y or y = ~ x 2 , z = 0. Then we can choose the parameter
x
=t
y = ~t2 . Since C also lies on the surface 3z = xy, we have z .= ~xy = ~(t)(~t2) = i t 3 . Then paramen:ic
=}
equations for C are x
= t, y = ~t2, z =
*t 3 and the corresponding vector equation is r (t) = (t, ~ t 2 , ~t3 ). The origin
corresponds to t = 0 and the point (6, 18,_36) corresponds tot = 6, so L=
f06
lr'(t)l dt =
]~6
1(1, t , lt
)1dt = J~ V12 + t 2 + (~t2 ) 2 dt = f 06 J1 + t 2 + i t4. dt
2
= f 06 J(1 + ~t2 )2 dt = j~(1 + 13. r (t) = 2t i + (1  3t) j + (5 + 4t) k
J;
8 = 8(t) = J~ lr' (u) Idu = have r (t(s)) =
~8 i +
(1
=}
2
tt ) dt = [t +
~t 3)~ =
r '(t) = 2 i 3j + 4 k and
6 + 36 = 42
"* = lr'(t) l :::::: v'4 +9 + 16 = .;29. Then
.;29 du = .;29 t. Therefore, t = ~ 8, and substituting for t in the original equation, we
~8)j +
( 5+
~8) k.
15. Here r (t) = (3 sin t , 4t, 3 cos t), so r ' (t) = (3 cos t, 4, 3 sin t) and lr '(t)l =
..)9cos2 t + 16 + 9 sin2 t = v'25 = 5.
The point (0, 0, 3) corresponds to t = 0, so the arc length function beginning at (0, 0, 3) and measuring in the positive direction is given by 8( t) =
J; lr' (u) Idu = J; 5 du = 5t.
8(t) = 5
5t = 5 .
=}
=}
t = 1, thus your location after
moving 5 units along the curve is (3 sin 1, 4, 3 cos 1). 17. (a) r (t) = (t, 3 cost, 3 sin t)
=}
r' (t) = (1,  3 sin t, 3 cost)
=}
lr' (t) l =
..)i + 9sin 2 t + 9 cos2 t = v'IO.
·( ) r' (t) 1 ( . ) / 1 ;l • 3 ) ThenT t = lr '{t)l = "710 1,  3 sm t,3cost or \"Tlii,'710smt,'711icost.
(b)
T' (t) =
fto (0,  3cost, 3sin t)
N (t) =
~~:~~~~ = ~~~ (0, 3 cos t , 3 sin t) =
~t(t) =
=}
IT '(t)l =
7io ..)o + 9cos
2
t + 9sin2 t =
#o· Thus
(0,cost, sin t).
IT '(t) l = 3/v'W = ~ lr'(t)l v'W 10
19. (a)r(t) = (v'2 t,et,e t)
=}
r '(t) = (v'2,et,et)
=}
Jr'(t)J=v'2 +e2L + e 2t =..J(et+et)2= et+et.
Then [after multiplying by :: ]
and
T '(t) = _1_ (v'2e' 2e2t 0)2e2t ( V'fe' e2t  1) e2'+ 1 ' ' (e2 ' + 1)2 ''
~
(e2'
~ 1)2
2 2 ((e2 ' + 1) (v'2e' ,2e ', 0)  2e ' ( .J2 e',e2 ',  1)) = (e2'
~ 1 )2
(v'2e' (1 .:... e
2
')
2 2 ,2e ',2e ')
® 20 12 Cengagc Lc11nting. All Righls Reserved. Muy no1be :scnnn<:d, copied, or duplicolcd, or poSicd 1o • publicly occcssiblc wcbsilc, in whole: ur In pa n.
SECTION 13.3 ARC LENGTH AND CURVATURE
d
Then
Therefore 2
e t +1 1 ( rn t( 2t) 2t 2t) T '(t) N (t) = !T '(t)! = J2 et (e2t + 1)2 v2e 1  e , 2e , 2e
= !T'(t)!
(b)
J2et 1 + 1 . et + e
~~:(t) = lr'(t)l =
21. r (t) = t 3 j + e k
1 (J2et(1 e2t) 2e2t 2e2t) J2et(e2t+1) ' ,
e2t
=}
J2et eSt+ 2et + et 
X
e2t+ 1
=?
lr '(t) ! =
,
r ' (t)
=?
Jr '(1)1 = v'1 + 4 +9 =
=
4
Jf4. r"(t) = (0, 2,6t )
,
29. f( x) = xe"',
~~:{x) =
J'(x) lf"(x) i
"( )
2 •
x
= 12x
= xe"' + e"',
f"(x)
[1 + (/'(x))2J312
31. Since y' = y"
.
3
f (x) = 4x , f
(3t2)2 + (2t)2 = v'9t4 + 4t2,
·6t2 _ 6t2 . ( ) . 9t '1 + 4t2 312 (v'9t4 + 4t2 ) 3
= 4sintj 
4cos~k,
lr' (t) x r~ (t)J = 2 ~ Jr '(t)[ · 5
= ..! . 25
(1, 2t, 3t 2). The point (1, 1, 1) corresponds to t = 1, and 1)1 (1) = (1, 2, 3)
II:
=
( ) x
xe"'
r"(1)
=?
·lr '(1) x r"(1)! = yf36 + 36 + 4 = J76. Then ~~:(1_}
27. f (x) = x ,
J2 et· J2 et) ,
r '(t) x r"(t) = 16 i + 12cos t j  12 sin t k,
= }256 + 144 cos2 t + 144sin2 t = J405 = 20. Then ~~:(t) =
25. r (t) = (t, t 2, t 3)
y'o2 +
r '(t) = 3 i + 4costj  4sin tk, r"(t)
Jg + 16 cos2 t + 16sin 2 t = .J9+T6 = 5,
,
J2e2t .J2e2t + 2e2t + 1 : (e2t + 1)2
" _ 2 _ lr '(t) X r "(t)i _ r (t)J  6t . Then ~~:(t)I () 3 r' t I
23. r (t) = 3t i + 4 sin t j + 4cost k
Jr' (t) x r " (t)!
e4t
r'(t) = 3e j + 2t k, r"(t) = 6t j + 2 k,
, r ' (t) x r"(t) = 6t2 i, Jr (t)
Jr '(t)! =
t 
= _1_ (1 e2t
= e"', the curvature is ~~:(x) =
x r"(1)
= lr'(~;,~1 ~'~( 1 )1 = ::;; = ~
= (6,  6,2),so
VSJ.
2
=
2
lf"(x) ! l12x 1 12x [l + (J'(x))2)3/2 = [1 + (4x3)2]3/ 2 = ( 1 + 16x6)3/ 2
+ 2e"',
!xe"'+2e"' l
= [1 + (xe"' + e"') 2)3/2
= (0, 2,6). r '(1)
=?
!x+2Je"'
= [1
+ (xe"' + e"')2 j3/2
Jy"(x) J [1 + (y'(x))2]312
= (1 + e"' e2"' ) 312
_ :r 2x 3/2  e (1 + e ) .
To find the maximum curvature, we first find the critical numbers of ~~:(x): 2:r 3 2x 1 2 2x '( ) _ "'( 2x)3/2 2x) 5/2( 2 2x) _ x 1 + e  e _ x e x  e 1+ e + e "'(;!)( . 2 1+ e e  e (1 + e2"')5/2  e (1 + e2x)5/2.
II:
~~:' (x) = 0 when 1  2e 2"' = 0,
so e2"' =
~ or x =  ~ ln 2. And since 1  2e2"'
> 0 for x <  ~ ln 2 and 1  2e2"' < 0
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163
164
D
CHAPTER 13 VECTOR FUNCTIONS
for x
>  ~ ln.2, the maximum curvature is attained at the point ( ~ ln 2, e( In 2 ) 12 )
= ( ~ ln 2, ~).
Since lim e"'(1 + e2 x) 3 12 = 0, ~~:(x) approaches 0 as x+ oo. :.~:  oo
33. (a) C appears to be changing direction more quickly at P than Q, so we would expect the curvature to be great~r, at P. (b) First we sketch approximate osculating circles at P and Q. Using the axes scale as a guide, we measure the radius of the osculating circle
.!.
at P to be approximately 0.8 units, thus p = "'=
c
=>
K,
.!.p ~ 01.8 ~ 1.3. Similar!y,'we estimate the radius of the
.!.p ~ ....!... ~ 0.7. 1. 4
osculating circle at Q to be 1.4 units, son, =
j6x'1 1
X
6
[1 + (2x3)z]3/2
= :z;4 (1 + 4x  6)3/ 2'
The appearance of the two humps in this graph is perhaps a little surpris ing, but it is explained by the fact that y = x  2 increases asymptotically at the origin from both
1
directions, and so its graph has very little bend there. [Note th~t r;,(O) is undefined.]
37. r(t)
= (tet , e
r' (t) x r" (t) r'(t)J
l
1
=>
.r '(t) = ((t +l)et,e'·,../2),
= (.J2et, ../2(t + 2)e1 , 2t + 3),
= . l(t + v
,../2t)
1)2e2t + e zt + 2,
( ) and
"'
t
r"(t) = ((t+2)et,et, O).
· Jr ' (t) x r " (t)l
=
lr'(t) x r "(t) l lr'(t)13
= y'2e 21 + =
Then
2(t + 2) 2 e 2t + (2t + 3)2,
2 2 y'2e t +'2(t + 2)2e t + (2t + 3)2 ((t + 1)2 e2t + e2t + 2]3/2
We plot the space curve and its curvature function for  5 :::; t :::; 5 below. K(l)
0.6
5
y
5
I
From the graph of r;,(t) we see that curvature is maximized fort = 0, so the curve bends most sharply at the point (0, 1, 0). The curve bends more gradually as we move away from this point, becoming almost linear. This is reflected in the curvature graph, where n,(t) becomes nearly 0 as /tl increases. 39. Notice that the curve b has two inflection points at which the graph appears alrnos,t straight. We would expect the curvature to
be 0 or nearly 0 at these values, bu't the curve a isn 't near 0 there. Th~s, a must be the graph of y = of curvature, and b is the graph .of y
f (x) rather than the graph
= K( x).
® 2012 Ccngngc l cnm ing. All Rights Reserved. Mny not be scanned, copied, or duplicated. or posted to t1 publicly accessible webs ite, in who le o r in part.
SECTION 13.3 ARC LENGTH AND CURVATURE
41. Using a CAS, we find (after simplifYing) ~~:(t) =
D
165
K(!)
6 v'4cos2 t  12cost + 13 ( _ cos t)3 12 • (To compute cross 17 12
products in Maple, use the VectorCa1cu1us or Li nearA1gebra package and the Cross Produc t (a, b) 0
command; in Mathematica, use Cross (a, b].) Curvature is
21T
41T
61r
I
largest at integer multiples of2:rr. 43.
X
= t2
=> :i; = 2t => X = 2, y = t 3 => iJ 2
\xy iJx\
Then ~~:(t) = [:i;2 + y2)3/2 45. x
= et cos t
=>
:i;
=
j(2t)(6t)  (3t )(2) j [(2t)2 + (3t2)2j3/ 2
= e1(cos t  sint)
y = et sint => iJ = et (cos t + sin't )
\:i;ij  1ixl . . [x2 + y 2 ] 312
x;(t) =
= 3t2
=>
=
=> ij = 6t.
j12t2 6t21 (4t2 + W)3/2
x = e1( 
6t2
= (4t2 + 9t4)3/2 ·
sint cost)+ et(cost sin t)
=> y = e1( sint +cost)+ et(cos t + sint) = 2et cost. Then 1
jet(cos t  sin t)(2et cost)  e (cos t + sin t)( 2et sin t}j = .:....__;..__ _ ___;_;_::_::':';;;;;~ 32 ([et(cos t sin t)J2 + [et(cos t +sin t)j2) /
j2e21(cos 2t  sin t cost + sin t cos t + sin2 t) j = [e21 (cos2 t  2costsint +sin2 t + cos2 t + 2 cos t sint +sin2 t)] 3 ' 2
47.
( 1,
)
2
=  2et sin t,
2
_
3 ,1 correspondstot  1.
_ r '(t) _ (2t,2t ,1) T (t)\r'(t)\  v'4t 2 + 4t 4 +
1
T '(t) =  4t(2t2 + 1) 2 ( 2t, 2t2, 1) + (2t2 + 1)1 (2,4t, 0) = (2t2 + 1) 2 ( 8t2
+ 4t
2
3
+ 2, 8t + 8t
3
j2e2 t(1) j
= [e2t(1 + 1)]312 =
2e2t 1 e31 (2) 3 / 2 = ..fiet
_ (2t,2t\ 1) _ (2 2 1 2t 2 + 1 ,soT(1)  3•3•3)·
[by Formula 3 of Theorem 13.2.3]
+ 4t, 4t) =
2(2t2 + 1)  2 (1 2t2 , 2t, 2t) ( 1  2t2, 2t, 2t)  (1 2t2, 2t,  2t) v'1  4t2 + 4t4 + 8t2 1 + 2t2
49. (0,7r,?) corresponds tot T(t) = r ' (t) \r' (t) \ T(1r)
=
='lr.
r(t) = (2sin 3t,t,2cos3t)
=>
(6 cos 3t,1,6sin3t) =:  1 (6 cos 3t, 1, 6 sin 3t ). } 36 cos 2 3t + 1 + 36sin2 3t V37
= ~ ( 6, 1, 0) is a normal vector for the normal plane, and so (6, 1, 0) is also normal. Thus an equation for the
plane ·is  6(x  0) + 1(y 1r) + O(z + 2)
= 0 ory ·
6x
= ~· 18
ffi=> N (t) =
~~:m,
= (sin3t, O,  cos 3t). So N(1r) = (0, 0, 1) and B (1r) =
~ (6, 1, 0) x
(0,0, 1) =
~ {1, 6, 0).
Since B(1r) is a normal to the osculating plane, so is (1, 6, 0) . An equation for the plane is 1(x 0) + 6(y 1r) ©
+ O(z + 2) =
0 or x
+ 6y =
61r.
2012 Ccngo&e Lcorning. All Rights Reserved. May not be scanned, copied, or duplicutcd, or posted to • publicly accessible wel>
166
0
CHAPTER 13 VECTOR FUNCTIONS
51. The ellipse is given by the parametric equations x ~~:(t) =
= 2 cost, y = 3 sin t, so using the result from Exercise 42,
l:i:ii xyl = l(2sint}(3sint) (3cost)( 2cost) 1 = 2 2 [:i:2 + 1?]3 / 2 ( 4 sin t + 9 cos2 t~S/ 2 ( 4 sin t
At (2, 0}, t = 0. N.ow ~~:(0} =
6
+ 9 cos2 t)3/ 2 • s
= ~.so the radius of the osculating circle is
6 27
1/ ~(0} = ~ and its center is ( ~, 0). Its equation is therefore
(x + ~ ) 2 + y 2 =
¥·
At (0, 3}, t = ~.and~~:(~) = ~ = ~· So the radius of the osculating circle is~ and its center is (0, ~). Hence its equation is x 2
+ (y 
lf.
~) = 2
53. The tangent vector is normal to the normal plane, and the ve~tor (6, 6,  8) is normal to the given plane.
II r '(t) and (6, 6, 8) II
But T (t)
r (t) = (t3 1 3t, t 4 )
(3, 3, 4), so we need to find t such that r' (t)
II (3, 3,  4) when t =
r ' (t) = (3t 2 , 3, 4t3 )
:::::?
II (3, 3,  4}.
 1. So the planes are parallel at the point (  1, 3, 1).
55. First we parametrize the curve of intersection. We can choose y = t; then x = y 2 = t 2 and z = x 2 given by r (t) = ( t 2 , t, t 4 ).
= t 4 , and the curve is
r ' (t) = (2t, 1, 4t3 ) and the point (1, 1, 1) corresponds tot= 1, so r ' (1)
:= (2, 1, 4) is a normal
vector for the norma l plane. Thus an equation of the normal plane is
2(x  1) + 1(y  1} + 4(z 1) T ' (t) = 
= 0 or 2x + y + 4z =
r' (t) 7. T (t) =  '( )I = · 1r t
1
J
4t2
. ( 2t, 1, 4t
+ 1 + 16t6
3
)
and
H4t2 + 1 + 1?t6 )  S/ 2 (8t + 96t5 ) (2t, 1, 4t3 ) + (4t2 + 1 + 16t6 )  l/ 2 ( 2, 0, 12t 2 ). A normal vector for
the osculating plane is B (l) = T (1) x N (1), but r'(1 ) = (2, 1, 4) is parallel to T (1) and T ' (1} =  ~ (21) 3 : 2 ( 104} (2, 1, 4} + (21) 1 12 (2, 0, 12)
=
21
3TI (31, 26, 22) is parallel to N (1) as is (31, .:...26, 22),
so (2, 1, 4} x ( 31,  26, 22} = (126, 168, 21) is normal to the osculating plane. Thus an equation for the osculating plane is 126(x  1)  168(y  1)  21(z 1)
57
=0
or 6x 8y  z = 3.
dTI dT  l dTI_IdT /dtl_ ldT / dtl d N  dT / dt N  ldt dt _dT /dt_dT b h Ch' R l . ~~: ds ds/dt  ds/dt an  ldT/dtl 'so~~:  ~dTids  ds / dtds yte am ue. dt
59. (a) IB/ = 1
(b) B = T
:
~
B ·B
xN =
=
=1
dt
d
~
dB J. B
ds (B · B )= 0
ds
:::::?
!
(T x N) = 1
[(
T
X
T' ) IT ' I
:~ (T +
(
T
x N}
X
N
ds~dt =
')]
1 , lr '(t)l
:t T
(T x N } X
N'
= Tr'Tt)'l
lr'~t)l = [(T ' x N ) + (T X N'))lr'~t)l
:::::?
dB J. T ds
(i) 2012 C.flll•S• !.=nina;. All RighiS Rc:sen'Cd. Moy not be scanned, copied. or duplicated. or posted to o publicly ucccssible website. in whole or in part.
SECTION 13.3 ARC LENGTH AND CURVATURE
=>
(c) B = T x N
0
167
T _L N , B _L T and B _LN. SoB, T and N form an orthogonal set of vectors in the three
dimensional space !R3 . From parts (a) and (b), dB fds is perpendicular to both BandT, so dB / ds is parallel to N . Therefore, dB / ds = r(s)N , where r(s) is a scalar.

(d) Since B = T x N , T _L N and both T and N are unit vectors, B is a unit vector mutually perpendicular to both T and N. For a plane curve, T and N always lie in the plane of the curve, so that B is a constant unit vector always perpendicular to the plane. Thus dB / ds = 0, but dB f ds =  r(s)N and N =/= 0, so r(s) = 0.
~~ s' =
=> r " = s" T + s' T ' = s" T + s'
61. (a) r ' = s' T
s" T
+ ~(s') 2 N
by the first SerretFrenet formula.
(b) Using part (a), we have
r'
X
r" = (s' T )
X
(s" T
+ ~cs')
2
N]
= ((si T ) x (s" T )] + ((s'T ) x (~(s'? N))
= (s' s")(T x T ) + ~~;(s')
3
(T x N )
[by Property 3 of Theorem 12.4. 11)
= 0 + ~~;(s'? B =
~~;(s') 3 B
(c) Using part (a), we have r 111 = [s" T
+ ~~;(s')
2
= s"' T + s"dT
ds
N ]'
=s
111
T
+ s" T ' ;t ~'(s'? N + 2~~;s' s" N + ~~;(s'? N '
s' + ~'(s'? N
+ 2~s' s" N + ~(s') 2 dN s' ds
= s"' T + s" s'~ N + K1(s') 2 N + 2~s' s" N + ~~;(s') (~ T + rB) = [s"'  K2 {s'?J T + (3~s' s" + ~~;'(s') ] N + ~~;r(s') B 3
[by the second formu la]
3
2
(d) Using parts (b) and (c) and the facts that B · T = 0, B · N = 0, and B · B = 1, we get (r ' x r ") . r 111 = ~~:(s') 3 B · { (s
111

~2 (s') 3 ] T
2
Jr ' x r " J 63. r = (t, T 

+ (3~s' s" + ~~ (s') 2 ] N + KT(s') 3 B} = ~~:(s') 3 ~r(s')3
=
(~~;(s')3] 2
J~~:(s')3 B J
2
·
T •
tt2, !t3 ) => r' = (1, t, t 2 ), r" = (0, 1, 2t), r 111 = (0, 0, 2) => r ' x r " = (t 2 ,  2t, 1) =>
(r 'x r ")· r "' Jr' x r 11 J2

(t2 ,  2t,1) ·(0,0,2) 2  ,....,..t 4 + 4t2 + 1  t 4 + 4t2 + 1
65. For one helix, the vector equation is r (t )
= (10 cost, 10 sin t, 34t/(2rr)) (measuring in angstroms), because the radius of each
helix is 10 angstroms, and z increases by 34 angstroms for each increase of 271' in t. Using the arc length formula, Jetting t go from 0 to 2.9 x 108 x 21r, we find the appmximate length of each helix to be L=
J;·
9
B
xlO x zrr
Jr'(t)J dt
= j~ ·
9
s
x 10
. /
x zrr
y ( 10 sint) 2 + {10cost) 2
+ U!)2 dt =
8 ] 2.9x l0 x2rr
..j100 + (~!)2 t
0
= 2.9 x 108 x 21r
J10o+ (g! ) ~ 2.07 x 10 2
10
A more than two meters!
® 2012 Ccngagc lc.'lrning. All Ri&llts RCSCt"Vcd. Mo.y not be scanned, copied, or duplicated, or posted ton publicly ncccssible website, in whole or in port
168
CHAPTER 13 VECTOR FUNCTIONS
D
13.4 Motion in Space: Velocity and Acceleration 1: (a) Ifr(t) = x(t) i + y (t) j + z(t) k is the position vector of the particle at timet, then the average velocity over the time .interval [0, 1] is v ave
? k s· . I _ r(l) r (O) _ (4.5 i + 6.0j + 3.0k) (2.7 i + 9.8j + 3.7k) _ . . h th  1.8 1  3 . 8 J  0 . . 1m11ar y, over t e o er 1 10 •

intervals we have [0.5, 1] :
[1, 2] : [1, 1.5]:
_ r(1) r (0.5) _ (4.5 i + 6.0j + 3.0 k) (3.5 i + i2j + 3.3 k ) _ ._ k ._ 1  0.5 0.5  2.0 1 2 . 4 J 0.6
Vave
_ r (2) r (1) _ (7.3i+7.8 j+2.7k)  (4.5i+6.0j +3.0k) _ . ._ k .  28 · 1 + 1·8 J 03 · 2_ 1 1
V ave 
_ r(1.5) r(1) _ (5.9i+6.4j+2.8k)(4.5 i +6.0 j+3.0 k ) _ . · ._ k  28 · 1 + 08 · J 04 • 1. 5 1 0.5 . .
Yavo
(b) We can estimate the velocity at t = 1 by averaging the average velocities over the time intervals [0.5, 1] and [1, 1.5]:
v(1) ~ ~{(2 i  2.4 j  0.6k) + (2.8 i +0.8j 0.4k)] = 2.4i  0.8j  o·. 5k. Then the speed is Jv (1)\
3. r(t)
~ )(2.4) 2 + (  0.8)2 + ( 0.5)2 ~ 2.58.
= ( . 21 t 2 ,t)
=>
Att = 2:
= (2, 1)
v{t)
= r'(t) = (t, i)
v{2)
a(t)
= r"(t) = (: 1, 0)
a(2) .= (1,0)
jv(t)\ = Jt2 + 1
5.r(t)=3costi+2sintj
=>
Att = 71'/3:
v (t) = 3sinti+2costj
v(~) =  ¥ i+j
a(t) = 3cost i 2sintj
a(i) = ~ i  J3j
/
(3,0) X
Jv (t) \ = )9sin 2 t + 4cos2 t;:::: .) 4 + 5sin2 t Notice that x 2 /9 + y 2j4 = sin 2 t + cos2 t = 1, so the path is an ellipse. z
Att = 1: v(t) = i + 2tj a(t)
v (1) = i + 2 j
= 2j
a(1) = 2j
Jv(t)l=~
Here x = t, y = t2
Y
=> y =
parabola in the plane z = 2.
©
2
~ and z = 2, so the path of the particle is a X
2012 Ccngugc Learning, AU Rights Rescr.·cd. Mny not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in pru1.
.
SECTION 13.4 MOTION IN SPACE: VELOCITY AND ACCELERATION
9. r (t) =
0
169
(t2 + t, e  t , t 3 ) => v(t) = r ' (t) = (2t + 1, 2t  1, 3t 2 ), a(t) = v' (t) = (2, 2, 6t),
lv (t)i = .j(2t + 1)2 + (2t 1)2 + (3t 2 )2 = v'9t4 + 8t 2 + 2. 11. r (t) = v'2t i + e' j + et k
= J(e' + e') 2 =
lv(t)l = v'2 + e2 ' + e zt 13. r (t) = e'(cost, sint, t)
v (t)
=> v (t) = r'(t) = v'2i + e' j  e' k, a(t) = v'(t) = e' j + e' k, e' + e' .
=>
= r'(t) = e'(cost,.sin t, t~ + et (sint,cost, 1) = et (cost 
sint,sint +cost, t + 1)
a(t) = v'(t) = 'e'(cost sint  sin t  cost, sin t +cost+ cost  sin t , t + 1 + 1) = et( 2sin t , 2 cost, t + 2)
lv(t)i = etJcos2 t + sin2 t  2cost sint + sin2 t + cos2 t + 2sin tcost + t 2 + 2t + 1 = e' .Jt2
15. a(t)
=i +
+ 2t + 3 =>
2j
v(t)
= Ja(t) dt = J(i +
.soC = k and v(t) = t i + 2tj + k. , r {t)
2j) dt
=ti+
2tj + C and k
= J v(t)"dt = J (t i +
2t j + k) dt
= v {0) = C, = ~t
2
i+
t2 j + t k + D .
But i = r (0) = D, soD= i and r(t) = (tt2 + 1) i + t 2 j + tk.
=> ·
17.. (a) a(t) = 2t i + sintj +cos 2t k
(b)
= J (2t i + sin t j + cos 2t k) dt = t2 i and i = v (O) = j + C, so C = i + j v(t)
and v(t) = (t 2 + 1) i + (1 cost) j + r (t) = J[(t 2 + 1) i + (1 cost) j +
= (tt3 + t) Butj = r(O) 19. r (t) = (t 2 ' 5t,
cost j
+ t sin 2t k + C
t sin2tk.
%
0.6 0.4 0.2 0
t sin2tk)dt
i + (t sin t)j ~cos 2t k + D
=  i k+ D , soD = j + {kand r (t) = (tt3 +t) i+ (t sint + 1)j + (! { cos2t) k.
e 16t)
=}
v(t)
= (2t, 5, 2t 
,16), lv(t)i = v'4t2 + 25 + 4t2  64t + 256 = v'8t2  64t + 281
and dd lv(t)i = t(se  64t + 281) 1 12 (16t  64) . This is zero if and only if the numerator is zero, that is, t . 16t  64 = 0 or t
= 4. Since
:t
lv(t) l < 0 fort < 4 a~d
:t
lv(t)l > 0 fort > 4, the minimum speed of /153 is atta.ined
at t = 4 units of time. 21 . IF (t)i = 20 N in the direction of the positive zaxis, so F (t) = 20 k. Also m = 4 kg, r{O) = 0 and v {O) Since 20k
= F (t) = 4 a(t), a (t) =
5k. Then v (t) = 5t k +
speed is lv (t)l = v 1 + 1 + 25t 2 = v'25t2
c1
=i
 j.
where c1 = i  j so v(t) = i  j + 5t k and the
+ 2. Also r (t) = t i 
t j + ~t2 k + cz and 0 = r (O), so C 2
=0
and r (t) = t i  t j + ~ t2 k.
®
2012 Ccngagc Learning. All Rights R..c:scrvld. Muy not be scanned, COilicll. or duplicated. or posted to a public:l)' accessible wcbsirc. in whole or in pan.
170
CHAPTER 13 VECTOR FUNCTIONS
0
23. lv(O) I = 200 m/s and, since the angle of elevation is 60°, a unit vector in the _direction of the velocity is (cos60°)i+ (sin60°)j
= ~i + 4!j. Thusv(O) = 2oo(~i +
4j)
= lOOi+ 100J3j and ifwesetuptheaxessothatthe
projectile starts at the origin, then r(O) = 0. Ignoring air resistance, the only force is that due to gravity, so F(t)
= ma(t) = m9 j where 9 ~ 9.8 m/s2. Thus a(t) = 9.8j and, integrating, we have v (t) = 
100 i + 100 V3 j
= v(O) =
C , so v( t)
r (t)
= lOOt i + (100 V3 t  4.9t2 ) j +
r(t)
= lOO t i +
'
= 100 i + (100 V3 
9.8t) j and then (integrating again)
D where·O = r(O) = D. Thus the position function of the projectile is
(100 V3t 4.9t2 ) j .
(a) Parametric equations for the projectile are x(t) y(t)
9.8t j +C. But
= O(and t > 0) =>
100 V3t 4.9t 2
= lOOt, y(t) =
100 V3 t  4.9t2. The projectile reaches the ground when
= t(100 V3 4.9t) = 0 => t = 104~f
~ 35.3 s. So the range is
x( 4°;0) = 100e 4~f3) ~ 3535 rn. 0
10
(b) The maximum height is reached when y(t) has a critical number (or equivalently, when the vertical component of velocity is 0): y' (t) = 0
=> 100 J3  9.8t = 0 => t
= 100~{3 ~ 17.7 s. Thus the maximum height is
v( 10~{!) = 100 V3 ( 10~{!)  4.9 ( 109~{!r ~ 1531 m. (c) From part (a), impact occurs at t =
°{!) = 1~0 i +
v ( 104
J.;f s. Thus, the velocity at impact is
10
[100 J3  9.8 ( 104
°f)] j = 100 i  100 J3j and the speed is
1.f) I= JIO,OOO + 30,000 = 200 mjs.
lv ( 10
25. As in Example 5, r (t)
= (vo cos45°)t i +
> .0)
velocity is vo
= J90g ~ 30 mjs.
t
= v5 s~ 2a. Thus
1502
(voJ2 t gt2 )
j] . The ball lands when
L.
27. Let a be the angle ,of elevation. Then v 0
d
= t[voJ2 t i +
voJ2 . . Iands 90 m away, . 90 = x = 1 vo vrn2' vo J2 = s. Now smce 1t  or v 02 = 2 9 g
y = 0 (and t
::::>:
[(vo sin45°)t t9t2] j
2 ;in a
Two angles of elevation then are a
'
= 150m/sand from Example 5, the horizontal distance traveled by the projectile is
= 800 => ~
909 and the .. . I m1t1a
sin 2a
10.2° and a
:=:::
= ~~~; ~ 0.3484 =>
2a
~ 20.4° or 180 
20.4
= 159.6°.
79.8°.
29. Place the catapult at the origin and assume the catapult is 100 meters from the city, so the city lies between (100, 0)
and (600, 0). The initial speed is v 0 = 80 m/s and let 0 be the angle the catapult is set' at As in Example 5, the trajectory of the catapulted rock is given byr(t) = (80cosO)ti + [(80sin0)t 4.9t2] j. The topofthe near city wall is at (100, 15), which the rock will hit when (80 cos 0) t = 100 =>
®
5 O and (80 sin O)t  4.9t2 4 cos
t=
= 15 =>
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D
SECTION 13.4 MOTION IN SPACE: VELOCin' AND ACCELERATION
171
2
80 sin 9 . 5  4.9 ( 5) 4cos9 4 cosfJ
=> 100 t a n ()  7.65625 sec2 fJ = 15. Replacing sec2 9 with tan 2 8 + 1 gives
= 15
7.65625 tan2 9  100 tanfJ + 22.65625 = 0. Using the quadratic formula, we have tan 9 ~ 0.230635, 12.8306
=>
B ~ 13.0°, 85.5°. So for 13.0° < 9 < 85.5°, the rock will land beyond the near city wall. The base of the far wall is located at (600, 0) which the rock hits if (80 cos fJ)t
= 600
15 = 2 COSu n and (80sin 9)t 
=> t
4.9t2
=0
=>
2
. 15n  4.9 (  15 (} ) 80sm0 · 2cOS!7 2 COS
= 0
275.625 tan2 9  600 t an0 + 275.625
=> 600tan9 275.625sec2 fJ = 0 =>
= 0. Solutions are tan(}~ 0.658678, 1.51819
rock lands beyond the enclosed city ground for 33.4°
=> (} ~ 33.4°, 56.6°.· Thus the
< (} < 56.6°, and the angles that allow the rock to land on city ground
< () < 33.4~, 56.6° < 9 < 85.5°. If you consider t~at the rock can hit the far wall and bounce back into the city, we
are 13.0°
calculate the angles that cause the rock to hit the top of the wall at (600, 15): (80 cos 9)t = 600 (80sin fJ)t 4.9t2 = 15 =>
600tan0 275.625sec2 (}
Solutions are tan 9 :::::: 0. 727506, 1.44936 13.0°
2 cos(}
=> 275.625 tan2 (}  600 tan(}+ 290.625
= o.
=> (} :::::: 36.0°, 55.4°, so the catapult should be set with angle fJ where
< (} < 36.0°' 55.4° < (} < 85.5° .
31. Here a (t)
=  4j 
32 k so v (t)
= 4tj 
32t k + v o =  4t j  32t k + 50 i +80 k
r (t) = 50t i 2t2 j + (BOt 16e) k (note that r o and t
= 15
=> t = ~ and
> 0: SOt  16t 2
= 16t{5  t) = 0
=>
= 50 i 4t j +
(80  32t) k and
= 0). The ball lands when the zcomponent of r (t) is zero
t = 5. The position of the ball then is
r (5) = 50(5) i  2{5) 2 j + [80(5)  16(5) 2 ] k = 250 i  50 j or equivalently the point (250,  50, 0). This is a distance of J250 2 + ( 50)2 + 02
= )65,000 :::::: 255ft from the origin at an angle oftan 1 (~)
::::::
11.3° from the eastern direction
toward the south. The speed of the ball is lv (5) l = I50 i  20j  80 k l = J50 2 + (  20)2 + ( 80)2 = v'9300:::::: 96.4 ftls. 33. (a) After t seconds, the boat will be 5t meters west of point A. The velo~ity
20
of the water at that location is 4~0 (5t)(40 5t) j . The velocity of the boat in still water is.5 i , so the resultant velocity of the boat is v(t) = 5 i + 4~ 0 (5t)( 40 5t) j = 5i + ( ~t fGt 2 ) j . Integrating, we obtain r (t)
= 5t i +
(~tl
 fGt3 ) j +C. lfwe place the origin at A (and consider j
to coincide with the northern direction) then r (O)
=0
reaches the east bank after 8 s, and it is located at r(8)
=> C
= 0 and we have r (t) = 5t i +
= 5(8)i +
( ~(8) 2

fG (8?) j
= 40 i +
(~t 2

1 16
e) j . The boat
16j. Thus the boat is 16 m
downstream. (b) Let Q be the angle north of east that the boat heads. Then the velocity of the boat in still water is given by
5(cos Q) i + 5(sin Q) j . At t seconds, the boat is 5(cos Q)t meters from the west bank, at which point the velocity of the water is .1 ~ 0 [5(cos Q)t][40 5(cos Q)t] j . The resultant velocity of the boat is given by
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172
D
CHAPTER 13 VECTOR FUNCTIONS
g
v(t) = 5(cos a:) i + [5 sin a:+
4 0 (5t cos a:)( 40
 5t cos a:)] j
Integrating, r (t) = (5t cos a:) i + (5tsin a:+ ~t2 cos a:
=
ftt3 cos2 a:) j
the origin at A). The boat will reach the east bank ~hen 5t cos a: = 40 In order to land at point B ( 40, 0) we need 5t sin a: + ~ 8 5(  ) sin cos a:
8 a:+~ (cos  ) a:
40 sin a: + 16 = 0
=?
2
8
3
sin a: =
 %. Thus a: =
sin 
e cos a:  ft t
ft (cos a:) .cos2 a:= 0
cos a:
l
(%)
fue cos 2 a:) j.
(5 cos a:) i + (5 sin a:+ ~t cos a: (where we have again placed =?
3
t
=~
cos 2 a: = 0
=?
8
=  . cos a:
5 cos a: =?
1   (40sina: + 48  32) cos a:
=0
=?
~ 23.6°, so the boat should head 23.6° south of
If
east (upstream). The path does seem realistic. The boat initially heads upstream to counte~act the effect of the current. Near the center of the river,
0 1=:::==::::::::;::>of'...::::::::=:=::::~ 40
the current is stronger and the boat is pushed downstream. When the boat nears the eastern bank, the current is slower and the boat is able to progress upstream to arrive at point B.
 12
35. Ifr' (t) = c x r(t) then r' (t) is perpendicular to both c and r(t). Remember that r ' (t) points in the direction of motion, so if r' (t) is always perpendicular to c, the path of the particle must lie in a plane perpendicular to c. But r ' (t) is also perpendicular to the position vector r(t) which confines the path to a sphere centered at the origin. Considering both restrictions, the path must be contained in a circle that lies in a plane perpendicular to c, and the circle is centered on a line through the origin in the direction of c. 37. r(t)
= (3t 
t 3) i
+ 3t 2 j
r'(t) = (3  3t2 ) i + 6tj,
=?
lr' (t)i = J(3 3t2 )2 + (6t)2 = v'9 + 18t2 + 9t4
=
J(3  3t 2 ) 2 = 3 + 3t
2
,
r" (t) ·= 6t i + 6j, r' (t) x r" (t) = (18 + 18t2 ) k. Then Equation 9 gives  r ' (t) . r" (t)  (3  3e)( 6t) + (6t)(6)  18t + 18t 3 aTlr '(t)/ 3 +3t2 3 + 3t2
aT = v
I
=
d [ 2] ] dt 3 + 3t = 6t
39. r (t) = cost i +sin t j + t k r" (t) =  cost i  sin tj, Then D.T
an
=?
dE


2
18t(1 + t ) 3(1 + t 2 )


6t
[or by Equation 8, 2
. LO .  lr'(t) X r "(t) i  18 + 18e  18(1 + t ) quat10n g1ves aN ir'(t)i  3 + 3 t 2  3 (1 + t 2)
r '(t) =  sint i + costj + k ,
ir'(t)i = Vsin 2 t + cos2 t + 1 =

6 ·
V'2,
r' (t) x r" (t) =sin t i  cos tj + k.
_ r'(t) · r"(t) _ sintcost  sintcost _ d _lr'(t) x r"(t)l _ Vsin2 t+cos 2 t+ 1 = .J2 = l 0 lr'(t)i  . .j2  an aN lr'(t)i /2 /2 ·
41. r (t)=et i +/2t j +et k
=?
r"(t) = et i + et k. Then aT =
®
r'(t)=et i+/2j e 1 k, e2t _ e 2t
et
+e
t
=
(et
+ e t)(et _ et + e t
ir(t) i=v'e21· +2+ e
e t)
2
t=J(e1· +e t)2=e'· +e t,
.
= et  e: t = 2sinht
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CHAPTER 13 REVIEW D
173
43. The tangential component of a is the length of the projection of a onto T, so we sketch the scalar projection of a in the tangential direction to the curve and estimate its length to be 4.5 (using the fact that a has length 10 as a guide). Similarly, the normal component of
a is the length of the projection of a onto N, so we sketch the scalar projection of a in the normal direction to the curve and estimate its lengttJ to be 9.0. Thus aT
:::::::
4.5 cmf s 2 and
X
45. _If the engines are turned off at timet, then the spacecraft wi ll continue to travel in the direction ofv(t) , so we need at such that for some scalars
4 so 7 t2
> 0, r (t) + s v (t)
24t 12t2
8(3 t)t
+ 1 + (t2 + 1)2
= 9
¢}
v(t) = r ' (t) = i
= (6, 4, 9).

4
 ;:;;: :::.;: = 2 (t2 + 1)2
¢}
t
4
+
+ ~ j + (t2 ~ 1) 2 k =>
+ 8t2 
·a is easily seen that t = 1 is a root of this polynomial. Al~o 2 ln 1 + ~
13
3
1
12t + 3
= 0.
= 4, sot= 1 is the desired solution.
Review CONCEPT CHECK
1. A vector function is a function whose domain is a set of real numbers and whose range is a set of vectors. To find the derivative or integral, we can differentiate or integrate each component of the vector funct ion.
2. The tip of the moving vector r (t) of a continuous vector function traces out a space curve. 3. The tangent vector to a smooth curve at a point P with pos ition vector r(t) is the vector r ' (t). The tangent line at p ' is the line through P parnllel to the tangent vector r ' (t). The unit tangent vector is T (t) =
,::m,.
4. (a) (a)  (f) See Theorem 13.2.3.
5. Use Formula 13.3.2, or equivalently, 13.3.3. 6. (a) TI1e curvature of a curve is"' =
(b) K.(t)
'(t) I = I. Tr '(t)
I~~
I
where Tis the unit tangent vector.
_
(c) K.(t) = lr' (t ) X r" (t)l
T' (t) 7. (a) The unit normal vector: N (t) = IT'(t) l'
The binormal vector: B (t)
if"(x) l
·
(d) K(x) [1 + (f'(x))2j3/2
[r'(t)[ 3
= T (t)
x N (t).
(b) See the discussion preceding Example 7 in Section 13.3. 8. (a) Ifr(t) is the position vector of the particle on the space curve, the velocity v (t) = r'(t), the speed is given by lv (t )i, and the acceleration a (t) = v ' (t) = r " (t). © 2012 Ccngugc Learning. All Rights Reserved. Mny not be SCAnned, copied. or duplicated, or posted to a publicly ncccssib\e website. in wltolc or in part.
174
D
CHAPTER 13 VECTOR FUNCTIONS
(b) a = a7:T +aN N where aT = v' and aN = 1w 2 .
9. See the statement of Kepler's Laws on page 892 [ET 868].
TRUEFALSE QUIZ 1. True. If we reparametrize the curve by replacing u
with direction vector i
= t 3 , we have r(u) = u i + 2u~ + 3uk, which is a line through the origin
+ 2 j + 3 k.
3. False. The vector function represents a line, but the line does not pass through the origin; the xcomponent is 0 only for t ;, 0 which co~esponds to the point {0, 3, 0) not {0, 0, 0). d 5. False. By Formula 5 ofTheorem 13.2.3, d (u(t) x v(t)]
.
t
7. False.
K
= u '(t ) x
v (t)
+ u (t) x v'(t).
is the magnitude of the rate of change of the unit tangent vector T with respect to arc length s, not with respect tot.
9. True. At an inflection point where f is twice continuously differentiable we must have !" (x) = 0, and by Equation 13.3.11, the curvature is 0 there. 11. False. If r(t) is the position of a moving particle at timet and lr(t)l = 1 then the particle lies on the unit circle or the unit sphere, but this does not mean that the speed lr'(t)l must be constant. As a counterexample, let r (t) = (t, Jl=t2), then r' (t)
= ( 1, t/J1 
t 2 ) and lr(t) l =
Jt 2 + 1 
t 2 = 1 but lr'(t)l
= yf1 + t 2 /(1 t 2 ) =
1/ ~ which is not
constant.
13. True. See the discussion preceding Example 7 in Section 13.3.
EXERCISES 1. (a) The corresponding parametric equations for the curve are x
y = cos 1rt, z = sin rrt. Since y 2 I
+ z2
= t,
= 1, the curve is contained in a
circular cylinder with axis the xaxis. Since x
= t, the curve is a helix.
= t i + cos ri j +sin rrt k => r'(t) = i  rrsin 1rt j + rrcos rrt k ~
(b) r(t)
r" ( t) = rr2 cos 1rtj 
1r
2
X
siri 7rt k
3. The projection of the curve C of interSection onto the xyplane is the circle x 2 x
= 4 cost, y = 4 sin t,
0
S t S 21r. From the equation of the plane, w~ have z
equations for Care x = 4 cost, y r(t) = 4 cost i
©
+ y2 =
+ 4 sin t j + (5 
= 4 sin t, z = 5 
4 cost) k , 0
4 cost, 0
=5
16, z = 0. So we can write x
=5
4 cos t, so parametric
S t S 2rr, and the corresponding vector function is
S t S 21r.
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CHAPTER 13 REVIEW· 0
5.
f~(t 2 i +tcos 1rtj
+sin 1rt k ) dt
= (!01 t 2 dt) i + (I~ tcos 1rtdt) j +(I; sin 1rtdt) k I
= [~t3 ]~ i + (~sin 1rt]~  f 01 ~sin 1rtdt) j +[~cos 1rt]~ k 1 •
t]l .
[ 1
= ]j l +
2
;1fCOS1l" 0 J+ ;r
k ;= 1 1•  ;:lf 2 • 2 k J+ ;r 3
where we integrated by parts in the ycomponent. 1. r(t)
= ( t 2 , t 3 , t 4) .
. L=
=?
1
r (t)
= ( 2t, 3e,4t3 )
=?
Jr (t) J = J 4t 2 + 9t4 + 16t6 and 1
J; Jr (t)J dt = J; J4t2 + 9t4 + 16t6 dt. Using Simpson's Rule with f(t) = J4t 2 + 9t4 + 16t 1
have tlt
6
and n = 6 we
= 3(i0 = ~ and L ~ ';' [f(O) + 4f(~) + 2f{1) + 4Ja) + 2/(2) + 4f(~) + /{3)]
=i
[vo
+o +o + 4 . ..j4(~/
+ 9 (~)
4
+
16(~)6 +
2 . .J4(1)2 + 9{1) 4 + 16(1)6
2 4 + 4. J4G) + 9G) + 16 Gt + 2 . .J4(2)2 + 9(2)4 + 16{2)6
+4· ~
4(~)2 +9(~)4 +16(~t+.J4(3) 2 +9{3)4+ 16(3)6 ]
86.631
9. The angle of intersection of the two curves, 8, is the angle between their respective tangents at the point of intersection.
For both curves the point {1, 0, 0) occurs when t r W)
=
sin t i +cos t j + k
r i {0) · r~ (0) = (j + k ) · i
=?
= 0.
rHO)
= 0.
=j +
k and r 2(t)
= i + 2t j + 3t2 k
=}
r~(O)
Therefore, the curves intersect in a right angle, that is, 8 =
= i.
~.
(b) T'(t) = Ht4 +t2 + 1) 3 12 (4t3 +2t) (t2,t, 1) + (t4 + t 2 + 1) 1 12 (2t, 1, 0)  2t3  t 2 1 (t4 + t2 + 1)3/ 2 (t , t , 1) + (t4 + t2 + 1)1 / 2 (2t, 1, 0)  ( 2t5 1
IT (t)l
=

t 3 ,  2t4

t 2 ,  2t3  t) + (2t5 + 2t3 + 2t, t 4 + t 2 + 1, 0) (t4 + t2 + 1)3/ 2 
Jt 6 + 4t4 + 4t2 + t 8
 2t 4 + 1 + 4t6 + 4t 4 + t 2 (t4 + t2 + 1)3/2
=
(t 3 + 2t,  t 4 + 1, 2t 3 (t4
JtB + 5t6 + 6t4 + 5t2 + 1 (t4 + t2 + 1)3/ 2
+ t2 + 1)3/2 and
(t 3 + 2t 1  t 4  2t3  t ) ' '  Jt!!. + 5t6 + 6t4 + 5t2 + 1 .
N (t) 
(c) "'(t)
=
JT 1 (t) \ Jrl(t) l
=
(t4 + t2 + 1)2
13. y I = 4X 3 , y II = 12 X 2 and It( X )
®
+ 4t2 + 1 + t2 + 1)3/ 2
Jt 4
J t!!. + 5t6 + 6t4 + 5t2 + 1 or
(t4
= (1 + lv"l (y')2)3/ 2
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t)
175
176
0
CHAPTER 13 VECTO'R FUNCTIONS
15. r (t)
= (sin2t, t,cos2t) =>
T '(t)
= ~ ( 4sin 2t, 0, 4cos 2t)
B = T xN =
=> T (t) =
r '(t) = (2cos2t, 1, 2 sin2t)
=> N (t)
js (2 cos2t, 1,  2sin2t) =>
= (sin 2t, 0, cos2t). SoN =
N ('rr) = (0, 0,  1) and
js (1, 2, 0}. So a normal to the osculating plane is ( 1, 2, 0) and an equation is
 l (x 0) + 2(y 1r) + O(z  1) = 0 or x  2y + 27r = 0. 17. r (t) = tlnt i +tj + et k , v (t) = r '(t) = (1 + lnt) i + j  e~ k,
lv (t)i =
J(l + ln t)2 + 12 + (e:t)2 =
vf2 +2 ln t + (ln t)2 +e 2 t,
a(t) = v'(t) =
t i + et k
19. We set up the axes so that the shot leaves the ath lete's hand 7 ft above the origin. Then we are given r (O) = 7j, lv (O)I = 43 ft/ s, and v (O) has direction given by a 45° angle of elevation. Then a unit vector in the direction ofv(O) is ~(i + j )
=> v (O) =
~(i + j ). Assuming air resistance is negligible, the only external force is due to gravity, so .as in
Example 13.4.5 we have a =  g j where here g ~ 32 ft/s 2 . Since v ' (t) = a (t), we integrate, giving v (t) =  gt j + C where C = v (O) = r(t) =
~(i + j ) => v (t) = ~ i + ( ~  gt) j . Since r '(t) = v (t) we integrate again, SO·
~t i + ( ~t  ~gt2) j + D. ButD =
(a) At 2 seconds, the shot is at r(2)
r (O) = 7j
=> r (t) = ~t i + ( ~t ~gt2 + 7)j.
= ~(2) i + ( ~ (2)  ~g(2) 2 + 7) j ~ 60.8 i + 3.8j, so the shot is about 3.8 ft. above
the ground, at a horizontal distance of 60.8 ft from the athlete. (b) The shot reaches its maximum height when the vertical component of velocity is 0: ~  gt = 0
=>
t = ;: ~ 0.95 s. Then r (0.95) ~ 28.9 i + 21.4j , so the maximum height is approximately 21.4 ft. v2g
.
(c) The shot hits the ground when the vertical component ofr(t) is 0, so ~t  ~gt 2 + 7 = 0 => 16t2 + ~t + 7 = 0 => t ~ 2.11 s.
r (2. 11) ~ 64.2 i  0.08j , thus the shot lands approximately 64.2 ft from the
athlete. 21. (a) Instead of proceeding directly, we use formula 3 ofTheorem 13.2.3: r (t) v = r'(t)
= t R (t) =>
= R (t) + t R' (t) = coswt i + sinwtj +t v a.
(b) Using the same method as in part (a) and starting with v = R (t) + t R '(t), we have
a= v ' = R '(t) + R '(t) + t R "(t) = 2 R '(t) + t R "(t)
= 2vd +
tad.
(c) Here we have r(t) = e' coswt i + e t sin wtj ,; et R (t). So, as in parts (a) and (b), v = r '(t) = et R '(t) et R (t) = et[R '(t)  R (t)) => a = v' = et [R"(t) R '(t)) e:t[R '(t)  R (t)) = e~ [R"(t) 2 R ' (t) + R (t)J
Thus, the Coriolis acceleration (the sum of the "extra" terms not involving aa) is 2e~ v .i + et R.
® 2012 Ccngage Learning. All rughts Reserved. MDy not be scanned, copied, or duplicated. or posted too publicly accessible website, in who It: or in pnrt.
CHAPTER 13 REVIEW 23. (a) r (t) = R coswt i
v
= wR(
+ R sin wtj =>
sin wti + cos wtj). v · r
v = r ' (t) = wR sinwt i
= wR 2 ( 
coswt sin wt
+ wR coswt j , so r
= R(coswt i
+ sinwtcoswt) = 0, so v
0
177
+ sinwtj) and
..L r . Since r points along a
radius of the circle, and v ..L r , vis tangent to the circle. Because it is a velocity vector, v points in the direction of motion. (b) In (a), we wrote v in the form wR u, where u is the unit vector  sin wt i + cos wt j . Clearly Ivi . Ie campI etes one revo I ut10n, . . T speed wR , the part1c a d.1stance 21r R,.m t1me
= wR lui = wR . At
= 2·nR R = 27r . w
. w
(c) a = dv =  w2 R coswt i  w2 Rsinwtj =  w2 R(coswt i + sinwt j ), so a =  w2 r. This shows that a is proportional
dt
to rand points in the opposite direction (toward the origin). Also, ja j = w 2 jr j
= w2R .
(d) By Newton's Second Law (see Section 13.4), F = ma, so jF j = m jaj = mRw2 = m
(~R)
2
= m ~~
2
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0
PROBLEMS PLUS
1. (a) The projectile reaches maximum vosina t= an.d y
. )(vosiiJ.a) = (vosma  g
g
!
he~ght when 0 = ~~ =
[(Vo sina)t
1 (vosina) g 2 g
2
2
2
°
Vo sm
2g
~gt2] = vo sin a 
gt; that is, when
a . Th'1s IS . the max1mum . 'ght attamed . 11e1 when .
the projecti le is fired with an angle of elevation a. This maximum height is largest when a = ~ . In that case, sin a = 1
. . . vfi and the maximum he1ght IS  . 2g (b) Let R = vfi
/g.
We are asked to consider the parabola x 2 + 2Ry  R 2 = 0 which can be rewritten as y = 
The points on or inside this parabola are those for which  R
~ x ~ Rand 0 ~ y ~ ;~ x 2 + ~.
__!._ x 2 + 2R
!!:. . 2
When the projectile is
fired at angle of elevation a, the points (x, y) along its path satisfy the relations x = (v 0 cos a) t and
tyt 2 , whereO ~ t
y = (v0 sina)t 
lxl
~ (2v0 sin a)/g(as inExan1ple 13.4.5). Thus
~ lvo cos aCvo ;ina) I= I~ sin 2al ~ I~; I =
IRI. This shows that R
~ x ~ R.
vo sin a ~ gt ) = t gt Cvo ;in a t) ;: : 0 and
For t in the specified range, we also have y = t (
y
, X 9 = (vosma)  vo cos a 2 0
 1
2
y ( 2Rx +
R)
2
(
X
 vo cos a
)2= (tana)x 
g
x = .
1 1 2 ( 2 = 2Rcos2 ax + 2Rx + tan a) x
=
~ ( 1  _1_ ) cos2
2R
2
a
1
2
2v 02 cos2 a
2
R x +(tan a) x. Thus 2 cos 2 a
R
2
2
2
+ (tana)x _ R = x (1 sec a) + 2R{tana) x  R 2 2R
2
_ {tan a) x + 2R (tan a) x  R 2 ·_ 2R 
:
2
[{tana) x  R]2 . 2R ~0
We have shown that every target that can be hit by the projecti le lies on or inside the parabola y = _ __!._ x 2 + R. 2R 2 Now let (a, b) be any point on or inside the parabola y = 
2~ x
2
+
~· Then·R ~a~ Rand 0 ~ b ~

2~ a
2
+
~
We seek an angle a such that (a, b) lies in the path of the projectile; that is, we wish to find an angle a such that
b= 
2~c~s2 a a
2
+(tan a) a or equivalently b =
~ (tan2 a ~1 1)a2 +(tan a) a . Rearranging this equation we get
;~ tan2 a a tan a+(;~+ b) = 0 or a 2 (tana? 
2
2aR(tana) + (a + 2bR) = 0 (*) . This quadratic equation
for tan a has real solutions exactly when the discriminant is nonnegative. N~w B 2

4AC ;::: 0 <=>
© 2012 Ccngage L""ming. All Rights Rc:scn'Cd. Mny not be scanned, copied, orduplicnu:d. or poslcd loa publicly nccessible \\'Cbsile, in whole or in pan .
179
180
0
CHAPTER 13 PROBLEMS PLUS
(2aR? 4a2 (a 2
b:::;
2~ (R
2
a2 )

y ·= 
2~ x
tan a
= 2aR ±
2
+ 2bR);:::: 0 <=?
<=?
4a2 (R 2

a2

2bR);:::: 0
<=?
 a2  2bR+ R 2
;::::
0
<=?
;~ a2 + ~ This C?ndition is satisfied since (a, b) is on or inside the parabola
b:::;
~ It follows that (a, b) lies in the path of the projectile when tan a satisfies(*), that is, when
+
J4a2(R2  a 2

2bR)
2a2
R ± ,jR2 2bR a2 = ''   a
If the gun is pointed at a target with height hat a distanceD downrange, then
(c)
tan a = h/ D. When the projectile reaches a distanceD downrange (remember we are assuming that it do.esn' t hit the ground first), we have D = x = (vo cos a)t, D
D
sot=  andy=(vosina)t~gt2 = Dtana
2
gD 2 2
2
s=so + 2t i ~gt 2 j=3.5 j +2t i tgt 2 j
=>
~~Q
.
~~Q
Meanwhile, the target, whose xcoprdinate is also D, has fallen from height h to height
h
tge =
3.(a)a=gj
2
D tan a 
{D . Thus the projectile hits the target. · 2 v0 cos2 a .
=> v=vogt j =2 i gtj =>
s = 2t i + (3.5
tgt2 ) j . Therefore y = 0 when t =
J7l9 seconds. At that instant, the ball is 2 J7l9 r::::: 0.94 ft to the
right of the table top. Its coordinates (relative to an ori'gi n on the floor direct ly under the table's edge) are (0.94, 0). At impact, the velocity is v
= 2i 
(b) The slope ofthe curve when t =
,f7gj, so the speed is !v i=
.J4+79 r::::: 15 ft./s.
dy dy jdt gt Vf!_g.'s. dx = dx jdt = 2 =
g
J779 =  ,j7g ,;rg  . Thus cot B =  2
2
2
and () r::::: 7.6°. (c) From (a), lvl = vl4+7g. So the ball rebounds with speed 0.8 .J4+79 r::::: 12.08 ft/s at angle of inclination 2 . 2 90° () r::::: 82.3886°. By Example 13.4.5, the horizontal distance traveled between bounc~s is d = Vo sm a, where
g
v 0 r::::: 12.08 ft./s and a~ 82.3886°. Therefore, d ~ 1.197 ft. So the ball strikes the floor at about 2
J779 + 1.197 ~ 2.13 ft to the right of the table's edge.
· 5. The trajectory oftbe projectile is given by r (t) = (vcosa)t i + [(v sin a)t ~gt 2 ] j , so v(t)
= r '(t) = vcosa i + lv (t)i
(vsina gt) j and
= J(v cosa) 2 + =g
(
v . g
(vsin a gt) 2
't   sm a )
2
2
= Jv 2 2

(2vgsin a) t
v . 2 + g2  vg2 sm a =g
.
( t
+ g2t2 = 2
g2
(
t2

2v (sin a) t g
2
+ vg2 )
2
2a ~sino:) +~cos 9 g2
@ 2012 Ccngttge Lcnming. All Rights Rc..crvec.J. Mny no1 tM: scnMcd, copied, or duplicated, or posted to 4 publicly ;K:ccssible web~'ilc, in whole or in pan.
CHAPTER 13 PROBLEMS PLUS
The projectile hits the ground when (v sina)t ~gt2 (2v/g)sin o
L(a) =
.9
10
=g
[
0
t
2 ;
=
181
sin a, so the distance traveled by the projectile is
2
1 (211/g)sin<>
lv(t)l dt =
= 0 =>
0
v ) v2 2 ( t  .9 sin a + 29 cos a dt
t  (v / g) sin o 2
+ [(vl g) cosa]2 ln 2
(t~sino + g
[using Formula 2 I in the Table of lntegrals]
[v. (~.rna)' + (~ oo.o)' +(~ '"'")' m( ~ •ina+ (~ •mo)' +(~=aY)
= g gsma
2
(~ •ina)' + (~=o)' (~oooo)' m(~•ina+ (~•ina)' + (~=oY)]
+~sin o g
[~sino · ~+ v cos2 aln(~ sino+~) +~ sino·~  v cos2 a ln(~ sino+~)] g g2 g g g g g2 g g 2
2
= g_ 2 g 2
2
+ cos v 2 . = v stno a g 2f1 We want to maximize L(a) for 0
vg
1n(  (vl(vIg)sin a+v/g) = sm v v a + cos an 1 (1+sino) g) sin a +vI g g 2g 1  sin a 2
2
.
2
s; a s; 1r/ 2.
. ( . a ln ( 1 + sin = v cos a + v cos a · 2  2 cos a sm . a)] g 2g coso , 1 sma 2
L'(a) =  coso+ v
2
[
2g
2
v
2
9
a

a)]
2cosa smo ln 1 + sin . . 1 sm o
2
•
=
a
2 cos cos2 a · 1  sin · . . 1 +sma (1smo) 2
[
2
2 v COSO! [ 1 SlDO! • COSO!+
g
2
] . ln (l+sina) 1  SlDO!
=
v
9
. •ln (1+ sin COSO! [ 2 SIDQ . a) 1  SJDO! ,
L( a ) has critical points for 0 < a < 1r12 when L' (a) = 0 => 2  sin a ln (
~
+ 
]
s~ a) = 0 [since cos a =1= 0].
SIDQ
Solving by graphing (or using a CAS) gives a~ 0.9855. Compare values at the critical point and the endpoints: L(O)
=
0, L(1r12) = v 2 I g, and £(0.9855) ~ 1.20v 2 I g . Thus the distance traveled by the projectile is maximized
for a
~
0.9855 or
~
56°.
7. We can write the vector equation as r (t) = ·at2 + bt + c where a= (a1, a2 , a a}, b = (b1 , b2 , ba }, and c = (c1 , c 2, c3). Then r ' (t) = 2t a+ b which says that each tangent vector is the sum of a scalar multiple of a and the vector b . Thus the tangent vectors are all parallel to the plane determined by a and b so the curve must be parallel to this plane. [Here we assume that a and bare nonparallel. Otherwise the tangent vectors are all parallel and the curve lies along a single line.] A normal
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182
0
CHAPTER 13 PROBLEMS PLUS
vector for the plane is a x b = (a2b3 a3~, a3b1  a1b3, a1b2  a2b1). The point (c1. c2, c3) lies on the plane (when
t
= 0), so an equation of the plane is (a2b3 a31h)(x c1) + (a3b1  a1 ~)(y  c2) + (al~  a2bl)(z  c3) = 0
or
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2012 Cengoge Learning. All Rights Reserved. May nol be sconncd, copied, or duplicated, or posted ton pub licly accessible wcbsilc, in whole or in p.vt.
14.1
PARTIAL DERIVATIVES
0
14
Functions of Several Variables
1. (a) From Table 1, f(  15, 40) = 27, which means that if the temperature is  15°C and the wind speed is 40 km/ h, then the air would feel equivalent to approximately 27°C without wind. (b) The question is asking: when the temperature is 2!tC, what wind speed gives a windchill index of  30°C? From Table 1, the speed is 20 km/h. (c) The question is asking: when the wind speed is 20 km/ h, what tem·perature gives a windchill index of 49°C? From Table I, the temperature is 35°C. (d) The function W = j( 5, v) means that we fix T at 5 and allow v to vary, resulting in a function of one variable. In other wor~s, the function gives windchill index values ~or different wind speeds when the temperature is 5° C. Fro111 Table 1 (look at the row corresponding toT
=
5), the function decreases and appears to approach a constant value as v
increases. (e) _The function W = j(T, 50) means that we fix vat 50 and allow T to vary, again giving a function of one variable. In other words, the function gives windchill index values for different temperatures when the wind speed is 50 km/h . From Table 1 (look at the column corresponding to v 3. P(120, 20)
= 50), the function increases almost linearly as T
increases.
= 1.47(120)0 '65 (20) 0 ·35 ~ 94.2, so when the manufacturer invests $20 million in capital and 120,000 hours of
labor are completed yearly, the monetary value of the production is about $94.2 million. 5. (a) !(160, 70) = 0.1091(160) 0 .425 (70) 0 · 725 ~ 20.5, which means that the surface area of a person 70 inches (5 feet 10 inches) tall who weighs 160 pounds is approximately 20.5 square feet. (b) Answers will vary depending on the height and weight of the reader. 7. (a). According to Table 4, f( 4·0, 15)
= 25, which means that if a 40knot wind has been blowing in the open sea for 15 hours,
it will create waves with estimated heights of25 feet. '
(b) h = f(30, t) means we fix vat 30 and allow t to vary, resulting in a function of one variable: Thus here, h
=
f(30, t)
gives the wave heights produced by 30knot winds blowing for t.hours. From the table (look at the row corresponding to
v
= 30), the function increases but at a declining rate as t
increases. In fact, the function values appear to be approaching a
limiting value of approximately 19, which suggests that 30knot winds cannot produce waves higher than about 19 feet. (c) h
= f(v , 30) means we fix tat 30, again giving a function of one variable. So, h = f(v, 30) gives the wave heights
produced by winds of speed "! blowing for 30 hours. From the table (look at the column corresponding to t = 30), the function appears to increase at an increasing rate, with no apparent limiting value. This suggests that faster winds (lasting 30 hours) always create higher waves. ©
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183
184
0
CHAPTER 14 PARTIAL DERIVATIVES
9. (a) g(2,  1) = cos(2 + 2(  1)) = cos(O) = 1 (b) x
+ 2y is defined for all choices of values for x andy and the cos ine function is defined for all input values, so the domain
of g is JR 2 . ·(c) The range of the cosine function is [ 1, 1] and x
+ 2y generates all possible input values for the cosine function, so the
range ofcos(x + 2y) is (1, 1].
(b) ft, .,fij, 4  x2

vz are defined o~ly when x ~ 0, y ~ 0, z ~ 0, and ln(4  x y2

z2
>0
{ ( x, y, z) 1 x 2 + y 2
x2
<=}
+ z 2 < 4,
2
y2


z 2 ) is defined when
+ il + z 2 < 4, thus the domain is
x ~ 0, y ~ 0, z ~ 0} , the portion of the interior of a sphere of radius 2, centered at the
origin, that is in the first octant.
13. ,.f2X=Y is defined only when 2x  y ~ O,or y::; 2x. So the domain off is { (x, y) I y ::; 2x}.
15. lll(9 x 2
9 x 2


9y2 ) is defined only when
9y 2
> 0, or ix 2 + y 2 < 1. So the domain off
is { (x, y) j ix2
+ y2 < 1}, the interior of an ellipse. y
, ,.,~,..    
 .........., 
,,~
17. J1  x 2 is defined only when 1  x 2 ~ 0, or x2
::;
1
¢:}
1 ::; x ::; .1, and
only when 1  y 2 ~ 0, or y 2
::;
\./1 1
19. Jy  x 2 is
y 2 is defined 1 ::;
¢*
y::;
ln addition, 1.
x
defi~ed only when y f
 x2
~
X
0, or y
is not defined if 1  x 2 = 0
<=}
= ±1. Thus the domain off is
{(x, y) I y ~ x 2 , x "I ±1}.
Thus the domain off is {(x,y)l1::;x::;1,  1 ::; y ::; 1}. y
1
1
0
I
X
 I
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~ x 2.
SECTION 14.1 FUNCTIONS OF SEVERAL VARIABLES
21 . We need 1  x 2
y2

soD= { (x, y , z) I x
2

z 2 ~ 0 or x 2 + y 2 + z 2 ~ 1,
+y +z 2
2
~
1} (the points inside
or on the sphere of radius 1, center the origin).
= 10 
4x  Sy or 4x + Sy + z
= 10, a plane with
185
23. z = 1 + y, a plane which intersects the yzplane in the line z = 1 + y, x
= 0. The portion of this plane for
x ~ 0, z ~ 0 is shown.
X
25. z
0
27. z = y 2
y
+ 1, a parabolic cylinder
intercepts 2.5, 2, and 10.
(0, 0, 10)
X
29. z = 9  x 2

9y 2 , an ell iptic paraboloid opening
downward with vertex at (0, 0, 9) .
31. z = )4 4x2 x
2
y2

y 2 so 4x2
+ y2 + z 2
= 4 or
z2
+ 4 + 4 = 1 and z ~ 0, the top half of an
ellipsoid .
.r
33. The point (  3, 3) lies between the level curves with zvalues 50 and 60. Since the point is a little closer to the level curve with
z = 60, we estimate that f( 3, 3)
~56.
The point (3, 2) appears to be j ust about halfway between the level curves with
zvalues 30 and 40, so we estimate !(3,  2) ~ 35. The graph rises as we approach the origin, gradually from above, steeply from below. © 2012 Ccnguge Lcnming. All Rights Rcscrwd. May no1 be scanned, copied. or duplicutcd. or posted to &I publicly ncccssiblc website, in Whole or in part.
186
0
CHAPTER 14
PARTIAL DERIVATIVES
35. The point (160, 10), corresponding to day 160 and a depth of 10m, lies between the isotherrnals with temperature values of8 and 12°C. Since the point appears to be located about threefourths the distance from the 8° C isothermal to the l2°C isothermal, we estimate the temperature at.that p~int to be approximately 11° C. The point ( 180, 5) lies betwe~n the 16 and 20°C isothermals, very close to the 20°C level curve, so we estimate the temperature there to be about 19.5° C. 37. Near A, the level curves are very close together, indicating that the terrain is quite steep. At B, the level curves are much farther apart, so we would expect the terrain to be much tess steep than near A, perhaps almost flat. 41.
39.
5
X
y
X
43. The level curves ~re (y  2x) = k or y = 2x ± 2
k
~
..Jk,
0, a fami ly of pairs of parallel lines.
45. The level curves are vfx + y
= k or y = vlx + k, a'
family of vertical translations of the graph of the root function y
=
vx. y
2
0 1
432 1
0
1234
2
47. The level curves are ye~
= k or y = ke x, a family of
49. The level curves are
J y2 
x 2 = k or y
2

2
x = k
2
k ~ 0. When k = 0 the level curve is the pair of lines
exponential curves.
y
= ±x. Fork > 0, the level curves are hyperbolas
with axis the yaxis.
0 X
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,
SECTION 14.1
51. The contour map consists of the level curves k = x 2 ellipses with major axis. the x axis. (Or, if k The graph of f(x, y) is the surface z = x
2
FUNCTIONS OF SEVERAL VARIABLES
0
187
+ 9y2 , a family of
= 0, the origin.)
+ 9y2 , an elliptic paraboloid.
X
If we visualize lifting each eUipse k = x 2
z
+ 9y2 of the contour map to the plane
= k, we have horizontal traces that indicate the shape of the graph of f . X
53. The isothermals arc given by k = 100/(1 + x 2
x2
+ 2y2 =
55. f(x , y)
(100 k)/k [0
= xy2 
+ 2y2 ) or
~ 100], a family of ellipses.
x3
The traces parallel to the yzplane (such as the leftfront trace in the graph above) are parabolas; those parallel to the xzplane (such as the rightfront trace) are cubic curves . The surface is called a monkey saddle because a monkey sitting on the surface near the origin has places fo r both legs and tail to res~.
4
X
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4
188
0
CHAPTER 14
59. z = sin(xy)
PARTIAL DERIVATIVES
(a) C
(b) ll
Reasons: This function is periodic in both x andy, and the function is the same when x is interchanged withy, so its graph is symmetric about the plane y = x. ln addition, the function is 0 along tl1e x andyaxes. These conditions are satisfied only by Can~
II.
61. z = sin(x y)
(a) F
(b) 1
Reasons: This function is periodic in both
X
andy but is constant along the lines y =
X
+ k, a condition satisfied only
by F and I .
(a) B
(b) VI
Reasons: This function is 0 a long the lines x
= ±1 andy = ±1. The only contour map in which this could occur is VI. Also
note that the trace in the x zplane is the parabola z
= 1
2
x and the trace in the yzplane is the parabola z
=1
y
2
,
so the
graph is B.
65. k
= x + 3y + 5z is a family of parallel plane$ w ith normal vecto r (1, 3, 5).
67. Equations for the level surfaces are k = y 2
+ z 2 . For k > 0, we have a family of circular cyl inders with axis the xaxis and
radius ,fk. When k = 0 the level surface is the xaxis. (fhere are no level surfaces fork
< 0.)
69. (a) The graph of g is 'the graph off shifted upward 2 units. (~)The graph of g is the graph
off stretched vertically by a factor of2.
(c) The graph of g i~ the graph off reflected about the xyplane. (d) T he graph of g(x, y) =  f(x , y) + 2 is the graph off reflected about the xyplane and then shifted upward 2 units.
71. f(x,y)
= 3x 
x4

4y
2

20r.
lOxy
10
0
y
y
Threedimensional view
Front view
It does appear that tlle function bas a maximum value, at the higher of the two "hilltops." From the front view graph, the maximum value appears to be approximately 15. Bolli h illtop~ could be considered local maximum points, as the values off there are larger than at the neighboring points. There doe~ not appear to be any local minimum point; although tlle valley sh(\pe between the two peaks looks like a minimum of some kind, some neig hboring points have lower function values.
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2012 Ccngage Learning. All Righls H.lscn'Cd. Mny not be scunncd, copied, or duplicated. or posted 10 a publicly accessible
wcbsil~.
in whole or in p;lr1.
SECTION 14.1
73.
FUNCTIONS OF SEVERAL VARIABLES
D
189
x+y +y
10
f (x, y) =   • As both x and y become large, the function values 2 2 X
5
appear to approach 0, regardless of which direction is considered. As
0
(x, y) approaches the ~rigin, the graph exhibits asymptotic behavior. From some directions, f(x , y)+ oo, while in others f(x , y) + oo.
5
(These are the vertical spikes visible in the graph.) If the graph is examined carefully, however, one can see that f(x, y) approaches 0 along the line y =  x.
y
75. f( x, y)
z=
2 eY
= ecx 2 +v2 • First, if c =
0, the graph is the cylindrical surface
(whose level curves are parallel lines). When c > 0, the vertical trace
above theyaxis remains fixed whi.le the sides of the surface in the x direction
z
"curl" upward, giving the graph a shape resembling an elliptic paraboloid. The level curves of the surface are ellipses centered at the origin.
c=O For 0
< c < 1, the ellipses have major axis the x axis and the eccentricity increases as c + 0.
y
1.2
c = 0.5 (level curves in increments of 1) For c = 1 the level curves are circles centered at the origin.
1.2
c
= 1 (level curves in increments of 1) [continued)
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190
0
CHAPTER 14
When c
PARTIAL DERIVATIVES
> 1, the level curves are ellipses with major axis theyaxis, and the eccentricity increases as c increases.
1.2
c = 2 (level curves in increments of 4) For values of c
< 0, the sides of the surface in the xdirection curl downward and approach the xyplane (while the vertical
trace x = 0 remains fixed), giving a saddleshaped appearance to the graph near the point (0, 0, 1). The level curves consist of a family of hyperbolas. As c decreases, the surface becomes flatter in the x direction and the surface's approach to the curve in the trace x = 0 becomes steeper, as the graphs demonstrate.
1.2
c =  0.5 (level curves in increments of0.25)
1.2
c 77. z = x 2
+ y 2 + =Y· When c <
= 2 (level curves in increments of0.25)
2, the surface intersects the plane z = k
I= 0 in a hyperbola. (See the following graph.)
It intersects the plane x =yin the parabola z = (2 + c)x 2 , and the plane x =  y in the parabola z = (2  c)x2 . These parabolas open in opposite directions, so the surface is a hyperbolic paraboloid. When c =  2 the surface is z = x 2
+ y2 
2xy = (x y) 2 . So the surface is constant a long each line x y = k . That
is, the surface is a cylinder with axis x  y = 0, z = 0. The shape of the cylinder is determined by its intersection with the plane x
+ y = 0, where z =
4x 2 , and hence the cylinder is parabolic with mini.ma ofO on the line y = x.
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SECTION 14.1
c =  5, z When 2
x2
< c:::;
=2
c =  10
surface and the plane z = k
y = 0 are parabolas z
191
c=  2
2xy = (x y) 2 ~ 0. IJthey have opposite signs, then cxy ~ 0. The intersection with the
> 0 is an ellipse (see graph below). The intersection with the s urface and the planes x = 0 and
= y 2 and z =
x 2 respectively, so the surface is an elliptic paraboloid.
> 0 the graphs have the same shape, but are reflected in the plane x = 0, because
x2 + y 2 + cxy = ( x) 2 + y 2 +
c
0
0, z ~ 0 for all x andy. Ifx andy have the same sign, then
+ y 2 + cxy ~ x 2 + 1i 
When c
FUNCTIONS OF SEVERAL VARIABLES
( c)(  x)y. That is, the value of z is the same for c at (x, y) as it is for cat ( x, y).
= 1, z = 2'
c= O
So the surface is an elliptic paraboloid for 0
79. (a) P = bLo: K 1  o:
=?
< c < 2, a parabolic cylinder for c =
= bL"' J< <>
p K
c = 10
::}
p = b(JL( ) J(
2, and ahyperbolic paraboloid for c
"'
ln : = lnb+ aln(~) (b) We list the values for ln(L/ K) and
ln(P/ K) for the years 18991922.
(Historically, these values were rounded to
2 decimal places.) Year
x = ln(L( K )
1899 1900 1901 1902 1903 1904 1905 1906 1907 1908 1909 1910
0  0.02 0.04 0.04 0.07 0.13  0.18  0.20 0.23  0.41  0.33 0.35
y
= ln(P/ I<.) 0 0.06  0.02 0  0.05  0.12  0.04  0.07  0.15 0.38  0.24 0.27
Year
19 11 1912 19 13 19 14 19 15 19 16 1917 1918 1919 1920 1921 1922
X =
ln(L/ K)  0.38  0.38  0.41 0.47 0.53 0.49 0.53  0.60  0.68  0.74  1.05  0.98
y
= ln(P/ K) 0.34  0.24  0.25  0.37  0.34  0.28  0.39 0.50 0.57  0.57  0.85 0.59
© 2012 Cengngc Lcnming. All Rights Rcscr\'cd. May not be sc:mncd, copied, or duplicoued. or posted 1o a publicly ncccssible website, in whole or in part.
> 2.
192
0
PARTIAL DERIVATIVES
CHAPTER 14
After entering the (x, y) pairs into a calculator or CAS, the resulting least squares regression line through the points is approximately y = 0.75136x
+ 0.01053, which we round toy= 0. 75x + 0.01.
(c) Comparing the regression line from part (b) to the equation y
a= 0.75 and lnb
p
= 0.01 =>
= ln b +ax with x = ln(L/ I<) andy= ln(P/ K), we have
b = e 0 ·0 1 ~ 1.01. Thus, the CobbDouglas production function is
= bL"' ](1a = 1.01£0.75 ](0.25.
14.2 Limits and Continuity 1. In general, we can't say anything about /(3, 1) I
lim
{x,y)•(3, 1)
= 6 means that the values of f(x, y) approach 6 as
f(x, y) ·
(x, y) approaches, but is not equal to, (3, 1). Iff is continuous, we know that lim
(x,y)(3,1)
=
f(x, y) = f(a, b), so
f(x, y) = f(3, 1) = 6. .
3. We make a table of values of
f(x,y)
lim
(x,y)<(a,b)
x2y3 + x3y2 _ 5 · fora set 2  xy
of (x, y) points near the origin.
l>z
 0.2
0.1
 0.05
0
0.05
0.1
0.2
0.2
2.551
2.525
2.5 13
2.500
2.488
2.475
2.451
0.1
2.525
2.513
 2.506
2.500
2.494
2.488
 2.475
0.05
2.513
2.506
2.503
2.500
2.497
2.494
2.488
0
2.500
2.500
 2.500
2.500
2.500
 2.500
0.05
2.488
 2.494
 2.497
2.500
2.503
 2.506
 2.5 13
0.1
2.475
 2.488
 2.494
2.500
2.506
2.513
 2.525
0.2
 2.451
2A75
2.488
 2.500
 2.513
 2.525
 2.551
As the table shows, the values off (x, y) seem to approach  2.5 as (x, y) approaches the origin from a variety of different directions. This suggests that
lim (:t:,!J)  {0,0)
f(x, y)
= 2. 5. Since f
defi ne d at (0, 0 ) , so we can use d1rect substltutwn to esta bl'ISI1 th at o
o
o
is a rational function, it is continuous on its domain. l'
1m (:r.,.y)>{0,0)
f
is
!( x, y ) = 0203 + 0302  5 =  5 , ven'fy'mg 2
2 0 0 0
our guess. 5. f(x, y)
= 5x3 
. 7. f( x, y ) =
x 2 y 2 is a poly nomial, and hence continuous, so
4  xy X2
+3y2
f(x, y)
+

f(x, y)
= f(l, 2) = 5{1) 3 
2
{1) (2)
2
= 1.
.tS a ratwna . .., . dh . . d . 1 ;unctiOn an encc contmuous on tts omam.
.
(2, 1) is in the domain off, so f is continuous there and 9. f(x, y) = (x4
' lim
(x,y) {1 ,2)
4y2 )/(x2
+ 2y 2).
lim
{x, y) < (2,1)
f(x, y) = !(2, 1) =
(~)~ +( 2~~~~ 2 4
First approach (0, 0) along the xaxis. Then.f(x, 0) = x /x
0. Now approach (0, 0) along theyaxis. For y =f. 0, f(O, y)
= 4y2 /2y2 =
2
= x2
2 7 for x
=f. 0, so
2, so f(x , y)+ 2. Since f has
two different limits along two different lines, the limit does not exist. ® 2012 Ccngoge Le3ming. All Rights Reserved. May nol he scanned, copied, urduplicated, or po5tcd ton publicly ncccssiblc website, 'in "'··hole or in p:m.
SECTION 14.2 11.
j(x , y) = (y 2 sin 2 x)j(x4
+ y 4 ).
On the x axis, f(x, 0) = 0 for x 2
.
x axis. Approaching (0, 0) along the hne y
lim sin x = 1, so f(x, y)+
x
:c o
·
13. f(x, y) = ~·
2
= x, f(x, x) = xx 4sin + x 4x
2
=
sin x x2 2
1 ( sinx) =2 ;
(0, 0) along the
2
for x :/: 0 and
has two different limits along two different lines, the limit does not exist.
·
~ I ~~~ lxl since lvl ~ x2 + y2 x2yeY X4
+ 4y 2
.
)x2
Then f(x, 0) = 0 for x
+.y 2 , and lxl + 0 as (x, y)+
'# 0, so f(x, y) +
(0, 0). So
f(x , y) + e0 /5
lim
(x,y)(o,o)
o:
f(x, y) =
0 as (x , y)+ (0, 0) along the x axis. Approaching 2
(0, 0) along theyaxis or the line y
= x also gives a limit ofO. But f (x, x 2 )
2
= x:: 4
::r2
4
tx)2 = x5: 2
::c2
T a:2
=
4
for x
'# 0, so
= t as (x,y)+ (0, 0) along theparabola y = x 2 . .Thus the limit doesn' t exist.
2
19. eY is a composition of continuous functions and hence continuous . xz is a continl.!ous function and tan t is continuous for
t '#
~
+ mr (nan integer), so the composition tan(a
f(x , y, z ) = ev~ tan{xz) is a continuous function for x z '# ~ + m r. Ifx = lim
(:r:,y ,z) (,..,0 ,1/ 3)
21.
f(x, y, z)
=
f(x, y, z)
+
193
= y 2 andy= x 2 . So we suspect that the limit exists and equals 0; we use the Squeeze Theorem to prove our
(0, O) such as x
15. Let f(x, y) =
'# 0, so f(x, y) + 0 as (x, y) +
0
We can see that the limit along any line through (0, 0) is 0, as well as along other paths through
xz +yz
assertion. 0
~·Since f
LIMITS AND CONTINUITY
x2
2
+ xz2 .
+y +z4 2
+ mr. Thus the product
and z
=i
then xz :j: ~
+ nw, so
= e02 tan(w · 1/ 3) = 1 · tan{w/ 3) = .;3.
f(x, y, z ) = f (w, 0, 1/3)
xy + yz
1r
~
Then f(x, 0, 0)
0. But f(x , x, 0) = x 2 /(2x 2 )
=t
= 0/x 2 = for x
0 for x
'# 0, so as (x, y, z)
'# 0, so as {x, y, z)
+
+
·
{0, 0, 0) along the xax1s,
{0, 0, 0) along the line y
= x, z =
0,
f( x,y,z) + ~· Thus the limitdoesn'texist. From the ridges on the graph, we see that as (x, y)
23.
+
(0, 0) along the
lines under the two ridges, f(x, y) approaches different values. So the limit does not exist. 0.5 y
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~·tB)' nol he scanned. copied. or duplicated. or posted
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194
CHAPTER 14
0
PARTIAL DERIVATIVES
25. h(x , y ) = g(f(x, y)) = (2x + 3y  6) 2 + y'2x + 3y  6. Since
is a. polynomial, it is continuo~s on JR2 and g·is
f
continuous on its domain { t I t ;::: 0}. Thus h is continuous on its domain.
D = {(x, y) I 2x + 3y 6;::: 0} = { (x, y) I y;:::  ~x + 2}, which consists of all points on or above the line y =  ~x + 2. From the graph, it appears that f is discontinuous along the line y = x.
27.
If we consider f(x, y) =
3 2
z
e 1 /(a:y)
as a composition offun.ctions,
g(x, y) = 1/(x y) is a rational function and therefore continuous except
1
where x y = 0
0 1
y = x. Since the function h(t) = et is continuous
=}
everywhere, the composition h(g(x,y)) = e 11(xv) = f(x,y) is 2
continuous except along the line y = x, as we suspected.
29. The functions xy and 1 + ex v are continuous everywhere, and 1 + ex v is never zero, so F(x, y) = xy is continuous . 1 + exy on its domain JR2 .
l +x2 +y2
31. F(x, y) =
1  x 2 y2
33. G(x, y) = ln(x 2 + y 2

is a rational function and thus is continuous on its domain
4) = g(f(x, y)) where f(x, y) = x 2 + y 2
4, continuous on ]~_2, and g(t) = In t, continuous on its

['t > 0}. Thus G is continuous on its domain { (x,_y) I x + y 2 2
domain {t
4

> 0}
= { (x , y ) I x 2
+ y 2 > 4 }, the_exterior
ofthe circle x 2 + y 2 = 4.
35. f(x, y, z ) = h(g(x, y, z )) where g(x, y , z) = x 2 + y 2 + z 2 , a polynomial that is continuous everywhere, and h(t) = arcsint, continuous on [ 1, 1). Thus f is continuous on its domain
{(x,y, z) l1 S x 2 + y 2 + z 2 S 1} = {(x,y,z)! x 2 + y 2 + z 2 S 1}, so f is continuous on the unit ball. if (x , y) =I= (0, 0) if (x,y)
The first piece off is a rational function defined everywhere except at the
= (0,0)
is continu~us on JR2 except possibly at the origin. Since x 2
origin, so
f
know that
jy3 \ + 0 as (x, y) + (0, 0). So, by the Squeeze Theorem,
s 2x
2
+
y2, we have jx2 y3/ (2x 2 + y 2 ) I.s jy3 j. We
.
2 3
lim (a;,y)  (0,0)
f(x, y) =
lim (x ,y)  (0,0)
~ y
2X
+ Y2
But f(O, 0) = 1, so f is discontinuous at (0, 0). Therefore, f is continuous on the set { (x, y) I (x, y) =/= (0, 0) }. 39 .
. 41.
3
3 3 3 + y (r cos 8) + (1' sin8) lim ( 3 . 3 B) .  2 = 1tm = r· cos 8 + rsm = 0 2 f y r> O+ 1' r~o+
lim

lim
e x2  y 2  1 x2 +y2
x
( x ,y)(0,0) X 2
(a;,y)> (0,0)
2
lim e r (  2r·) r  o+ 2r
[using !'Hospital's Rule]
2
= lim  e r =  e 0 =  1 r·o+
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= 0.
SECTION 14.3
43. f( x, y) =
{
sin(xy) xy
if (x, y)
1
if (x, y) = (0, 0)
PARTIAL DERIVATIVES
0
195
:f. (0, 0)
From the graph, it appears that f is continuous everywhere. We know
xy is continuous on R2 .and sin t is continuous everywhere, so
1
d sin(:cy) . . R2 . R sin (xy ) is contmuous on 2 an    IS contmuous on xy except possibly where x y = 0. To show that f is continuous at those points, consider any point (a, b) in R 2 where ab = 0. Because x y is continuous, x y + ab == 6 as {x, y ) + (a, b). If we lett= x y , then t+ 0 as (x , y) + (a, b) and lim (x,y)(a, b)
sin(xy). = Jim sin(t) = l by Equation 2.4.2 [ET 3.3.2). Thus xy to t
lim
f (x, y) = f(a , b) and f is continuous
(:z:,y)(a, b)
on R 2 • 45. since t
>
Jx  aJ2 = lxl2 + Jal 2 .  2JxJ ial cos O ~ Jxl2 + Jal2  2Jx.JJal = (lxJ Ial) 2 , we have llxl  lall
0 be given and set
o=
t. Then ifO
< lx  al < o, llxJ IaJI
$
Jx  aJ < o=
t . Hence
$
limxa Jxl =
lx aJ. Let Jal and
f (x) = Jxl is continuous on R ". 14.3 Partial Derivatives 1. (a) 8TI 8x represents the rate of change ofT when we fi x y and t and consider T as a function of the single variable x, which describes how q uickly the temperature changes when longitude changes but latitude and time are constant. 8T18y represents the rate of change of T when we fix x and t and consider T as a function of.y, which describes how quickly the temperature changes when latitude changes but longitude and time are constant. 8TI 8t represents the rate of change ofT when we fix x andy and consider T as a function oft , which describes how quickly the temperature changes over time for a constant longitude and latitude. (b) f .,(158, 21, 9t represents the rate of change of temperature at longitude 158°W, latitude 21 °N at 9:00AM when o nly longitude varies. Since the air is warmer to the west than to the east, increasing longitude results in an increased air temperature, so we would expect .f:r: (158, 21, 9) to be positive. / 11 (158, 21 , 9) represents the rate of change oftemperature at the same time and location when only latitude varies. Since the air is warmer to the south and cooler to the north, increasing latitude results in a decreased air temperature, so we would expect / 11 (158, 21, 9) to be negative. ft(158 , 21, 9) represents the rate of change of temperature at the same time and location when onJy time varies. Since typically air temperature increases from the morning to the afternoon as the sun warms it, we would expect f t(158, 21, 9) to be positive.
. 4, f T 3. () a By Definitlon
(
15 , 30)
lim f( = h0
15 + h,30) . by cons1'dermg . h h  f(  15,30) , w h'1c h we can approx 1mate
and h = 5 and using the va lues given in the table:
©
201 2 Ccngoge J.cnming. All Righ t" Rt:scr,.cd. May not~ .sc
=5
196
D
CHAPTER 14
PARTIAL DERIVATIVES
! 7'(15 ' 30) ~ f (10, 30) 5 f ( 15, 30) =  20  5( 26) = ~5 = 1.2' " f{  20, 30) f (15,30) h(  10, 30) ~ , _ 5
=
 33  { 26)
_5
.
7
= _5 = 1.4. Averaging these values, we estimate
fr( 15, 30) to be approximately 1.3. Thus, when tl1e actual temperature is  15°C and t he wind speed is 30 km/ h, the apparent temperature rises by about 1.3°C for every degree iliat the actual temperature rises.
. . I Stml 1ar y, and h
=
f ( 15 30) v , =
. . . f (  15, 30 +h)  f( 15, 30) 11m h wh1ch we can approxunate by considering h
l•  0
lO: f .., ( 15, 30) ~ f ( 15, 40)  f(  15, 30) 10
f( 15 20)f( 15 30) f v{15, 30) ~ ' _ ' 10
=
=
 24  (26)
_10
 27  (26) 10
=
_210
= 10
=  1 =  0.1, 10
.
=  0.2. Averaging these values, we estimate
f..,(  15, 30) to be appr~ximately  0.15. Thus, when the actual temperature is  15°C and ilie wind speed is 30 km/ h, ilie apparent temperature decreases by about 0.15°C for every km/ h that the wind speed increases. (b) For a fixed wind speed v, the values of the windchill index W increase as temperature T increases (look at a column of the table), so
a;
is positive. For a fixed temperature 1', the values of W decrease (or remain constant) as v increases
(look at a row of the table), so
~: is negative (or perhaps 0).
(c) For fixed values of T , the function values f (T , v) appear to become constant (or nearly constant) as v increases, so the corresponding rate of change is 0 or near 0 as v increases. This suggests that lim ( awI ov) = 0. V> 00
5. (a) If we start at (1, 2) and mo~e in the positive xdirection, the graph off increases. Thus f ,(1, 2) is positive. (b) If we start at (1, 2) and move in the positive ydirection, tbe graph off decreases . Thus fv (1, 2) is negative. 7. (a) f zx
= g, (!,), so f xx is the rate of change of f ,
in tl1e x direction. fx is negative at (  1, 2) and if we move in the
positive x direction, the surface becomes less steep. Thus the values of f., are increasing and !:ex (  1, 2) is positive. (b)
f 1111
is the rate of change of f 11 in tlle ydirection. / 11 is negative at (  1, 2) and if we move in the positive yd.i rection, the
surface becomes steeper. Thus the values of f 11 are decreasing, and / 1111 (  1, 2) is negative. 9. F irst of all, if we start at the point (3,  3) and move in the positive ydirection, we see that both b and c decrease, while a increases. Both b and c have a low point at about (3,  1.5), while a is 0 at this point. So a is defi nite ly the graph of j 11 , and one of b and c is the graph off . To see which is which, we start at the point (  3,  1.5) and move in the positive xdirection.
b traces out a line with negative s lope, while c traces out a parabola opening downward. This tells us that b is tlle xderivative of c. Soc is the graph of f, b is the, graph of f ,, and a is tbe graph of
f 11 •
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SECTION 14.3 PARTIAL DERIVATIVES 11. f(x, y) = 16  4x2
y2

0
197
=> f ,., (x , y) =  8x and fv(x , y) = 2y => f ,., (1, 2) =  8 and fv (1, 2) =  4. The graph
off is the paraboloid z = 16  4x2

y 2 and the vertical plane y = 2 intersects it in the parabola z = 12  4x 2 , y = 2
(the curve C 1 in the first figure). The slope of the tangent line to this parabola at (1, 2, 8) is f x(1, 2) =  8. Similarly the plane x = 1 intersects the paraboloid in the parabola
z ~ 12  y 2 ,
X
= 1 (the curve C2 in the second figure) and
the slope of the tangent line at (1 , 2, 8) is f v(1, 2)
=
4. X
JC
20 0
 20
2
2
Note that traces off in planes parallel to·the xzplane. are parabolas which open downwru:d for y and the traces of ! r:: in these p lanes are straight lines, which have negative slopes for y
< 0 and upward for y > O,
< 0 and positive slopes for y > 0. The
traces off in planes parallel to the yzplane are cubi_c curves, and the traces of fu in these planes are parabolas. 15. f( x, y) = y 5
19.
...:..
z = (2x + 3y) 10
by 23· f (x,y) = ax+ ex +dy x
)
aazX = 10{2a: + 3y) 0
=>
21. f(x , y) = x f y = x y
f Y ( ' 1Y
f x(x, y) = 0  3y =  3y, / y(x, y) = 5y4
3xy =>
1
=> fx( x, y ) =
=>
= (ex+ dy)(b} 
f  (x • )  ,y
=
y
0
1
2 = 20{2x + 3y) 9 ,
azy =
3x
10(2x + 3y) 0 3 = 30(2x + 3y) 0 0
.
= 1/ y, fv (x , y ) = xy 2 =  x / y 2
(ex + dy )(a)  (ax+ by)(c) (cx +dy) 2
(ax+ by)(d} (ex+ dy) 2

=
(ad · bc)y ' (ex + dy )2
= (be 
ad)x (ex + dy} 2
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198
D
PARTIAL DERIVATIVES
CHAPTER 14
25. g(u,v)
= (u2 v v3 ) 5
gu(u,v)
=?
gu(u, v) = 5(u2 v v 3 ) 4 (u2
= 5(u2v 
3v2 ) = 5(u2

v 3 ) 4 • 2uv
0 F11 (x,y) = o·y
Fx(x,y)= :xl" cos (e
=?
1"' cos(e) . dt = oy0 [ I.( cos(e) dt 11
.
t
v3 ) \
3v2 )(u2 v v 3 ) 4

. 29. F(x,y) = l " cos(et)dt
= 10u·u(u2 v 
t
]
1
'
dt=cos(e"' ) by theFundamentaiTheoremofCalculus, Partl;
)
0 =  oy
frv .. cos(et) dt = 
cos(e11 ).
11
33. w
= ln(x + 2y + 3z)
35. u = xysin ~(yz)
ou = xy . az
37. h(x, y, z, t)
1
= x + 2y + 3z'
ow oy
ow oz
2
= x + 2y + 3z'
3
= x + 2y + 3z
=>
1
J 1 (yz)2
xy2 (y) = ;:::.:=:::::::::==:::
J 1 y2z2
= x 2 y cos(z/t) =>
hz(x, y, z, t) 39.
ow ox
=>
h.,(x; y, z, t)
= 2xy cos(z/t),
= x2 y sin(z/t)(1/ t) = ( x2 y/t) sin(z/t),
u = Vx 21 + x 22 + · · · +
2 Xn·
For each t·
h11 (x, y, z, t)
ht(x, y, z, t)
= x2 cos(z/t),
= x2 y sin(z/t)( zC 2 ) = (x2 yz/t2 ) sin.(z/t)
= 1, ..., n, u,, = 21 ( x 21 + X22 + · · · + Xn2) 1/2 ( 2x; ) =
= 1n(x+ Jx2 + y2) '*
41 . f(x, y)
f., ( x,y) _
1 [ 1+21 ( x 2 +y2)  1/ 2(2x)] _ ~
x + Y x2
. so f.,(, 34= )
+ y2
1
3+~
(
3
1+
...f32+42
y
43. f(x,y , z)
= x+y + z
=?
fv(x,y,z)
2+(1) soj11 (2,1,  1) = (2 + 1 +( 1 )) 2
= x1l
45. f(x, y)
 x3 y
=
)
x
1 ( 1+ ~
+ Y x2 + y2
X ) ~ , Y x2 y2
= lim
.
l(x + y + z) y(l ) (x + y + z )2
x +z
= (x+y+ z )2'
1
4·
=>
h(y2

3x2 y  3xyh yh2 ) h
,...... o
f (
) _ li
x ,y 
h5
f(x, y + h)  f(x;y) _ h

· = lim {2xy + xh x /a.+0
+
=11 l . s ( + !s)! = s·
_ . f(x + h,y)  f(x,y) _ 1. (x +h)y2  (x + h) 3 y (xy 2 liD f "' (x, y )  ,.1liD .....o h h0 h .
11
Xi
y X~ + X~ + · · · + x?,
.
3
)
U:U h >0
= 2xy  x

x 3 y)
= lim (y 2 h ..... o

3x 2 y 3xyh yh 2 ) = y 2
x(y + h) 2

x 3 (y + h) (xy 2 h


3x2 y
x 3 y) _ 1'
 "~
h(2xy + xh x 3 ) h
3
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.
PARTIAL DERIVATIVES
SECTION 14.3
47. x 2 + 2y2 + 3z2
=1
=?

8
ax
8 (x 2 + 2y2 + 3z 2) =  (1) ox
8z 2x + 0 + 6z ox = 0 8z
8z =  2x =  x , and 8 (x 2 + 2y 2 + 3z 2) = 8 (1) 6z 8y 0y 3z 8x
ez '8 z = y ( X 8z
= xyz =>
49. ez
• 8z (e  xy) 8x
0 + 4y + 6z 8y = 0
ax
ax
=?
8z 6z 0y
_ 8z 8z e   xy OX ox
=?
= 4y
=?
=?
= yz =>
yz = yz, so 8z = e_.  x:y 8x • 8z e· oy
=> 8z
+ Z · 1)
8z 6z ax =  2x
=?
D

xy
8z
( e  xy ) 8 z = xz , so 8y
oy = xz =>
xz
ay = e= xy 51. (a) z = f( x ) + g(y)
(b) z = f(x + y).
. az 8y 53. f( x , y)
=
~:
=> Let u
=x+
df au= df (1) du 8y du
= x 3 y 5 + 2x 4 y
= ,q'(y)
. 8z df QU y. Then ox = du ox
=
df du (1)
=f
I
1
(u) = f (x + y) ,
= J'(u) = f'(x + y).
y) = 3x2 y 5 + Bx3 y , fu(x, y) = 5x 3 y 4 + 2x 4 . Then l rr::c (x, y) = fux(x , y) = 15x2 y 4 + Bx 3 , and f uu (x, y) = 20x3 y 3 • f,(x,
=?
f xu(x, y) = 15x 2 y 4 + 8x 3 ,
~
= /'(x),
6xy 5 + 24x2 y,
55. w = Ju2 + v2
uv (u2 + v 2)3/ 2 ' 2
1 · ~  v · ~(u + v 2 )  1 (2v) 2
W uv
=
(Ju2 +
x+y 57. z = arctan  
1  xy
Zx
v2)2
1
'
=
W uu
= v (21) ( u 2 +v2)  3/ 2 ( 2u) =. (u 2 +uvv2)3I 2'
Juz + v 1  v 2 j .,Ju2 + v 2 u2 + v2
u 2 + v2
=
v2 (u2 + v2)3/2
u2

=
(u2 + v2)3/2.
=?
_ 1 (1)(1xy)(x+y)( y)_ l+y2 1+y2 = 1 + (..:.±1L..) 2 . (1  x y)2  (1  x y)2 + (x +y)2  1 + x2 + y2 + x2y2 1  xy
1 + y2
1 = (1 + x2)(1 + y2) = 1 + x2 '
©
201 2 Cengogc Learning. All Righls R=rved. Moy nol be scanned, copied. or duplicnlcd. or posted loa publicly accessible wcbsilc. in whole or in part.
.
199
200
CHAPTER 14
0
PARTIAL DERIVATIVES
(1)(1xy)(x+y)(x) _ 1+x2 1 + ( x+u )2 . (1 xy)2 (1 xy)2 +(x + y)2
1 1 +x 2 (1 + x 2)(1 + y2) = 1 + y2 ·
1
=
zu
1xy
(
Thus U:cy = 61.
u
= 0, Zyx = 0,
Zyy
2y
= (1 + y 2) 2 . 2y =
=
U:c
sin(x2 y) · 2xy
= 2xysin(x2y),
= 2xy · cos(x2 y) · x 2 +sin(x 2y) · (2x) = 2x 3 ycos(x2y) 2xsin(x2 y) and
Uxy
= sin(x2y) ·x2 =  x 2 sin(x 2y),
63. f(x, y)
= x 4 y 2  x3 y =>
= 8x3 y 3x2 ,
f xv
67. u=er8 sin(J 2 ;
0
1t11 x
=
x 2 ·cos(x2y) ·2xy+sin(x2 y) · ( 2x)
= 2x 3 ycos(x2 y) 2xsin(x 2 y).
= Uux.
Thus Uxy
;
Zxy
Uy:c.
= cos(x2 y) =>
uu
+2xx 2 )2, 1
= 4x3 y 2 
fx
3x2 y, fxx
= 12x2 y 2 
6xy, fxxx = 24xy2

6y and '
2
fxvx = 24x y 6x.
=> :~ . =er 8 cosfJ+sinfJ·er 8 (r)=e'. 0 (cosO+rsinO),
= er8 (sinfJ) +(cosO+ rsinO) erO (0) = ero (sinO+ OcosB + rOsinO),
{)~:~O = ero (0 sin 0) +(sinO+ 0 cosO+ rOsinO) · e''8 X
69.w=y+2z=x(y+2z)1
83w {)z{)y{)x
OW 82 w . 2 2 ax=(y+2z)  I, 8yax = (y+2z) (1)=(y+2z) ,
=>
. .
= (2)(y +
2z) 3 (2)
(0) = Oero (2sin0 + Ocos 0 + rO sinO).
= 4(y +
4
2z) 3 = (y + z)3 and 2
ow {)y
2
= x(1)(y + 2z) (1) = x(y + 2z)
71. Assuming that the third partial derivatives off are continuous (easily verified), we can write
f(x, y, z)
= :J;y 2 z3 + arCSin ( X vz)
=?
73. By Definition 4, fx(3, 2) = lim !(3 + h, h 0
= 2xyz3 + 0, fvx = 2yz3 , and fyxz
fu
fxzy
= 6yz 2 =
=
fyxz·
2
,
Then
fxzy·
l
2
2 !(3 , ) which we can approximate by considering h = 0.5 and h = 0.5:
f (3 2 ) ~ !(3.5, 2)  f(3, 2) _ 22.4  17.5 _ 9 8 f (3 2 ) ~ !(2.5, 2)  f(3, 2) _ 10.2 17.5 _ 14 6 A . "' ' ~ 0.5 0.5  · ' "' ' ~ 0.5 0.5 · · veragmg these values, we estimate f,.(3, 2) to be approximately 12.2. Similarly, f x (3, 2.2) _
3 = h"""""+O lim !( +
22 3 22 h , · } !( , · ) which L
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SECTION 14.3
·
· · . . '
dI
we can approximate by cons1denng h = 0.5 an
f x(3, 2.~)
1
f (
f(2.5, 2.2)  f(3, 2.2) 9.3  15.9 _ _ = _ _ 05 05
~
To estimat~
)
=  0.5: "' 3, 2. 2 ::::::
PARTIAL DERIVATIVES
0
201
!(3.5, 2.2)  !(3, 2.2) 26.1 15.9 = = 20.4, 0 .5 0.5
. I I h f ( )' = 13. 2. Averagmg t 1ese va ues, we ave "' 3, 2.2 ~ 16.8.
f, 11 (3, 2), we first need an estimate for .fx(3, 1.8):
~ /(3.5, 1.8)  /(3, 1.8)  20.0  18.1  3 8
.fx ( 3 • 1.8 ) ~
0.5

0.5

.
f (3 1 8) ~ /(2:5, 1.8) !(3, 1.8)  12.5  18.1  1 2 . ' x ' . ~ 0.5  0.5  1. .
.
Averaging these values, we get f,(3, 1.8) ~ 7.5. Now fx 11 (x, y)
..
()a[
variables, so Defimtton 4 says that fx 11 x, y .
= ;::;uy
(
fx x, Y
))
=
a
= ;::;[f.,(x, y)) and fx(x, y) is itself a function of two uy li f,(x,y+h)fx (x,y ) m I · ho a
=>
. I· · 1 · · k ·11 f , 11 (3, 2) = lim f x(3, 2 +h) fx(3, 2) . We can estimate t liS va ue usmg our prev1ous wor Wit 1 a = 0.2 and h = 0.2: h
h o
) ~ j,(3, 2.2) f x(3, 2) = 16.8 12.2 = f ( 2 ) ~ f x (3, 1.8) J,(3, 2) 23 0.2 0.2 ' xy 3 ' 0.2
f xy (3 • 2 ~
= 7.5 
12.2 0 .2
= 23 5
· ·
Av~raging these values, we estimate f, 11 (3, 2) to be appr~ximately 23.25.
=
U:r:x
(x2
+ y2 + z2 )  3/2  x( ~) (x2 + y2 + z2)  5f2(2x) = .l
2y2  x2  z·2
By symmetry, u uu
= .(x2 + y2 + z2 )5/2
Thus Uxo: ·+ u,,y + Uzz .
79. Let v
= x +at,
=
2x2  y2  z2
and
U ::z
=
2z 2  x2  y2 (x2 + y 2 + z2 )5/2 .
+ 2y2  x2  z2 + 2z2 (X 2 +y + Z 2)"/2 "
x2  y2
w = x  at.
Then Ut =
8[f(v) + g(w)] f)t
df(v) 8v
u,
= f'(v) + g'(w) and Ux x = 8z
= in(e"' + e!l) => '8x
f"(v)
e"'
+ g"(w) . Thus Utt 8z 8y
=    and 
e"'
+ e!l
eY
2
= a
·
Ux:r.·
8z
8z
= , so  +e"' + eY Dx 8y
D2z e"'(e"' + eY) . e"'(e'") e"'+ 11 82z 0  e11 (e"') 8x 2 = (e +eY) 2 = (e"' + e11 )2' 8x8y=(e"' + ev) 2 = 8y2
dg(w) 8w
= ~ ot +~7ft = af'(v)  ag'(w) and
o[af'(v)a~ ag'(w)J = a[af"(v) + ag"(w) ) = a 2 [f"(v) + y"(w) ). Similarly, by using the Chain Rule we have
=
82 z
= 0.
?
uu
81. z
2x2  y2 z2 . (x2 + y2 + z2) 5/ 2
.
e11 (e"'
+ e 11 )  e 11 (e11 ) + eY )2
(e"'
(
e" +eY) 2
•
e"
=  e"' + e11
e·u
+ = e"' + eY
e"' e"'
+ eY + eY = 1.

ex+y (e"' +eY)2' and
T hus
e"'+Y (e"' + eu) 2
)2 =
(e"'+u? (e"'+Y? (e"' + eY)1  (ex + eu )1 = 0
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202
0
CHAPTER 14
PARTIAL DERIVATIVES
.83. By the C hain Rule, taking the partial derivative of both sides with respect to R 1 gives
8R 1 8R 8R 8R1 85. If we fix K
dP p
=
8[(1/Rl) + (1/R2) + (1/R3)]
=
8R1
or R
 2 8R 8R1
_
= Ko , P.(L,Ko) is a function of a single variable L, and ~~ =a
dL =>aL
Jp = Ja L dP
dL
R2 _Rf'
8R
=  R 1 2. Thus 8R1
I
=
is a separable differentia; equation. Then
=> In IPI = a In 141 + C (Ko), where G_(Ko) can depend on K 0 • Then .

IPI = e<> lniLIf;C(Ko), andsinceP > 0 andL >0, we have P = ec:dnLeC(Ko) = eC(T
87. ( P+
n a) V2 (~ .:...nb) = nRT => _
We can also wnte P
+
2
1 ( n a) 8T = nR P+ "'if'2 (V nb),so p = 8
T
n2a nRT V 2 = V _ nb
=> P
1
nR (1)(V  nb)
n RT n2 a _ nb  V2 = nRT(V  nb)  1
=V

V nb = ~
n 2 av 2 , so
2
8P ( )_ 2 ( ) 2n a nRT _3 2 1 +2naV = _V 3  (Vn'b)2' 8 v=  nRTV _nb 89. By Exercise 88, PV = mRT
.
PV
=> P =
8P 8V
mRT 8P mR v• so fJT = v· Also, PV = mRT
PV mR mR
Smce T .= mR' we have T fJT fJT = mR · V
8K
91. ;:;um
·
8K
= 21 v 2 , 8v = mv,
93. f:x:(x, y)
= x + 4y =>
everywhere but f:x:v(x, y)
82 K
~
uv
= m.
f xv(x, y)
·p
8K &2 K ·.8 m 8v 2
Thus 
= 4 and
fv(x, y)
__.._ . V _ mRT d 8V _ mR ..p an 8T p .
= mR.
= 21 v 2 m = K.
= 3x y =>
fvx(x, y)
= 3. Since f xy and f vx are continuous
f f 11.,(x, y), Clairaut's Theorem implies that such a function f(x, y) does not exist.
95. By the geometry of partial derivatives, the slope of the tangent line is f x(l , 2). By implicit differentiation of
4x2
+ 2y2 + z 2 = 16, we get8x + 2z (8zj8x) = 0 =>
8zf8x
=  4x/z, so when x = 1 and z =
8zj8x = 2. So the slope is fx(1, 2) = 2. Thus the tangent line is given by z  2 = 2(x  1), y parameter to bet = x  1, we can write parametric equations for this line: x 97. By C lairaut's Theorem, f xvv = (f:x:y) 11
= (/11x}y = fv xv =
Uv).,'ll
= 1 + t,
y
= 2, z = 2 
2 we have ·
= 2. Taking the 2t.
= Uv)yx = fvv x·
= x(x 2 )  3 12 e0 = ·x ix J' 3 • But we are using the point (1, 0), so near (1, 0), g(x) = x  2 • Then g'(x) =  2x 3 and g'(1) = 2, so using (l) we have fx(l, 0) = g'(l) =  2.
99. Let g(x)
101. (a)
=
f(x, 0)
(b) For (x, y)
f
(0, 0),
f x(x,y) =
(3x2y y3)(x2 + y2)  (x3y xy3)(2x) (x2+y2)2 x4y + 4x2y3 _ ys (x2 + y2)2
@) 2012 Ccngagc LC!Ulling. All Rights Reserved. ~·by not be scanned, copied, or duplicated, or posted too publicly ~cccssiblc website. jn whole or in pnrt.
SECTION 14.4
TANGENT PLANES AND LINEAR APPROXIMATIONS
D
203
(c) f (0 0) = lim f(h, 0)  f(O, 0) = lim {O/ h2)  0 = 0 and jy(O, 0) = lim j(O, h)  f(O, 0) = '0. "' ' J•O h h o h ho h 5
Y
4
(0 0) = of, = lim f.,(O, h) f , (O , O) = lim (  h  O) / h = 1 while by (2), ' "'Y ' oy h o h h o h
(d) B (3) f
f ,(0, 0) = ofy = lim jy(h, 0)  jy(O, 0) = lim h5/ h4 = ox
!I
(e) For (x,·y)
h
h .....o
,,..... o
h
1.
i= (0, 0), we use a CAS to compute x6
f xy(x, y)=
+ gx4y2 _ gx2y4 _ y6 · (x2+y2)3
Now as (x , y) > (0, 0) along the xaxis, fxv(x, y)> 1 while~
(x, y) > (0, 0) along theyaxis, fxy(X , y)
>
 1. Thus fx 11 isn't
continuous at (0, 0) and Clairaut's Theorem doesn't apply, so there is no contradiction. The graphs of f, 11 and fy x are identical except at the origin, where we observe the discontinuity.
14.4 Tangent Planes and Linear Approximations 1. z = f(x, y) = 3y 2  2x2 + x
=?
f .,(x, y) = 4x + 1, / 11 (x, y) = 6y, so f,(2, 1) = ~7, / y(2, 1) = 6.
By Equation 2, an equation of the tangent plane is z  (3) = f, (2 ,  1) (x 2) + f v(2, 1)[y ( :1)]
=:
z+ 3=  7(x  2)6(y + 1) or z=  7x  6y + 5.
3. z = f(x , y) =
.,fXY =>
f,(x, y) = ~(xy)  112 · y = ~gx, fu(x , y) = ~(xy) 1 1 2 · x = ~.../XlY, so f.,(1 , 1) = ~
and fv(1 , 1) = ~Thus an equation of the tangent plane is z  1 = f ,(1, 1)(x 1) + jy(1, 1)(y 1)
z 1 = ~(x 1) + Hv  1) or x + y  2z 5. z = f( x, y) = xsin(x + y)
=?
=?
= 0.
f , (x , y) = x · cos(x + y)
+ sin (x +
y) · 1 = x cos(x + y) + sin(x + y),
fv( x, y) = xcos(x + y), so f,( 1, 1) = (1) cosO+ sin 0 =  1, / 11 ( 1, 1) = ( 1) cosO =  1 and an equation of the tangent plane is z  0 = (1) ( x + 1) + (1) (y  1) or x + y 7. z = f(x , y)
= x2 +
xy + 3y2, so f,.(x , y)
= 2x
+y
+z
= 0.
=> f,(1, 1)
= 3, fv(x, y) = x + 6y
=?
fv(1, 1) = 7 and an
equation of the tangent plane is z  5 = 3(x 1) + 1(y 1) or z = 3x + 7y  5. After zooming in, the surface and the
® 2012 Ccngogc Le;uning. All Rights Res
204
D
CHAPTER 14
PARTIAL DERIVATIVES
tangent plane become almost indistinguishable. (Here, the tangent plane is below the surface.) If we zoom in farther, the surface and the tangent plane will appear to coincide.
_ xy sin {X  y) 2 2 1+ x +y
9· !( x,y) 
fv(x, y) =
. f ( ' ) ysin(x  y) + xycos(xy) ACAS g1ves "' x,y = 1 + x2 + y2
x sin (x  y)  xy cos ( x  y) 1 + x 2 + y2


2x 2 ysin {x y) d an ( 1 + x2 + y2)2
2xy2 sin (x  y) · ( 2 . We use the CAS to evaluate these at (1 , 1), and then 1 + x2 + y2)
substitute the results into Equation 2 to compute an equation oqhe tangent plane:
z= tx  tY· The surface and tangent
plane are shown in the first graph below. After zooming in, the surface and the tangent plane become almost indistinguishable, as shown in the second graph. (Here, the tangent plane is shown with fewer traces than the surface.) If we zoom in fart her, the surface and the tangent plane will appear to coincide.
z 0 1
11. f(x, y)
= 1 +x Jn(xy  5) :
and f v (x, y)
xy
The partlal derivatives are f , (x , y)
.
+ f v (2, 3)(y 
functions for y
=I
= l(x
t
y) )~'t( x +y
1
+ 4y
 23.
) = y/ (x + y) 2 and
+ y) 2 , so f x(2, 1) = ~and / 11 (2; 1) =
~. Both f, and j 11 are continuous
 x , so f is differentiable at (2, 1) by Theorem 8. The linearization of f at (2, 1) is given by
= !(2, 1) + f x(2 , 1)(x ®
xy +ln(xy 5) xy  5
and / 11 are continuous functions for
3) = 1 + 6( x  2) + 4(y  3) = 6x
_ x _ . The partial !;lerivatives are f x(x, y) x+y
fv(x , y) = x ( 1){x + y) 2 • 1 = x/ (x
L(x, y)
(y) + Jn (xy  5) · 1 =
is differentiable at (2, 3). By Equation 3, the linearization off at (2, 3) is given·by
L(x, y) = /(2, 3) + f x (2 , 3)(x 2)
~
5
1 = x · x y5 (x) = __£_ , so f x(2 , 3) = 6 and / 11 (2 , 3) = 4. Both f , xy 5 ,
> 5, so by Theorem 8, f
13. f(x, y)
= x · xy 1
2) + / 11 (2;1)(y  1)
= 1+ ~ (x  2) 
~(y  1) = ~ X  ~ y
+ ~·
20 12 Ccngage Learning. All Rights Reserved. May not be scanned, cop ied, or duplicated, or posted to D publicly ncccss iblc we bsite. in whole o r in part.
SECTION 14.4 15. f(x , y) = e"' 11 cosy.
TANGENT PLANES AND LINEAR APPROXIMATIONS
0
205
The partial derivatives are f,.(x , y) = e"' 11 ( y) cosy .= ye"'!1 cosy and
J11 (x , y) = e  "'V( siny) + (cosy)e"' 11 (  x) = e"'11 (siny + xcosy), so f.,(1T, 0) = 0 and / 11 (1T, 0) = 1r. Both f., and fv are continuous func;tions, so I is differentiable at (1r, 0), and the linearization of I at (1r, 0) is
L(x, y) = j(1r, 0) + /:r(1r, O)(x 1r) + j 11 (1r, O)(y  0)
= 1 + O(x 1r) 
2x + 3 17.Letf(x,y) = .Thenj.,(x,y)=  2 . andj11 (x,y) 4y + 1 4y + 1
1r(y  0)
= 1 1ry.
 12 = (2x+3)( 1 )(4y+1) _2 (4)=( 48x )2 .Bothf.,andl11 y+ 1
are continuous functions for y =I=  ~. so by Theorem 8, f is differentiable at (0, 0) . We have f:,,(O, 0) = 2, f 11 (0, 0) = 12 and the linear approximation off at (0, 0) is f(x , y) ~ f(O, 0) + f:,(O, O)(x  0) + / 11 (0, O)(y  0) = 3 + 2x 12y. 19. We can estimate /(2.2, 4.9) using a linear approximation off at (2, 5), given by.
f(x, y) ~ /(2, 5) + f,.(2, 5)(x  2) + / 11 (2, 5)(y  5) = 6 + 1(x  2) + ( 1 )(y 5) !(2.2, 4.9)
~
9. Thus
2.2 4.9 + 9 = 6.3.
21 .j(x,y,z)"=Jx2 +y2 +z2 =>
f:(x, y, z)
= x y +
f.,(x,y,z)= J 2 +x 2 + 2 ,f11 (x,y,z)= X y Z
= ,jx2 +y2 z , so f.,(3, 2, 6) = ~. +z2
J X 2 +y y2 +Z
2
,and
fv(3, 2, 6) = ~; f,(3, 2, 6) = ~·Then the linear approximation off
at (3, 2, 6) is given by l(x, y, z) ~ /(3, 2, 6) + j .,(3, 2, 6)(x  3) + / 11 (3, 2, 6) (y 2) + f z(3, 2, 6)(z'  6) = 7 + ~(x 3) + ~(y 2) +
Thus J(3.02) 2 + (1.97)2 + (5.99) 2 23. From the table, /(94, 80)
¥
= /(3.02, 1.97, 5.99) ~ ¥(3.02) +
~ (1.97) +
¥(5.99) ~ 6.9914.
= 127. To estimate /T(9~, 80) and /H(94, 80) we follow the procedure used in Section 14.3. Sine~
/(94 + h, 80)  /(94, 80) . . h" . . hh li Jr (94, 80) = "~ . h , we approximate t IS quantity w1t
= ± 2 and use the values g1.ven .m the
table: fr(94, 80) ~ /(96, 80)  /(94", 80) 2
Averaging these values gives /T(94, 80)
/H(94 , 80) ~ /(94, 85) ~ /(94, 80) Averaging these values gives /H(94, 80)
= 135 127 =
4,
2
fr(94, 80) :::::: /(92, 80)  /(94, 80) = 119 127 = 4  2  2
~ 4. Similarly, /J.I(94, 80) =
= 132 ~ 127 = 1, ~
lim /(
h~ o
94 80 • +
h~ /(94 •80 ), so we use h = ±5:
!u(94, BO) ~ /(94, 75) /(94, 80) = 122 127 = 1 5 5
1. The linear approximation, then, is
f(T, H) ~ /(94, 80) + fr(94, 80)(T  94) + /H(94, 80)(H 80) ~
©
127 + 4(T  94) + 1(H 80)
[or 4T + H  329]
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206
0
CHAPTER 14
PARTIAL DERIVATIVES
Thus when T = 95 and H = 78, !(95, 78) ~ 127 + 4(95 94) + 1(78  80) = 129, so we estimate the heat index to be approximately 129°F. 25. z = e 2 "' cos 27rt
=>
dz = aaz dx + f)z dt = e 2"' ( 2 ) cos .21rt dx . X 0t "3
27. m = p0 q ·
29. R
+ e2"' (
.
2
sm2?rt)(27r) dt = 2e "'cos 27rt dx 21re
x
.
sin 21rt dt
=> dm = am dp + om dq = 5p43 q dp + 3pa"2 q dq ap
= o:(32 cos 1
=>
oq
aR oR oR 2 2 . dR = oo: do: + (3 d(3 + O"f d"f = (3 cos '"Y do: + 2o:(3 cos 1 d(3  o:(3 sm 1 d1 0
31. dx = ll.x = 0.05, dy = ll.y
= 0.1, z
•
= 5x
2
+ y 2 , z., = lOx, Zy =
2y . .Thus when x. = 1 and y = 2,
dz
= zx(1, 2) dx + z 11 (1, 2) dy = (10)(0.05) + (4)(0.1) = 0.9 while
ll.z
= ! ,(1.05, 2.1) /(1 , 2) =
33. dA
2
5(1.05? + (2.1) 2
8A = 8A ox dx + oy dy = y dx + x dy and Jll.xJ

5 4 = 0.9225.
:::; 0.1, Jll.yJ :::; ~.1. We use dx
the maximum error in the area is about dA = 24(0.1) + 30(0.1)
. = 0.1, dy = 0.1 w1th x=
30, y
= 24; then
= 5.4 cm2 •
35. The volume of a can is V = 1rr 2 h and ll. V ~ dV is an estimate of the amount of tin. Here dV
= 27rr h dr + 1rr~ dh, so put .
dr = 0.04, dh = 0.08 (0.04 on top, 0.04 on bottom) and then ll. V ~ dV = 27r( 48)(0.04) + 7r(16)(0.08) ~ 16.08 cm3 . Thus the amount of tin is about 16 cm3 .
37. T =
";'gR
2r +R2
,
so the differential ofT is
dT = aT dR 8T d oR + or . r mg(2r2 =
=
2
2
2
2
(2r + R )(mg)  mgR(2R) dR + (2r +R )(0)  mgR(4r) d (2r 2 + R 2 ) 2 (2r2 + R2)2 r
R2 ) 4mgRr (2r2 + R2)2 dR (2r2 + R2)2 dr 
Here we have ll.R = 0.1 and ll.r
= 0.1, so we take dR = 0.1, dr = 0.1 with R = 3, r =
0.7. Then the change in the
tension T is approximately
dT
=
mg [2(0.7? (3) 2 ] 4mg(3)(0.7) (2(0.7) 2 + (3) 2 ]2 (O.l)  (2(0.7) 2 + (3)2)2 (0. 1)
0.84mg 1.642 0.802mg (9.98)2  (9.98) 2 =  99.6004 mg ~  0.0 165mg Because the change is negative, tension· decreases.
@) 2012 Ccngoge Learn ing. All Rights Re.scn•ed. May not be scanned, copied , or duplicated, or posted ton publicly accessible website, in whole or in p:m.
SECTION 14.5 THE CHAIN RULE
:~
39. First we find
0
207
implicitly by taking partial derivatives of both sides with respect to R 1 :
oR . R 2 oR = Ri. Then by symmetry, 1
2 2 1 17 oR R oR = R2 . When RI = 25, R2 = 40 and Ra = 50, R = 2 00 oR2  ~· oR3 R3
¢::}
R = 21°7°
n. Since the possible error
for each Ri is 0.5%, the maximum error of R is attained by setting t:J.R, = 0.005R.. So
oR oR oR 2( 1 1 1 ) 1 !:l.R ~=:::: dR = oR !:l.R1 + oR !:l.R2 + oR !:l.R3 = (0.005)R R + R + Ra = (0.005)R = l7 ~=:::: 0.059 st. 2 1 2 3 1
~ so l!:l.wl $ 0..02 and l!:l.hl $ 0.02. The relative . . error m . the calculated surface 41 . The errors in measurement are at ~ost 27o,
w
T
area is
0 425  1 )h0·725 dw + 0.1091w0.4 25 (0.725h0·725 0.1091w0.426h0.725 .
dS 0.1091(0.425w · s!:l.S s= 1'::::
To estimate the maximum relative error, we use :
1
}
dh
dw dh = 0.425:;:; + 0.725/i:
I~w I= 0.02 and ~~i = I~hI = 0.02
=
=>
dS = 0.425 (0.02) + 0. 725 (0.02} = 0.023. Thus the max imum percentage error is approximately 2.3%.
s
43. !:l.z = f(a + !:l.x, b + !:l.y)  f(a, b) =(a+ !:l.x? + (b + !:l.y) 2  (a2 + b2 )
= a 2 + 2a!:l.x + (!:l.x) 2 + b2 + 2b!:l.y + (!:l.y? a 2

b2 = 2a !:l.x + (!:l.x? + 2b!:l.y + (!:l.y?
But f,(a, b) = 2a and /y(a, b) = 2b and so !:l.z = f x(a, b) !:l.x + /y(a, b) !:l.y + !:l.x !:l.x + !:l.y !:l.y, which is Definition 7 with e: 1 = !:l.x and e: 2 = !:l.y. Hence f is differentiable. 45. To show that f is continuous at (a, b) we need to show that
equivalently
lim
(6x ,6y) (O,O )
lim
(z,y)~(a,b)
f(x, y) = f(a, b) or
f(a + !:l.x, b + !:l.y) = f(a , b). Since f is differentiable at (a, b),
f(a + !:l.x, b + !:l.y)  f(a, b)= !:l.z = f,(a, b) 6x + /y(a, b) !:l.y + e:1 !:l.x + e:2 !:l.y, where €1 and (!:l.x, !:l.y}
>
€2 +
0 as
(0, 0). Thus f(a + !:l.x, b + !:l.y) = f(a, b)+ f,(a, b) !:l.x + / 11 (a, b) !:l.y + e:1 !:l.x + e:2 !:l.y. Taking the limit of
both sides as (.C:..x, .C:..y) ___.. (0 , 0) gives
lim
( 6 x,61J) (0 ,0)
f (a
+ .C:..x, b + .C:..y) =
f (a , b). Thus f is continuous at (a, b).
.
14.5 The Chain Rule 1. z =
x? + y 2
+ xy,
3. z =·.J1+ x 2 + y 2 ,
x =sin t, y = et x = In t, y
~
&~
&~
=> d =,.d +,. d =(2x+ y}cost+(2y +x)e t
uX
t
uy
t
t
= cost =>
8zdy= 1 (1+x2 + y) 2 1/ 2 . 2 1/2 dz =oz  dx + (2x)·1 + 21 (1+x2 + y) (2y)(smt) = 2 dt ox dt 8y dt t ·
1
.J1 + x2 + y2
(x . )   ysmt
© 2012 Cc:ngagc Lc:aming. All Righ rs Reserved. Mny not be sc.mncd, copied. or duplicalcd, or posrOO 10 a publi~ly occcsslblc website, in whole or in part.
t
208
D
CHAPTER 14
PARTIAL DERIVATIVES
5. w=xevl =, x = t2, y = lt, z=1+2t =>
(.!) .( 1) +
dw = OW dx + ow dy + ow dz = ey/ z. 2t + xevl= dt ox dt /Jy dt oz dt z
(J!...) .2 = evl=(2t z
xeYI=
2
~  2xy) z
z2
7. z = x 2 y 3 , x = scost, y = ssint => az oz ax oz 8y 3 2 2 • = !i"'"' <> + a a = 2xy cost+ 3x y sm t us uxus y s .
!i"'"
az . 8z Ox az 8y s . . 2 2 3 2 2 ot = ox at+ oy = (2xy )( ssmt) + (3x y )(scost) = 2sxy smt + 3sx y cost
at
9. z = sin() cos, () = st2 , ¢ = s 2 t
=>
~; = ~; ~~ + ~~ ~! =(cos 8 cos q))(t~) + (~; = ~; ~~ + ~~~~ 11. z = er cos 0, r = st, 8 =
dz dt
=(cos() cos ¢>)(2st) + (sinO sin )(s
J s2 + t 2
/Jf dx
= OX dt
/Jf dy
I
I
•
'
& u& 8y u8y au = e ' au = cos v, au = e ' av =
·
Sill v.
g..,(O, 0)
=
a1
= 62.
=> ~
·
M&
By the Cham Rule (3), au = ox au
Yu(O, 0) = f ,., (x(O , 0), y(O, 0)) Xu(O, 0} + fv(x(O, 0), y(O, 0)} y,.(O, 0)
a1 ax
= 2stcos0 cos¢ s 2 sinO sin
(6)(5) + (  8)( 4)
15. g(u, v) = f(x(u ,v) , y(u, v)) wherex = eu + sinv, y = eu + cosv
. . ag Similarly, OV
)
2stsin0 sin
By the Chain Rule (2),
= f ,., (2, 7)g (3) + fv(2, 7) h (3) =
+ oy dt
2
= t 2 cosO cos¢ 
=?
= 3, x = g(3) = 2 andy = h(3) = 7.
13. When t
sinO sin q.,)(2st)
+
M8y
/Jy au. . Then
= / :c (1, 2)(e0 ) + fv(l, 2)(~
0
) = 2(1)
+ 5_{1) = 7.
av
= OX av + oy av . Then
f ., (x(O, 0), y(O, 0)) x.,(O, 0) + fv (x(O , 0), y·(o, 0)) Yv (0, 0)
=
f,. (1, 2)(cos O)
+ fv(l, 2)( sin 0)
= 2(1} + 5(0) = 2 17.
u
II
/~y
= f(x , y), x = x (r, s , t), y = y(r, s , t) =>
X
/ 1\ /1\
r
s
I
r
s
I
© 2012 Ccng.:~gc Lcuning. All Right. Ro= rvc:d. May not_be scaruu:d, copied, or dupllcntcd , or posted too publicly accessible wcb:ritc, in whole or in p:lrt.
SECTION 14o5 w
19o
r
/Is ~
210 z
J
X
y
= r(x, y),
ow
ow as
ow or
s
D
209
= s(x, y), t = t(x, y) =>
ow at
aw
aw ar
aw as
ow at
fu = &ofu+asax+8tax·~=aroy+asay + Bt~
r
/\ /\ 1\
X
w = f(r, s, t), r
THE CHAIN RULE
J
X
= x 4 + x 2 y, x = s + 2t 
u, y
= stu2 =>
oz = az ox+ az.ay = ( 4xa + 2 xy)( 1) + (x2)(tu2), OS OX OS ay as ' 0
oz
=
oz ax+ az ay
m &m
~m
= (4x3 + 2xy)(2) + (x2)(su2)
·
oz = az ax+ az ay = (4x3 + 2xy)(  1} + (x2)(2stu). au ax au ay a·u  When s = 4, t = 2, and u = 1 we have x = 7 and y = 8, so a z = (1484)(1} + {49)(2} = 1582, aaz = {1484) (2) + (49){4) = 3164, aaz = (1484)(1) + {49)(16} = 700. as t u 23o w = xy + yz + zx, x = r cos 0, y = r sin 8, z = r8 => aw = awax + aawaay + aawaaz ar axar yr zr
~(y +z}(cos8} + (x+z)(sin8}+(y+x)(O), .
aw = aw ax+ aaw aay + aaw aaz8 = (y + z)(  rsinO) + ao ox o8 . y 8 z
(~ + z )(rcosO) +
(y + x}(r·)o
aw When ro = 2 and()= ?r/2 we have x = 0, y = 2, and z = 1r, so ar = (2 + rr}(O) + (0 + rr){1) + (2 + 0)(1r / 2) = 27r and
~~ ..
= (2 + 1r) (  2) + (0 + 1r)(O) + (2 + 0)(2) = 2rro
25o N = p + q, p = u + vw, q = v + uw, r = w + uv =9 p+r aN = aN ap au ap au
'
+ aN aq + aN ar . aq au
8r au
= (p + r){1) (p + q){1) (1) + (p + r)(1)  (p + q)(O) (w) + (p + r) (O)  (p + q)(1) (v) (p+r)2 (p + r)2 (p + r)2 =
(r q) + (p+r)w  (p+q)v (p+r)2
oN = aN ap + aN oq + aN or = r  q (w) + p + ro (1) + (p + q) (u) = (r  q)w + (p + r}  (p + q)u ' av op ov 8q 8v or ov (p+r)2 (p +r) 2 (p+ r)2 . (p+1')2 0
aN _ aN ap ow  ap ow
+ 8N
aq aq ow
+ 8N !!!.._ = or OW
ro  q (v) + p+r (u)+ (p +q) (1) (p + r )2 (p + r) 2 (p + r) 2
· When u = 2, v = 3, and w = 4 we have p = 14, q = 11, and r = aN _ ( 1)(4)+24(25)(2) _ 30 __ ~ d 8N av (24)2  576 96' an aw
~0,
= (r q)v +(p+r)u (p + q) (p + r )2
aN 1 + (24)(4)  (25)(3) 20 5 so ou = . ( )2 . = = , 24 576 144
= ·( 1)(3) + (24)(2)25 =~ = ~
(24)2
576
144 °
© 2012 Ceng;~gc LCJming. AJI Rights Reserved. Mn)' nol be scanned. co ph.d, or duplicated, or posted lo o publicly accessible websitt:, in whole or in p;~rt.
l
210
CHAPTER 14
0
PARTIAL DERIVATIVES
= x 2 + y 2 , so let F(x, y) = ycosx  x 2 y sin x  2x 2x + y sin x = :'"
27. y cosx
dy
dx
cos x  2y
y
2
= 0. Then by Equation 6
cosx 2y .
+
[2xy  (1 + y 2 ) (1 xiy 2 )]/(1 + x 4 y 2 ) (x2  2xy(1 + x4y 2 ))/(1 + x4y 2 )
and
(1
_ 
+ y 2 )(1 + x 4 y 2 ) 
2xy x 2  2xy(1+ x4y2)
 1 + x4y2 + y2 + x4y4  2xy x 2  2xy  2x 5 y3
31. x 2
+ 2y2 + 3z 2 = 1, so let F(x, y, z) = x 2 + 2y 2 + 3z2 
= .:_ Fx = _ 2x = ~
f)z ox _ 33. e·
Fz
6z
and
3z •
= x yz, so let F(x, y , z) = e· 
8z
 xz
F11
= e=xy =
 Fz
oy
oz oy xyz
= _ Fy Fz
= _ 4y QZ
oz
1
= 0. Then by Equations 7
= _ 2y
=. 0. Then~ =ux
3z .
F. .
F 'z
yz
yz
= .=e·  xy ez  xy
xz
ez___x_y
35. Since x andy are eac~ functions oft, T( x, y) is a function oft, so by the Chain Rule, 3 seconds, x
Then~~
dx
1
~ = ~: + ~~~;.After
= .Jl+t = .,;r+3 = 2, y = 2 + ~ t = 2 + H 3) = 3, d t = 2 .J1+t 1 +t
= Tx(2, 3)
and
~~ + Ty(2 , 3) ~; = 4(t) + 3(~)
37. C = 1449.2 + 4.6T 0.055T 2
.!.
1
2 v'1 + 3  4,
and dy .!. dt  3 .
= 2. Thus the temperature is rising at a rate of2°C/ s.
ac = 4.6 + 0.00029T 3 + 0.016D, so 8T
ac = 0.016..
O.llT + 0.00087T and oD 2
According to the graph, the diver is experiencing a temperature of approximately 12.5°C at t = 20 minutes, so
.c:;, = 4.6  0.11 (12.5) + 0.00087(12.5) dD
dt
~
1
dT
'2 and (jj
~
2
~ 3.36. By sketching tangent lines_at t = 20 to the graphs given, we est~ate
1 . dC . Then, by the Cham Rule, (jj 10
dT 80 dD = 80 &T (jj + &Ddt ~ (3.36) (Iii) + (0.016) ( 2) r::::: 0.33. 1
1
.
Thus the speed of sound experienced by the diver is decreasing at a rate of approximately 0.33 m/s per minute. 39. (a) V = l!wh, so by the Chain Rule, dV dt
= 8V dl! + 8V dw + oV dh = wh dl! + l!h dw + l!w dh = 81! dt
aw dt
&h dt
dt
dt
dt
2 . 2 . 2 + 1 . 2 . 2 + 1 . 2 . ( _ 3)
= 6 m3/s.
© 2012 Ccngagc Learning. All Rights RescrYcd. May not be scanned. copied, or duplicated. or, posted toP publicly accessible website, in whole or in part.
SECTION 14.5
THE CHAIN RULE
D
211
(b) S = 2(£w + £h + wh), so by the Chain Rule,
~~ = ~~: + ~! :~ + ~~ ~~ = 2(w +h):+ 2(£ + h):~ + 2(£+ = 2(2 + 2
2
2
(c) L = £ + w + h
dP
2£
::::?
~~ = 2f. ~ +
2w
:~ +
2h
~: =
~~
= 10 m fs 2(1)(2) + 2(2)(2) + 2(2)(3) = 0
::::?
= 0 mjs.
dL/ dt
41.
2
2)2 + 2(1 + 2)2 + 2(1 + 2)(  3)
w)
2
dT
T
dt = 0.15, V = 8.31?
dt
= 0.05,
dV dt
= 8 31 [ 0.15 .
20
and
dV
8.31 dT
T dP 8.31 p 2 Yt· Thus when P
dt = Pdt 
= 20 andT =
320,
_ (0.05)(320)] ~ _ 0 .27 Lf s. 400
43. Let x be the length of the first side of the triangle andy the lengtl1 of the second side. The area A of the triangle is given by A= ~xy sin B where 0 is the angle between the two sides. Thus A is a function of x, y, and (}, and x, y, and 0 are each in tum
functions of timet. We are given that
dA
dt
8A dx = ax dt
and 0 =
+
8A dy ay dt
+
8A dO ao dt
~~ = 3, ~~ ·= 2, and because A is constant, :~ = 0. By the Chain Rule, dA
::::?
.
1 0· dt = 2YSI.Il
dx dt
.
1 0· + 2xsrn
dy
dt
1 + 2xycos(} ·
d8
dt. When x = 20, y
/6 we have
1r
0=
~(30)(sin ~)(3) + ~(20)(sin ~)(2) + ~(20)(30)(cos ~) ~~·
= 45 · l2
+ 300 · vf3 · dO = ll2 + 150 v,) '3 d(} 2 dt dt
 20 · l2
. .or ., d dO g1.ves ddO = ;;;  25/2 = :r.i· 1 · at a rate o f So Ivmg sot he angIe between the s1·des 1·s decreasmg t t 150 v 3 . 12 v 3
1/ (12
~ 0.048 radj s.
V3)
45. (a) By the Chain Rule, :
or = ({Jz) 2
(b)
2
2
(
 r sin B)
+ ~~ r cos 0.
2
2
47. Letu = x 
~= cos(} + ~; sin B, ~; = ~=
2 ({)z . ( {)z ) • 2 ax ) cos 0 +.2 8{)zx {)z 8 y cosO ~mO + By sm 0,
8 z ) + r2 1 (az 80 ) ( 8r
®
=
2
=
2
[(az 8x ) + (az) 8y
dz dz 8u y.Then 88 zX = d  = d U8X U
and
2 ]
2
2 2 . (cos 0 + sm 0) = (az 8x ) + (oz 8y ) .
8z 8z a8zy = ddzU (1). Thus a + 8 y = 0. X
2012 Ccngogc lc:uning. All Rights Res
= 30,
212
D
CHAPTER 14
49. Let u
PARTIAL DERIVATIVES
=X + at, v =X  at. Then z = f(u)
+ g(v), so az jau ~ f'(u) and azjav ~ g1 (v) .
az f}z aU az av 1 1 Thus at = au at + av at = af (u) ag (v) and 2
= a at 2._ [/1( u ) 
8 z [)t2
. ,arly .,8z S. tmt uX
51 .
1
= a 2/"(u ) + a 2g" (v ) .
I( )] = (df'(u) 8u  dg (v ) au ) g v a du 8t dv 8t 2
2 2 8 z = !"( u) + g "(v )' . Thus ";5'2 8 z =a2 <:12· 8 z = fl( u ) + g I(v ) and <:12 uX ut ux
8z 8z .,=aazX 2s ·+ .,2r. Then uS vy
2
. . . 8 z By the contmutty of the parttals, "' 8 rus
2
z 8z + (4r 2 + 4s 2 ) 8axuy "' + 2 · 8y 2
2
8 z + 4rs  2 8y
= 4rs , 88 x z2
az .
8z az [}z 8z ' 8z 53.  = ., cos (} +a sm9and {)(} = .,r smO +a rcosB. Then 8r uX y vX y 2 z a iJr2
= cos 0
(8
2
tPz
z
.
8:r:2 cos (} + ay 8x sm ()
= cos2 (} 8~Xz2
)
. + sm ()
(8
2 z . f):y2 sm (} +
2
8 z ) ax 8y cos ()
. ~z . ~z + 2 cos 8 sm (J + sm 2 8 a 2 8X 8y 1J
and
a.,. + ( rsinO) (a<:12z (r sin(J) + ~a 8 z ) rcos9
2
Dz
~ a(J
2
=
rcos(J "'
uX
2
uX
az + r cos f) r s~ (J .,vy
uy
(8 a y 2
2
z r cos 8 + ~ 8 z ( r sin 9)) 2 uxuy
2
8z . 8z 8 z . =  r cos (J .,  r sm (} ., + r 2 sm 2 f) a2 vx
uy
,
X
X

·
~z 8 2z . 27·2 cos 8 sm (} ~ + r 2 cos 29 <:12 v X vy
uy
Thus
a2
2
z 1 a z 1 fJz <>2 + 2 <:i2 +  .,vr r 'v8 r v1·
= (cos2 0 +
2
2
. a z ( . 2 ) 8 z sm2 0)  2 + sm 2 (} + cos (J  2 8y 8X
.
1 . 8z 1( az . az ) cos(} {)z  1 sm 7' ax r B + r cos(}  + sm (J ay
ay
az 2
= aX2
ax
az . a as destred. 1J2 2
+
55. (a) Since f is a polynomial, it has continuous secondorder partial derivatives, and
f(tx, ty) = (tx) 2 (ty ) + 2(tx)(tv? + 5(ty) 3 = t 3 x 2 y + 2t3 xy2 + 5t3 y3 = e(x2 y + 2xy2 + 5y3 ) = t 3 f (x, y). Thus, f is homogeneous of degree 3. ® 20 12 (engage Ll.'itming. All Rights RCSc.'f\'Ctl.
M:~y
nna be SC3.Mcd. coried, or duplicated, or postld to a publicly accessible website, in whole or in pan.
SECTION 14.6
DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR
0
213
(b) Differentiating both sides of f(tx , ty) = t" f(x , y) with 'respect tot using the Chain Rule, we get
at [t" f(x,y) ]
8 8 8
t f(tx , ty) =
¢:>
8
8 8(tx) (tx) f(tx , ty) · {j't
Setting t = 1: x
8
+ 8 (ty)
f(tx, ty) ·
8 8 f(x , y) + y f (x, y) .8 y 8X
57. Differentiating both sides of f(tx , ty)
8
ax f(tx,ty)
8 = 8x [t"'f(x , y)]
8 8(tx) a (tx) f.(tx , ty) . a;;
8(ty) 8 a at = x a(tx ) f(tx, ty) + y 8 (ty) f(tx , ty) = nt"
1
f(x , y).
=.nf(x, y).
= tn f(x , y) with respect to x using the Chain Rule, we get
¢:>
8
+ 8 (ty)
f(tx , ty).
8(ty) n a;:=t
8 8x f(x, y)
tfx(tx, ty)
¢:>
= tn f,.(x, y).
Thus f ,. (tx , ty) = t"· 1 J,.(x , y) . 59. Given a function defined implicitly by F(x, y) = 0, where F is differentiable and F11 =J: 0, we know that
.
G(x , y) =  F,. so ddy Fy X 2
d y dx2
=  F,. . Let ~
= G(x, y). Differentiating both sides with respect to x and using the Chain Rule gives
= 8G dx + 8G dy 8x dx
~~
~
ay dx
where 8G 8x
= ,E_ ( 8x
Fx) Fy
= _ F11 Fxx 
F,.F
11 ., ,
FJ
8G ay
= ,E_ 8y
(F.,)
=
Fy
T~us 2
d y= ( dx 2
FuF:.,,.  F,.Fyx) (l) F112
+ ( FyFxll 
FxF,m)
F112
(Fa;) F11
Fxx F;  FyxFxFy  F,.uF11Fx + FyyF; FJ But F has continuous second derivatives, so by Clauraut's Theorem, F11 ,
d211 F.,.,F;  2F.,11 F, Fu dx2 = F.3
+ FuuF;
= F.,11 and we have
d . ed as es1r .
II
14.6 Directional Derivatives and the Gradient Vector 1. We can approximate the directional derivative of the pressure function at K in the direction of S by the average rate of change
of pressure between the points where the red line intersects the contour lines closest to K (extend the red line slightly at the left). In the direction ofS, the pressure changes from 1000 millibars to 996 millibars and we estimate the distance between these two points to be approximately 50 km (using the fact that U1e distance from K to S is 300 km). Then the rate of change of pressure in the direction given is approximately
3. Du /{ 20, 30) /T(  20, 30)
= 'V j
{ 20, 30) · U
996
1000 ; 0
= 0.08 millibar/km.
= /T{ 20, 30) ( 7z) + /v{ 
20 ~ hlim !( + h, 30)t  !( +0
20, 30) (
?z).
20 30 • ), so we can approximate /T( 20, 30) by considering h
= ±5 and
© 2012 Ccngage Learning. All Rights Rc>en'Cd. Muy not be scnnncd. copied, or duplicoted, or posted to a publicly oe<:C>!Iiblc website, in whole or in part.
214
0
CHAPTER 14
PARTIAL DERIVATIVES
using the values given in the table·:
fr( 20, 30) ~
!( 25 30)  f( 20 30) '
Similarly, f,( 20, 30)
f (20 1J
,
_
'
= ye "'
_
'
10
=
34  (  33) 10
=  30_10( 33)
(  33) 5
= 1.4,
= 1.2. Averaging these values gives fr( 
20, 30) ~ 1.3.
=
 01 . '
= 0.3. Averaging these values gives
f ,( 20, 30)
~ 0.2.
~ 1.3(}z) + (0.2>(}z) ~ 0.778. fx (x, y)
:::>
= ye_., and fu(x, y) = e"'. If u is a unit vector in the direction of 8 = 21r / 3, then
from Equation 6, Du /(0, 4) = f x (0, 4) co~ 7. f(x, y)
~· f ( 20 • 30) =  26 
20 30 20 = h0 lim /(  • + h~ !(  • JO), so we can approximate f,( 120, 30) with h = ±10:
!(  20 20)  J( 20 30)
Then Duf( 20,30)
15 30 • )
39  (  33) _5
=
'
5
30) ~ !(  20, 40)  J(  20, 30) 10
f,(  20, 30) ~
5. f(x, y)
h ( 20, 30) ~ f (
en + f v(O, en 4) s in
= 4 · ( ~)
+ 1 · {1
= 2+
4.
= sin(2x + 3y). • I
8 (a) 'il f (x, y) = . 1 i 8X
(b) 'il f( 6,4)
+ ~! j = [cos(2x + 3y) · 2] i + [cos(2x + 3y) · 3]j = 2 cos {2x + 3y) i + 3 cos (2x + 3y)j •
~J
= (2cos0) i + (3cosO)j = 2 i + 3j
(c) By Equation 9, Du f(  6, 4) = 'il f( 6, 4) · u = (2 i
+ 3j ) · Hv'3 i 
= x 2 yz xyz 3 (a) 'il f( x, y, z ) = (f,. (x, y , z), fv(x, y, z), f:(x, y, z)) = (2xyz 
j ) = ~ (2v'3 3) = v'3  ~·
9. f(x , y, z)
(b) 'il/(2, 1, 1) = (4 + 1, 4 2,4 + 6) (c) By Equation 14, Duf(2,  1, 1) 11. f(x , y)
= e"' si.ny =>
= (3, 2, 2)
= 'il /(2,  1, 1) · u = (3, 2, 2) · ( 0, ~.  ~) = 0 + ~ 
unit vector in the direction.ofv is u =
J<:>+8 2
2
( 6, 8)
v '<"7f(O , 1r. / 3) · U _D u f(o , Tr / 3) 
(:.2 1) ( 5•5 3 4) _
vector in the direction ofv is u =
Vl;H2 (i + 3 j )
15. f(x, y, z)
2 •2
= (28 i 
= xeV + ye= + ze"'
:::>
·
=
= 1~ (6, 8) = M

10
i, V. so
+ 10il _ 4313 10 ·
12 j ) · ~(i + 3j) = ~ (28  36) '\1 f(x, y, z)
(
'ftu Ci + 3j), so
= (ell +
vector in the direction ofv is u = ..;2s.!.I+4 (5, 1, 2) = D u f(O, 0, 0)
~ = ~·
= (e"' s iny, e"' cosy}, '\1 f(O , Tr/3) = ( 4, ~ ),and a
'il f(x, y)
Dug(.2, 1) = 'ilg(2, 1) · u
yz 3 , x 2 z xz3 , x 2 y  3xyz2 )
ze"', xeV
=  ?to or 4 ~.
+ e=, ye= + e"'), 'ilf (O, 0, 0) = (1, 1, 1), and a unit
7:io (5, 1, 2), so
= "V/(0, 0,0) · u = (1, 1, 1) · ~ (5, 1,  2) = ~·
© 2012 Ccngagc learning. All Rights Reserved. Moy not be seaMed, copied, or duplicated, or posled to u publicly accessible websile, in whole or in pott.
SECTION 14.6
17. h(r, s, t) = ln(37· + 6s + 9t)
+; 4+3& (4i + 12j + 6k) = 4
Duh (1,1, 1) = '<'7h( v 1, 1, 1)
19. f(x, y)
= VxY =>
0
215
=> 'ilh(r, s, t) = (3/(3r + 6s + 9t), 6/(3r + 6s + 9t), 9/(3r + 6s + 9t)),
(k, ft, ~ ), and a unit vector in the direction ofv =
'ilh(1, 1, 1) = is u = .;16
DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR
'il f(x , y)
~ i .+ ~ j + ~ k, so
( 6•3•2 l 1 1) 2 6 3) 1 · ( 7•7•7  21
· U =
=
+ 12j + 6 k
4i
+ 72 + 143
yt=' ~)·so 'il!(2, 8) = ( 1, ~ ) . \2 yxy 2yxy
(Hxy) 1 (y), Hxy) 112(x)) 1 2

 23 42'
=/ .
The unit vector in the direction of PQ = (5  2, 4  8) = (3, 4) is u = ( ~,  ~ ), so
.
. 1 (3  2 Duf(2,8)='il/(2,8)· u= ( 1,4)· 5•6" ) 5.
21. f(~,y) = 4yy'X => 'ilf(x,y) = ( 4y · ~x 1 1 2 ,4v'x) = (2yjy'x,4.,fi).' · 'il f(4 , 1) = {1, 8) is.the direction of maximum rat~ ofch~ge, and the maximum rate is·/'il /(4, 1)/ = v'1 + 64 =
=> 'il f(x, y) = (y cos(xy), x cos(xy)), 'il!(1 , 0) = (0, 1). Thus the maximum rate of change is
23. J(x, y) = sin(xy) /'il f(1, 0)/
v'65.
= 1 in the direction {0, 1).
25. f(x,y ,z) =
Jx2 + y2 + .z2
'il f(3, 6, 2) = (
=>
7,w. ~· ~) =
( ~.
¥, ¥ ). Th~s the maximum rate ofchange is
27. (a) As in the proof of Theorem 15, Du f = /'il !I cos 0. Since the minimum value of cos(} is  1 occurring when(} =
minimum value of Du f is  /'il f/ occurring when(} = (assuming \7f
(b) f(x , y)
= x 4y 
x2 y3
y = x
that is when u is in the opposite direction of 'il f.
=> 'il f(x, y) = ( 4x3 y  2xy3 , x 4
=
¢'>

3x2 y 2 ), so f decreases fastest at the point (2, 3) in the
{12, 92) = (12, 92).
29. The directionoffastestchange is 'ilf(x,y)
+j
the
"f 0).
direction 'il /(2 ,  3)
parallel to i
11',
11',
= (2x 2) i + (2y 4) j , so we need to find all points (x,y) where 'ilf(x,y) is
(2x  2) i + (2y  4)j = k (i + j)
+ 1, so the direction of fastest change is i + j
¢'>
k
= 2x 2 and k =
at all points on the line y = x
2y  4. Then 2x 2 = 2y 4
+ 1.
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=>
216
0
CHAPTER 14
PARTIAL DERIVATIVES
k
k
Jx2 + y2 +z2 and 120 = T(1, 2, 2) = 3 so k = 360.
31. T =
 (1,  1, 1) (a ) uv'3 , DuT(1, 2, 2) .
= 'VT(1, 2, 2) · u = [.:._360( x 2 + y 2 + z2) S/2 (x, y, z)]. {1,2,2) . u = 
(b) From (a), 'VT = 360(x
2
+ y 2 + z 2 )  S/ 2 (x, y, z), and since (x, y, z)
.12 (1, 1, 1) 3 (1, 2, 2). 4.. v3
= ~ 3v3
is the position vector of the point (x, y, z), the
vector  (x, y, z), and thus 'VT, always points toward the origin. 33. 'VV(x, Y •.z)
= (lOx 3y + yz, xz 
3x, xy), 'VV(3, 4, 5)
{a) Du V(3,4,5) = {38, 6, 12} · ~(1, 1, 1} =
= (38, 6, 12)
'7a
(b) 'VV(3, 4, 5) = (38, 6 , 12), or equivalently, (19, 3, 6).
(c) IV'V(3,
4, 5)1 = v'382 + 6 + 12 = v'l624 = 2 v'406 2
2
t
t
35. A unit vector in the direction o(AB is i and a unit vector in the direction of AC isj. Thus D_..... f(1, 3) = f.,(1, 3) ~ 3 and AB
D f(l, 3) = / y(1, 3) AC
= 26. Therefore \7/(1 , 3) = (!.,(1, 3), /y(l , 3)) = (3, 26}, and by definition, t
DAD f (1 , 3) = V' f · u where u is a unit vector in the direction of AD, which is (fa, H). Therefore, DAD f (1, 3) = (3, 26) . .( 153, H) = 3. fa 37. () a
'["7(
v
+ 26. H =
312;.
au;
b)=(o(au~+bv) ' 8(au+bv)) =\a au bav au bav)= a (au a ~ + {) , a ~ + ~ a , au; ~ + b\av ~ 'a
au + v
y
uX
ux
X
uy
au , nu · au)  nu
'["7
uy
X
uy
ux
y
.= a 'Vu + b 'Vv
n  ( (d) '["7 vu 
39. f( x, y) = x
3
o(un) , 8(u")) ax
oy
+ 5x 2 y + y 3
.
( nun 
1
n 1
ox
n 1
{)y
vu
"*
Duf(x,y)
= 'Vf(x , y) · u = (3x2 +
D~.J(x,y)
= Du (Duf(x ,y)] =
lOxy, 5x2 + 3y2) · (~, ~)
\7 [Duf(x,y)] · u
=
~x 2 + 6xy+ 4x2 + ¥v 2
= ~x
2
2 + 6xy + ¥y . Then
= (¥x + 6y, 6x + \1Y) · (~, ~)
= l;s4x + Jfy + ¥x + ~y = 2is4x + ~ar,oY and D~f(2 , 1) =
2iu4
(2) +
12866 (1)
= 72754 .
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SECTION 14.6 41. Let F(x, y, z) = 2(x  2) 2
+ (y 
1?
+ (z 
DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR
3) 2 . Then 2(x  2)
2
+ (y 
F:.(x, y,z) = 4(x  ·2)
=?
F,(3, 3, 5) = 4, F 11 (x,y,z) = 2(y 1)
Fz(x,y, z) =2{z  3)
=?
Fz{3,3, 5) = 4.
1f
=?
+ (z 
3)
+y +z =
217
= 10 is a level surface of F.
F 11 (3,3,5) = 4, and
(a) Equation 19 gives an equation of the tangent plane at {3, 3, 5) as 4{x  3) + 4{y  3)
4x + 4y + 4z = 44 or equivalently x
2
0
+ 4{z 5)
= 0
<=>
11.
. 20, the normaI I'me has symmetnc . equat10ns . x 3 y  3 (b) By EquatiOn   =  4 4
. Ientl y = z 4 5 or eqUiva
x  3 = y  3 = z  5. Corresponding parametric equations are x = 3 + t, y 43. Let F(x , y, z ) = xyz2 • Then xyz2· = 6 is a level surface ofF and 'V F(x, y, z )
=;
3 + t, z = 5 + t.
= (yz2 , xz2 , 2xyz).
(a) 'V F(3, 2, 1) ~ (2,3, 12) is a normal vector for the tangent plane at {3, 2, 1), so an equation of the tange nt plane is 2(x  3)
+ 3(y 
2) + 12(z  1)
= 0 or 2x + 3y + 12z = 24.
(b) The norma l line has direction (2, 3, 12), so parametric equations are x
.
.
= 3 + 2t, y = 2 + 3t,
z = 1 + 12t, and
x3 . y2 z 1 =   = l 2. 3
symmetriC equations are  2
45. Let F (x , y, z) = x
+y +z 
e"'Yz. Then x
+y +z =
e"'11 :; is the level surface F(x, y , z )
= 0,
and 'V F(x, y, z) = (1  yze"'Y:;, 1  xze"'11"", 1  x ye"'11 z).
= (1, 1, 1) is a normal vector for the tangent plane at {0, 0, 1), so an equation of the tangent plane is 1(x 0) + 1{y  0) + 1(z  1) = 0 or x + y + z = 1.
(a) 'V F(O, 0, 1)
(b) The normal line has direction (1, 1, 1), so parametric equations are x
= t, y = t, z = 1 + t, and symmetric equations are
x = y =z 1. 47. F(x, y , z) = xy + yz
+ zx, 'V F (x, y, z) = (y + z , x + z, y + x), 'VF{1, 1, 1) = (2, 2, 2), so an equation of the tangent plane is 2:r; + 2y +2z = 6 or x+ y + z = 3, and the normal line is given by x  1 = y  1 = z  1 or x = y = z. To graph
3  xy the surface we solve for z : z =   . x+y
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218
CHAPTER 14 PARTIAL DERIVATIVES
D
49. f(x, y)
= xy
=> '\1 f(x , y)
= (y, x}, '\1 !(3, 2) = (2, 3). '\1 !(3, 2)
is perpendicular to the tangent line, so the tangent line has equation
=> (2, 3) · (x 3, x 2) = 0 =>
'\1](3, 2) · (x  3, y 2} = 0
2(x  3) + 3(y  2) = 0 or 2x + 3y = 12.
51. '\1F(xo, yo, zo)
2xo (i2 x xo
2 x
2xo 2yo 2zo J . . = I\ (i2 • 1)2• C2 . . Thus an equatiOn oflhe tangent plane at (x 0 , y0 , zo) 1s
2yo 2zo + 1)2 Y + C2 z
( x5
= 2 a2
() .
. .
. .
z~ ) + y~ b2 + c2 = 2 1 = 2 smce (xo, yo, zo) IS a pomt on the elhps01d. Hence
fth . . + b2 y + 2zo z = 1 1s an equation o e tangent p1ane. c
· Yo
a
(
53. '\1 F xo, yo, zo
2xo 2yo  1 J ' . . 2xo 2yo ) = I\ (i2 • 1)2, c ,so an equatiOn of the tangent plane 1s (i2 x 1: 1)2 y 
2xo 2yo or  2 x + b2 y
a
2
2
x =zc + 2 ( % + yb~ ) a
zo
zo
c
c
  . But 
2
2x5 2y5 zo c1 z = (i2 + 1)2  c
2
x . . = % + yb~, so the equat1on can be written as a I
2xo 2yo z + zo  2 x+  2 y =   . a b c 55. The hyperboloid x 2

y2

z2
= 1 is a level surface of F(x, y, z) = x2 
y2
 .z
2
and '\1 F (x, y , z)
= (2x, 2y, 2z) is a
nonnal vector to the surface and hence a nonnal vector for the tangent plane at (x, y, z). The tangent plane is parallel to the
= x + y or x + y z = 0 if and only if the corresponding normal vectors are parallel, so we need a point (xo, yo, zo) on the hyperboloid wh~re (2xo, 2yo,  2zo) = c (1, 1·, ....,1) or equivalently (xo,  yo, zo) = k (1, 1, 1) for some k ¥: 0. plane z
Then we must have x o = k , yo =  k , zo = k and substituting into the equation of the hyperboloid gives k2

( k)
2
~ k2 = 1 {:}
 k 2 = 1, an impossibility. Thus there is no such point on the hyperboloid.
57. Let (xo, y 0 , z 0 ) be a point on the cone [other than {0, 0, 0)]. The cone is a level surface of F(x, y, z) = x 2
+ y2 
z 2 and
= (2;t, 2y,  2z), so '\1 F(xo, yo, zo) = (2xo, 2yo, 2zo) is a nomial vector to the cone at this point and an equation of the tangent plane there is 2xo (x  xo) + 2yo (y  y0 )  2zo (z  zo) = 0 or xox + yoy  zoz = x~ + y5 z5. But x& + v5 = za so the tangent plane is given by xox + YOY  ZoZ = 0, a plane which always contains the origin. '\1 F(x, y, z)
59. Let F(x, y , z) '\1 F(1, 1, 2) y
= x 2 + y2 
z. Then the paraboloid is the level surface F (x, y, z)
= 0 and '\1 F (x , y ; z) =
(2x,2y, 1), so
= '(2, 2,1) is a nonnal vector to the surface. Thus the normal line at (1, 1, 2) is given by x = 1 + 2t,
= 1 + 2t, z = 2 t. Substitution into the equation of the paraboloid z = x 2 + y~ gives 2 t = {1 + 2t)2 + {1 + 2t?
2 t
= 2 + 8t + Bf
{:} 8t2 + 9t = 0 {:}
t(Bt + 9) = 0. Thus the line intersects the paraboloid when t
corresponding to the given point (1, 1, 2), or when t
=  ~,corresponding to the point (%, ~, ¥).
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= 0,
{:}
SECTION 14.6
DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR
0
219
61. Let (xo, y0 , zo) be a point on the surface. Then an equation of the tangent plane at the point is
y 2..;Yo
X
 +  + 2y'Xo
= + ..;Yo + _]f_
y'Xo
Z
JXO+Jy0 + /Zo
2yZo
2
r.;:: =
. But
. · · JXo + ..;Yo + v'Zo = vr;.c, so the equation IS
.;c. The x, y, and zintercepts are foO, ICYo and .jCZO respectively. (The X intercept is found by
~=
v zo
setting y = z = 0 and solving the resulting equation for x, and they and zintercepts are found similarly.) So the sum of the intercepts is .jC( JXo + ffo + v'Zo) = c, a constant.
63. If f(x, y, z ) = z  x 2
y 2 and g(x, y, z) = 4x2

+ y 2 + z 2 , then the tangent line is perpenoicular to both "il f
and "ilg
at (  1, 1, 2). The vector v = "il f x "ilg will therefore be parallel to the tangent line. We have "il f(x , y, z ) = (  2x, 2y, 1)
=?
"il f ( 1, 1, 2) = (2,  2, 1), and "il g(x, y , z ) = (8x, 2y, 2z) k
j
"Vg( 1, 1, 2)= (8, 2,4).Hencev="Vf x "Vg =
=?
2  2 1 =  10 i16 j12k.
8
2 4
Parametric equations are: x = 1  lOt, y = 1  16t, z = 2  12t.
65. (a) The direction of the normal line ofF is given by "il F, and that of G by "VG. Assuming that
"il F 1= 0 1= "VG, the two normal lines are perpendicular at P if "il F · "VG = 0 at P (8Fj8x, 8F/8y,8F/8z) · (8G j 8x, 8G/ 8y, 8G/ 8z ) = 0 at P (b) Here F = x 2
+ y2

z 2 and G = x 2
+ y 2 + z2
"il F · "VG = (2x, 2y,  2z) · (2x, 2y, 2z) = 4x2

#
F_..,G,
¢:?
+ FvGv + FzGz =
0 at P.
r 2 , so
+ 4y2 
4z 2 = 4F = 0, since the point (x , y, z ) lies on the graph of
F = 0. To see that this is true without using calculus, note that G = 0 is the equation of a sphere centered at the origin and F = 0 is the equation of a right circular cone with vertex at the origin (which is generated by lines through the origin). At any point of intersection, the sphere's normal line (which passes through the origin) lies on the cone, and thus is perpendicular to tl1e cone's normal line. So the surfaces with equations F 67. Let u
= (a , b) and v
= (c, d). Then we know that at the given point, Du f
= 0 and G =
= "il f
0 are everywhere orthogonal.
· u = afx
+ bfv and
+ dfv· But these are just two linear equations in the two unknowns fx and fv, and since u not parallel, we can solve the equations to find "V f = (f.,, f v) at the given point. In fact,
Dv I = "il I · v = clx
"Vf =
I d Du f \
 b Dv f , a Dv f  c D u f ad  be ad  be
).
@) 2012 CcnJ:!ISC Lc:uning. All Rights Rosel"\·ed. May nol be SC4Med, cop kd, or duplicated, or postod lo n publicly accessible wobsite, in wbolc or in port.
and v are
220
D
14.7
CHAPTER 14
PARTIAL DERIVATIVES
Maximum and Minimum Values
1. (a) First we compute D(1, 1) = f,x(1, 1) fvv(1, 1) [f:cy(1, 1W = (4)(2) f o:x (1, 1)
> 0, f
(1? =
7. Since D(1, 1)
> 0 and
has a local minimum at (1, 1) by the Second Derivatives Test. 2
(b) D(1, 1) = f.xx (1, 1) fuy(1, 1) [fxu(1, 1W = (4)(2) (3) =  1. Since D(1, 1)
< 0, f
has a sadd le point.at (1, 1) by
' the Second Derivatives Test.
3. In the figure, a point at approximately (1 , 1) is enclosed by level curves wh'ich are oval in shape and ind icate that as we move away from the point in any direc~ion the values off are increasing. Hence we would expect a local minimum at or near (1, 1) . The level curves near (0, 0) resemble hyperbolas, and as we move away from the origin, the values off increase in some directions and decrease in others, so we would expect to find a saddle point there. To verify our predictions, we have f(x, y)
=4+
x3 + y 3

have critical points where these partial derivatives are equal to 0: 3x2 first equation into the second equation gives 3(x2 ) 2

3x = 0
=> f x (x , y) = 3x 2
3xy
3y = 0, 3y

=> 3x(x 3

2

1) = 0

3y, f v (x, y) = 3y 2

3x. We
2
3x = 0. Substituting y = x from the
=> x = 0 or x = 1. T hen we have
two critical points, (0, 0) and (1, 1). The second partial derivatives are f xx (x, y) = 6x, fxu(x, y) = 3, and fvu(x, y) = 6y, '
'
2
2
so D(x, y) = fxx(x , y) fuv(x, y) [fxu(x, y)] = (6x)(6y) (3) = 36xy 9. Then D(O, 0) = 36(0)(0)  9 =  9, and D (1, 1) = 36(1)(1)  9 = 27. Since D(O, 0) D (1, 1)
> 0 and fxx(1, 1) > 0, f
5. f(x, y) = x 2 + x y + y 2 + y
< 0, f has a saddle point at (0, 0) by the Second D erivatives Test. Since
has a local minimum at (1 , 1).
=> f x
= 2x + y, fv = x + 2y + 1, fxx = 2, fx 11 = 1, fvv = 2. Then fx = 0 implies
y=2x,andsubstitutionintofy=x + 2y + 1=0 gives x+2(~2x) +1=0 Then y =~and the only critical point is
D(x, y) = fxxfvv  (fru) 2 = (2)(2) 
=> 3x=1 => x = i·
(t, ~).
(1? =
3, and since_
D(~,  D = 3 > 0 and fxx(i, ~) = 2 > 0, !(~ , ~) = ~ is a local minimum by the Second Derivatives Test.
7. f(x,y) = (x y)(1 xy) = x  y  x 2 y+ xy 2
=> f x = 1  2xy +y2 , fv =  1 x 2'+ 2xy, fx:n = 2y,
f xv =  2x + 2y, fvv = 2x. Then fx = 0 implies 1  2xy + y 2 = 0 and / 11 = 0 implies 1  x 2 + 2xy = 0. Adding the two equations gives 1 + y 2
©

1 x 2 = 0
=>
y2 = x2
=>
y=
±x, but ify = x then f x = 0 implies
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SECTION 14.7
1 '+ 2x 2 + x 2
=0
3x2
=}
then substitution into f,
MAXIMUM AND MINIMUM VALUES
0
221
= 1 which has no real solution. l fy = x
= 0 gives 1 
2x
+ x2 = 0
2
x
=}
2
=1
=}
x = ±1, so the critical points are (1, 1) and ( 1, 1). Now D (1, 1) = (  2)(2)  02 =  4 < 0 and
D(  1,  1) = (2)( 2) 0 2 = 4 < 0, so (1, 1) and ( 1,  1) are saddle points.
9. f(x, y)
= y 3 + 3x 2 y 
fw
= 6y 
3y2

12y
6x 2
12. Then f z
=0
6y 2
+2
=}
f ,. = 6xy  12x, ! 11 = 3y 2 + 3x2  12y,
f x:t
= 6y 12,
f xu
= 6x,
= 0 implies 6x(y 2) = 0, sox = 0 or y = 2. If x = 0 then substitution into j y = 0 gives
3y(y 4)
=}
substitution into jy

=0
=}
= 0 gives 12 + 3x2 
y = 0 or y
24 = 0
=}
= 4, so we have critical points (0, 0) and (0, 4). If y
x2
=4
= 2,
=}
x = ±2, so we have critical points (±2, 2).
= ( 12)( 12)  02 = 144 > 0 and fu(O, 0) =  12 < o•.so f(O, 0) = 2 is a local maximum. D{O, 4) = (12){12)  02 = 144 > 0
D(O, 0)
and f xx(O, 4) = 12
D( ±2, 2)
11. f(x, y)
4y 4
= {0)(0) {±12)2 = 144 < 0, so (±2, 2) are saddle po ints.
= x3 
implies x 2
= 4y
12xy + _8y3
=}
fx
= 3x 2 
12y, f u =  12x + 24y2 , fxx
y
= 6x, fx 11 =  12,/1111 = 48 y . Then fx = 0
·= 4y and / 11 = 0 implies x = 2y 2 • Substituting the second equation into the first gives {2y 2 } 2 = 4y =}
x = 0 and if y D(O, 0)
> 0, so f(O , 4) = 30 is a local minimum.
4y(y 3  1)
=0
=}
y
= 0 or y
= 1. lfy
=}
= 0 then
= 1 then x = 2, so the critical points are (0, 0) and {2, 1).
= {0){0) 
(  12) 2 =  144 < 0, so (0, 0) is a saddle point.·
D(2, 1) = (12)(48) ( 12?
= 432 > 0 and fxx(2 , 1) =
12 > 0 so
!(2, 1) = 8 is a local .minimum .
13. f(x,y)=e"'cosy
=}
f x=e"'cosy, / 11 =e"'siny.
= 0 implies cosy = 0 or y = f + mr for nan integer. But sin ( f + mr) =1 0, so there are no critical points.
Now f a:
© 2012 Cengage Le:sming. All Rights Rcscn·ed. Mny not be seaMed, copied~ or duplicated, or posted to a {iublicly accc.s.~iblc wc:bsilc, in whole or in pan.
4
222
D
CHAPTER 14
PARTIAL DERIVATIVES
l" = ( x 2 + y 2) e'~~ 2  :r2 (  2x) + 2xe'~~2 :r2 = 2xe'~~2  :r 2 (1  x. 2  y 2 ),
fzx = 2xe'fl2  o:2 (  2x) + (1 x 2  y 2 ) ( 2x ( 2xe'~~2 :r2) + 2e'fl2 f xv
= 2xe'112 2 o: (
2y) + 2x(2y)eY
22 :r
2
2
= 0 implies y = 0, and substituting into fx 2
2xe"' (1  x 2 )
=0 ~
=
22
(1  x  y ) = ~4xyeY  x (x 2
f 11 v = 2yeV2  o:2 (2y) + (1 + x 2 + y 2 ) ( 2y ( 2yeV2 :r2) + 2eY2 fv
:r2)
:r
2)
2e'~~2 :r2 ((1
x2

y 2 )(1  2x2 )

2x2 ),
+ y 2 ),
= 2e 712 
:r
2 ((1
+ x 2 + y 2 )( 1 + 2y 2 ) + 2y 2 ).
= 0 gives
= o.or x = ± 1. Thus the critical points are (0, 0) and (±1, 0). Now D(O, 0) = (2)(2)  0 > 0 and f :r:x (O , 0) = 2 > 0, x
so f(O, 0) = 0 is a local minimum. D(±l, 0) = ( 4e 1 )(4e 1 )  0
<0
so (±1, 0) are saddle points.
17.f(x,y)=y2  2ycosx
~
j ; =2ysinx,fv=2y 2cosx,
= 0 or x = 0, 1r, or 211" for 1 :::; x :::; 7. Substituting y = 0 into
fxx = 2y cos x, f ., 11
sinx = 0
~
= 2 sin x, f v11 = 2. Then/:.
f v = 0 gives cosx = 0
~
x = ~or
3 ;,
into f 71 = 0 gives y = 1, and substituting x
= 0 implies y
substituting x = 0 or x = 21r
= 1r into fv = 0 gives y =
 1.
Thus the critical points are (0, 1), ( ~. 0) , (1r, 1), ( 3; , 0), and (21r, 1).
D (~ , 0) = D
e
2"", 0)
= 4
< 0 so
(~, 0) and
e;,0) are saddle points. D(O, 1) = D(1r, 1) = D(21r, 1) = 4 > 0 and
fa:x(O , 1) = fxx('Tr,  1) = fxo:(27r, 1) = 2 > 0, so f(O, 1)
19. f(x, y) = x 2
+ 4y2 
4xy + 2
~
= j(1r, 1) =
f(27r, 1) =  1 are local minima.
fx = 2x 4y, fv =By  4x, f x:r: = 2, f, v = 4, / 1171 = 8. Then fx = 0
and fv = 0 each implies y = ~x, so all points of the form (xo, ~xo) are critical points and for each ofth~se we have
D(xo, ~xo) = (2)(8)  ( 4) 2 = 0. The Second Derivatives Test gives no information, but f(x,y) = x 2 +4y2  4xy + 2 = (x 2y} 2 + 2 ;:: 2 with equality if and only ify
=
~x. Thus f(x 0 , ~xo)
= 2 are all local
(and absolute) minima.
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SECTION 14.7
MAXIMUM AND MINIMUM VALUES
0
223
8· 7 6
8
z 5
6
2.5
2.5
4
4
3
2
2
 2.5 2
 I
0
2
y
From the graphs, there appear to be local minimaofabout/(1, ±1) = f(  1,±1) ~ 3 (and no local maxima or saddle points).
= 0 implies 2x4 y 2 
f,
2x2 y 4
y
f, = 2x  2x 3 y2 , fv = 2y  2x 2 y 3 , f xx = 2 + 6x 4 y  2 , fx11 = 4x 3 y 3 ,

2 = 0 or x 4 y 2
2 = 0, and with y 2
=2+
6x 2 y4 • Then
= 1 or y 2 = x  4 • Note that neither x nor y can be zero. Now f v =
= x 4 this implies 2x 6 
= ±1. So the critical points are (1, 1), (1, 
fxx
/ 1111
2 = 0 or x 6
0 implies
= 1. Thus x = ±1 and if x = 1, y = ± 1; if x = 
1),(  1, 1) and (  1, 1). Now D(1, ± 1) = D( 1, ±1) = 64  16
1, ·
> 0 and
> 0 always, so f(1, ± 1) = f (1, ± 1) = 3 are local minima.
23. f(x,y) = sin x +siny+sin(x +y), 0 :S: x :S: 2rr, 0 :S: y :S: 2rr
y
From the graphs it appears that f has a local maximum at about (1, 1) with value approximately 2.6, a local minimum at about (5, 5) with value approximately 2.6, and a saddle point at about (3, 3). f :z: = cos x + cos(x + y), fv = co:;y + cos(x + y), f xx =  sin x sin(x f xy
=  sin(x + 1J). Setting fx
or x = 2rr  y. l fx cos x
= y, f , =
+ y ), / 11 y = 
siny sin(x
= 0 and / 11 = 0 and subtracting gives cos x  cos y = 0 or cos i
0 becomes cosx + cos 2x
= 0 or 2 cos2 x +
cosx  1
+ y),
= cos y. Thus x = y
= 0, a quadratic in cosx. Thus
=  1 or ~ and x = rr, i , or r; , giving the critical points (rr, rr), ( i, i) and ( 5; , 5; ) . Similarly if
x = 2rr  y , f :z: = 0 becomes (cos x) + 1
= 0 and the resulting critical point is (rr, rr).
Now
D(x, y) = sin x sin y + sinx sin(x + y) +sin y sin(x + y) . So D(1r, 1r) = 0 and the Second Derivatives Test doesn't apply. However, along the line y = x we have f(x , x) = 2sin x + sin2x = 2 sinx + 2sin x cos x
©
= 2sin x (1 +
cosx), and
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224
CHAPTER 14
0
f(x , x)
PARTIAL DERIVATIVES
> 0 for 0 < x < 1r while f(x , x) < 0 for 1r < x < 21r. Thus every disk with center (1r, 1r) contains points where f
positive as well as points where f is negat~ve, so the graph crosses its tangent plane (z
f xx ( 5; ,
4x(x2 ( 0, 

> 0, so f ( 63"
2y)
= 0, sox= 0 or x 2 = 2y. lfx =
=
We have x 2
= 2y
= 0.259
=>
( 0, ~) and D
0 then substitution into f 11 = 0 gives 4y 3 = 2
= 0 gives 4y3 
8y + 2
D(±1.592, 1.267)
:::::}
•
y=
1  1i2, so
= 0. Using a graph, solutions are
~
x
= ±.J2Y, soy =
± 0.720 andy = ,1.267
1.526 gives no realvalued solution for x, but
=>
x
~
±1.592. Thus to three decimal places, the critical points are
~ (0,  0.794), (±0.720, 0.259), and (±1.592, 1.267). Now since f xx = 12x2
= (12x 2 
= ~ > 0 and
1.526, 0.259, and 1.267. (Alternatively, we could have used a calculator or a CAS to find these roots.)
=> x
5 6 3", ; )
is a local minimum.
is a critical point. Substituting x 2 = 2y into fv
approximately y
y
6 , ;)
= ¥
5 ;)
~)
= 0) there and (1r , 1r) is a saddle point.
*> 0 and fxx (~,·?f) < 0 so/(~,~) = .tf! is a local maximum while D(
D(~, ~) =
is
8y)(12y2 )  64x2 , we have D(O, 0.794)

8y, f ., 11 • =  8x, f 1111 = 12y2 ,
> 0, fx x(O,  0.794) > 0, D(±0.720, 0.259) < 0,
> 0, and f xx (± 1.592, 1.267) > 0. Therefore /(0, _ 0.794)
~  1.191 and /(±1.592, 1.267) ~  1.310
are local minima, and (± 0.720, 0.259) are saddle points. There is no highest point on the graph, but the lowest points are approximately (±1.592, 1.267, 1.310).
20
10
0
6 y
27. f(x , y) = x 4 +y3

3x2 +y2 +x  2y + 1
graphs, we see thatto three decimal places,
=> f x (x , y) = 4x 3
f., = 0 when x
~

6x + 1 and f 11 (x , y) = 3y2 + 2y 2. From the
1.301, 0.170, or 1.131, and / 11 = 0 when y
~
1.215 or
0.549. (Alternatively, we could have used a calculator or a CAS to find these roots. We could also use the quadratic formula to find the solutions of / 11 = 0.) So, to three decimal places, f has critical points at (  1.301,  1.215), (  1.301, 0.549), (0.170, 1.215), (0.170, 0.549), (1.131,  1.215), and (1.131, 0.549). Now since / zz = 12x2 and D = (12x2

6)(6y + 2), we have D( 1.301, 1.215)
6, fx 11.= 0, f 11 y = 6y + 2,
< 0, D(  1.301, 0.549) > 0, f xx( 1.301, 0.549) > 0,
D(0.170,  1.215)
> 0,
/ xz (l.131, 0.549)
> 0. Therefore, to three decimal places, f( 1.301, 0.549)
f xx (0.170,  1.215)

< 0, D(0.170, 0.549) < 0, D(l.131, 1.215) < 0, D(l.131, 0.549) > 0, and ~
3.145 and /(1.131, 0.549)
~
@) 2012 Ccngage Leruning. AU Rights Rc:ser\ro. May nol be scanned., COflied, or duplicated, or po~ed to n publicly accessible website, in whole or in p.vL
 0.701 are
SECTION 14.7
MAXIMUM AND MINIMUM VALUES
0
225
local minima, !(0.170,  1.215) ::::: 3.197 is a local maximum, and ( 1.301, 1.215), (0.170, 0.549), and (1.131 , 1.215) are saddle points. There is no highest or lowest point on the graph. 3
z
29. Since f is a polynomial it is contint,Jous on
D, so an absolute maximum and minimum exist. Here f x = 2x 2, jy = 2y, and
setting fx = jy = 0 gives (1, 0) as the only critical point (which is inside D), where f (1, 0) =  1. Along £ 1 : x = 0 and
f(O, y) = y 2 for 2 ~ y ::=:; 2, a quadratic function which attains its minimum at y =
b, where f( O, 0) = 0, and its maximum
·aty = ±2, wheref(0,±2) = 4. Along £2: y = x 2 forO ::=:; x ~ 2, and f(x,x 2) = 2x 2  6x + 4 = 2 (x _ ~) 2 _ ~ .
a quadratic which attains its minimum at x = Along £3: y = 2  x for 0
f(x, 2 x) = 2x 2
~
£, where f (~,  ~) =

~, and its maximum at x = 0, where f (0,  2) = 4. y
x ::::; 2, and
6x + 4 = 2(x ~)  ~.a quadratic which attains 2

its minimum at x =~~where f(~, ~) =~,and its maximum at x = 0,
(0,2)
(2,0)
L,
X
where f (O, 2) = 4. Thus the absolute maximum off on D is f(O, ±2) = 4 (0,  2)
and the absolute minimum is !(1, 0) = 1. 31. f x(x,y)
= 2x +2xy,
jy(x,y) = 2y +x2 , and setting f.,= f" = 0
y
On £ 1: y = 1, j(x,  1)
L•
= 5, a constant.
On £2: x = 1, /(1, y) = y 2
L,
(1 , 1)
gives (0, 0) as the only critical point in D , with .f(O, 0) = 4.
(I, I)
L2 0
+ 11 + 5, a quadratic in y which attai ns its
maximum at (1, 1), /(1 , 1) = 7 and its minimum at (1,  ~) , f(1 , 
t)
(1,1)
X
L,
(1, 1)
=.!f .
On £ 3: f(x , 1) = 2x 2 + 5 which attains its maximum at ( 1, 1) and (1, 1) with f(±l , 1)
=
7 and its minimum at (0, 1), f(O, 1)
= 5.
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226
D
CHAPTER 14 PARTIAL DERIVATIVES
On .i4 : f( 1, y)
= y 2 + y + 5 with maximum at ( 1, 1 ), f( 
= 7 and minimum at ( 1, D. f(  1, ~) = Jf.
1, 1)
Thus the absolute maximum is attained at both (±1, 1) with /(±1, 1) = 7 and the absolute minimum on Dis attained at (0, 0) with f(O, 0) = 4. 33. f(x , y) = x 4 +y 4
4xy + 2 is a polynomial and hence continuous on D, so

it has an absolute maximum and minimum on D. fx(x, y)
fv(x, y) = ·4y 3
=>
fv = 0

x
4x; then f x
= y 3 gives x 9 
x = 0
=>
x(x8  1) = 0
(1, 1) with f(1, 1) = 0 is inside D . On £1: y = 0, f(x, 0)
= 3, /(3, y) = y
On £ 2 : x /(3,

12y
+ 83, 0 ~ y
= 2, f( x, 2) =
x4

=> x
= 0 (0,0)
(3,0)
X
= x 4 + 2, x
= 3, !(3, 0)
= 83, and its minimum at x = 0, f (O, 0) = 2.
~ 2, a polynomial in y which attains its m inimum at y
{13) = 83 9 .{/3 ~ 70.0, and its maximum at y =
On L 3 : y
(0, 2)+   L  '3'1 (3, 2) .
1,  1), but only
~ 3, a polynomial in x which attains its maximum at 4
4y and
= 0 implies y = x 3 , and substitution into
±1. Thus the critical points are (0 , 0), (1, 1), and ( 
or x =
0~ x
= 4x 3 
Y
0, ! (3, 0)
= {/3,
= 83.
Bx + 18, 0 ~ x ~ 3, a polynomial in x which attains its minimum at x = q'2,
f( q'2, 2) = is 6 ij2 ~ 10.4, and its maximum at x
= 3, /(3, 2) =
75. On £ 4: x
= 0, f(O, y) =
polynomial in y which attains its maximum at y = 2, f(O , 2) = 18, and its minimum at y
y
= 0, f(O, 0)
4 + 2, 0 ~ y ~ 2, a = 2. Thus the absolute
maximum off on Dis /(3, 0) = 83 and the absolute minimum is /(1, 1) = 0. 35. f,(x, y) = 6x 2 and fv(x, y) = 4y3 • And so f x = 0 and fv = 0 only occur when x = y = 0. Hence, the only critical point
= 0. Now on the circle x 2 + y 2 = 1, y2 = 1  x 2 so let 2 2 f(x, y) =·2x 3 + (1 x 2 ) 2 = .x 4 + 2x 3  2x + 1,1 ~ x ~ 1. Then g'(x) = 4x3 + 6x  4~ =
.inside the disk is at x = y
g(x) = 2, or
~
/(0,±1) =
:::o 0 where /(0, 0)
g(~) = 1, !(~, ±1 ) = g(~) = ~· and ,(2,3)
0
=> x
= 0,
is not in D. Checking the endpoints, we get
.f(1, 0) = g( 1) = 2 and f(1 , 0) = g(1) = 2. Thus the absolute maximum and minimum off on Dare /(1, 0) = 2 and
=  2.
f(  1,0)
A nother method: On the boundary x 2
+ y 2 = 1 we can write x = cos (;I, y =
sin 8, so
f (cos 8, sin 8)
= 2 cos3 8 + sin4 (;I,
0 ~ () ~ 27T. 37. f(x, y)
=  (x2 
jy(X, y)
=
1) 2

(x 2 y x  1) 2
2(x2 y  x  1)x2 . Setting j y(x, y)
There are no critical points for x so f ,
( x,
f,(x, y)
=> f,(x, y)
x~ 1 ) = ·2(x2 
=  2(x 2

1)(2x)  2(x 2 y x  1)(2xy 1) and
= 0 gives either x = 0 or x 2y 
= O,.since f:.,(O, y)
1){2x) 2(x 2 x ~
= .:_2, so we set x 2y  x 1 1
 x  1) (2x x ~
1
x 1 = 0.
=0
{:}
y= x
~1
X
[x =/= 0],
 1) = 4x(x 2  1). Therefore
= fv(x, y) = 0 at the points (1, 2) and ( 1, 0). To classifY these critical points, we calculate
®
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MAXIMUM AND MINIMUM VAL4ES
SECTION 14.7
f :u:(x, y)
=
and f zv(x , y)
12x 2

12x2 y 2
= 8x3 y
+ 6x + 4x. In order to use the Second Derivatives
Of1~  I
D(1,0) = fxx(1 , 0) / 7171 (1,0)  [f:c 71 ( 1 , 0)]~ = 16 > 0,
2
= 10 <
0, D (1, 2) = 16 > 0, and f:r:c{ 1, 2) = 26 < 0, so
both ( 1, 0) and (1, 2) give local maxima.
X
39. Let d be the distance fr~m {2, 0, 3) to any point (x, y, z ) on the plane x where z
227
+ 12xy + 4y + 2, f vv(x, y) =  2x",
2
Test we calculate
/ :r:c (  1, 0)
0
= 1 x y, and we minimize d2
j;,(x,y) = 2(x 2) + 2{4 x 
= (x 
+y+ z =
1, sod= J(x  2) 2
+ y 2 + {4  x y) 2 • Then y)( 1) = 4x + 2y 12, f 71 (x,y) = 2y + 2(4  x y){1) = ~ f(x, y)
2)
+ y 2 + (z + 3)2
2
2x + 4y  8. Solving
4x + 2y 12 = 0 and 2x + 4y 8 = 0 simultaneously gives x = ~. y = ~.so the only critical point is(~,~). An absolute minimum exists (since there is a minimum di.s tance from the point to the plane) and it must occur at a critical poi nt, so the shortest distance occurs for x =
~. y = ~ for which d =
J(~ 
2)
+ (j) + (4  ~  ~) 2 = ji = ~2
2
41. Let d be the distance from the point {4, 2, 0) to any point {x, y, z) on the cone, sod = ,j(x  4)2
+ y 2 , and we minimize d 2 = (x  4) 2 + (y fz(x, y) = 2 (x  4) + 2x = 4x  8, f 71 (x , y) = 2 (y 
z2 = x 2
f.,
= 0 =>
x = 2, fu
= 0 =>
+ (y 2)2 + z2 where
+ x 2 + y 2 = f(x, y). Then 2) + 2y = 4y 4, and the critical points occur when
2)
2
y = 1. Thus the only critica l point is (2, 1). An absolute minimum exists (since there is a
minimum distance from the cone to the point) which must occur at a critical point, so the points on the cone closest to ( 4, 2, 0) are (2, 1, ±v'5).
43. x
+y+z
= 100, so maximize f(x , y) = xy(l OO  x y) .
f x = 100y 2xy y 2 , f 71 = 100x  x 2

2xy,
=  2y, f vu = 2x, fxv = 100 2x 2y. Then J,r; = 0 implies y = 0 or y = 100 2x. Substituting y = 0 into fv = 0 gives x = 0 or x = 100 and substi~ting y = 100  2x into fv = 0 gives 3x2  100x = 0 sox = 0 or ~ . f, :r.
1
1 0
1 0
Thus the critical points are (0, 0), (100, 0), (0, 100) and ( ~ , ~ ) .
D(O, 0) = D{100, 0)
= D {O, 100) =
10,000 while D(1~ 0 , 1 ~ 0 )
=
10
·~ 00 and f xx (1~0 , 1 ~0 ) = 
2
~ < 0. Thus (0, 0),
(100, 0) and {0, 100) are saddle points whereas f( 1 ~ 0 , 1 ~0 ) is a local maximum . Thus the numbers are x 45. Center the sphere at the origin so that its equation is x 2
+ y2 +z2
=
2 1· ,
V.,
y 2 ) = 8xy Jr 2  x 2
= (Bxy) . ~(r2 x2 y2)1 f2( 2x) + ,jr2 2
Setting Vx = 0 gives y = 0 or 2x
+ y2

~ .
+ y2 + z 2 = r 2 , so take (x, y, z) to be the vertex
in the fi rst octant. Then the box has length 2x, width 2y, and he ight 2z = 2 ,jr2  x 2 
1 0
and orient the inscribed rectangular box so that its
edges are parallel to the coordinate axes. Any vertex of the box satisfies x 2
V(x , y) = (2x)(2y)(2 Jr 2  x 2
= y =z =

y 2 with volume given by
y 2 forO < x < r, 0 < y < r. Then
x2  y2. By=
8y(r2

,jr2 _
2x 2  y 2 ). ·and Vv = 8x(r 2  x 2  2y 2 ) . x2 _ y2 . J r2 _ x2 _ y2
= r 2 , but y > 0 so only the latter solution applies. Similarly, V71 = 0 with x > 0
© 201 2 Ccngoge lcruniny. All Rights Rcscn"Cd. May nol be scwliiCd. copied. or duplicaled. or posted loa publicly accessible website, in \\bole or in p<>n.
228
0
PARTIAL DERIVATIVES
CHAPTER 14
implies x 2
3x = r
2
+ 2y2 =
2
::::>
r
x=
2
Substituting, we have 2x
.
.jr"i]3 =
2
+ y2 = x 2 + 2y2
::::>
x 2 = y2
f .,
y)
= y = r 1V3 and the maximum
= x; (6 x 2y), then the maximum volume is V = xyz .
= i(6y ~ 2xy y 2 )
= ~x (6  x 
= h(6 2x 2y) and fv
4y). Setting fx =:= 0 and j 11 = 0 gives the critical point
(2, 1} which geometrically must give a maximum. Thus the volume of the largest such box is V 49. Let the dimensions be x, y, and z; then 4x
V
'*
+ 2y2 = r 2
r I vf3 = y. Thus the oni,Y critical point is (r I vf3, r I ,/3 ). There must be a maximum
volume and here it must occur at a critical point, so the maximum volume occurs when x
.47. Maximize f(x,
y = x. Then x 2
::::>
= xyz = xv(ic x 
y)
+ 4y + 4z
= ic:cy x 2 y 
so Vx = 0 = Vy when 2x + y = ic and x
= (2)(1) ( ~) = t·
= c and the volume is
xy 2 , x > 0, y > 0. Then V,
= ~cy 2xy y 2 and Vy
+ 2y =~c. So lving, we get x =fie, y =
1 12 c
= ~ex x 2

2xy,
and z = ic x y = fie. From
the geometrical nature of the ·problem, thi s critical point must give an absolute maximum. Thus the box is a cube with edge length f2c. 51. Let the dimensions be x, y and z, then minimize xy + 2(xz
f(x,y)
+ yz) if xyz = 32,000 cm3 • Then = xy + [64,000(x + y)fxy] = xy + 64,000(x 1 + y 1 ), fx = y 64,000x 2 , f v = x 64,000y 2 .
And fx
= 0 implies y =
= 64,000 or x =
64,000/x2 ; substituting into / y = 0 implies x 3
D(x, y) = [(2)(64,000Wx  3y 3 dimensions of the box are x

1
40 and then y = 40. Now
> 0 for ( 40, 40) and 'f.,, (40, 40) > 0 so this is indeed a minimum.
Thus the
= y = 40 em, z = 20 em.
53. Let x, y, z be the dimensions of the rectangular box. Then the volume of the box is xyz and ' L
= J xz + y2 + z2
::::>
L2
= x 2 + y 2 + z2
::::>
Substituting, we have volume V(x,y) = xy JL2 x 2
2
Vy = xJL2 x2 y2 2x 2
+ y2 =
L 2 (since y

xy
JP  x2 
> 0), and
y2
= L2
::::>
x
= J L2 
x2

y 2 (x,y
> 0).

2 y .
. V., =Oimpliesy(L 2  x 2 y2 )=x2 y ::::> y(L 2 2x 2 y2 ) =0 ::::>
v;1 = o ·implies x(L2 
x 2 + 2y 2 = L 2 (since x > 0). Substituting y 2 = L 2 3x2
z

x2

y 2 ) = xy 2
2x2 into x 2
= L/vf3 (since x > 0) f!nd then y =
::::>
x( L 2
x2


+ 2y2 = L 2 gives x 2 + 2L2 
= 0 ::::> 4x 2 = L 2 ::::>
2y2 )
JL2 2(L/vf3) = L/vf3. 2
So the on ly critical point is (Lf,/3, Llvf3 ) which, from th~ geometrical nature of the problem, must give an absolute maximum. Thus th~ maximum volume is V(LI,/3, Ll,/3) = (LI,/3)
2
j L2 (Livf3)
2
2

(LI ,/3) = L 3 I (3 ,/3)
cubic units .
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2012 Cen,SD.Gc Learning. All Rir)lts Rcsco·cd. May not be scanned. cop ied. or duplicaLCd, or poster.! to a publicly accessible website, in whole or in pnrt.
SECTION 14.8 . .
n.
:L:
n
·
(x•Y• mx;  bxi) = 0 or .:L: XiYi = m
i= l
n
t. = l
n
.~ Yi = m i~l Xi
1
n
:L:
Now fmm =
n + i~l b=
2x~, fbb
m
n
·
•=1
229
2xi [Vi  (mxi + b)J = 0
n
x~ + b .:L: Xi and fb =
\= 1
:L:
2[yi  (mxi +b)] = 0 implies
1.=1
n
n
2 = 2n and fmb =
i= l
equations do indeed minimize
n
.:L:
0
(ni~t Xi) + nb. Thus we have the two desired equations.
= :L:
i =l
n
n
:L: [y, (mxi +b)]2 . Then /m. = i=l :L: i =l
55. Note that here the variables are m and b, and f{m, b) = implies
LAGRANGE MULTIPLIERS
:L:
.
2Xi. And fmm{m, b)
> 0 always and
i= l
f= d;.
i = 1
14.8
Lagrange Multipliers
1. At the extreme val';les off, the level curves off just touch the curve g(x, y) and the accompanying discussion.)
= 8 with a common tangent line. (See Figure I
We can observe several such occurrenc~ on the contour map, but the level curve
= c with the largest value of c which still intersects the curve g(x, y) = 8 is approximately c =59, and the smallest value of c corresponding to a level curve which intersects g(x, y) = 8 appears to be c = 30. Thus we estimate the maximum f(x , y)
value off subj ect to the constraint g(x j y) = 8 to be about 59 and the minimum to be 30. 3. f(x , y) = x 2
+ y2 , g(x, y) =
xy = 1, and \1 f = >..\lg =>
=>
From the last equation, x =I= 0 andy =I= 0, so 2x = >.y y2 = x2
=>
y
= ±x. But xy = 1, so X= y =
(2x, 2y)
= (>..y, >..x), so 2x = >..y, 2y = >..x, and xy = 1.
>.. = 2xjy. Substituting, we have 2y = (2xjy) x
=>
± 1 and the possible points for the extreme values off are {1, 1) and
( 1,  1). Here there is no maximum value, since the constraint xy = 1 allows x or y to become arbil:(arily large, and hence
f(x , y) = x 2 5. f(x, y) = y 2 and ix2
+ y 2 can be made arbitrarily large. The mi~imum value is !{1 , 1) = 
+ y2 =
x 2 , g(x, y)
= 1, and \1 f = >.\lg =>
{2x, 2y) = (t>..x, 2>..y) , so 2x = t>.x, 2y = 2>..y,
+ >..) = 0 => x = 0 or>. =  4. If x = 0 then the third equation  4 then the second equation gives 2y = 8y => y = 0, and substituting into the third equation,
1. From the first equation we have x{ 4
gives y = ±1. If), = we have x
= tx2 +y 2
f(  1, 1) = 2.
= ±2. Thus the possible extreme values off occur at the points (0, ± 1) and (± 2, 0). Evaluating fat these points,
we see that the maximum value is f{O, ± 1) = 1 and the minimum is !{±2, 0) = 4. 7. f(x, y, z)
= 2x + 2y + z, g(x, y, z)
2>.y = 2, 2)..z = 1, and x 2
+ y2 + z 2 = 2
the fourth equation gives
= x2
+ y2 + z 2 =
9, and \1 f = >.. \1g
9. The first three equations imply x = 2
(l) + ( l) + (
2
\ ) 2
© 2012 Ccngogc Learning. All Rights Rcsc:rvcd. May not be =
= 9 =>
9
 2 4)..
= 9
=>
(2, 2, 1) = (2>..x, 2>..y, 2>.z), so 2>.x = 2,
l• ·l· y =
and z =
\ . 2
~ut substitution into
=> ).. = ±t, so f has possible extreme values at
ned. copi~; or duplicated, or posted to a publicly occ~ssiblc website, in whole or in p3r1.
230
0
CHAPTER 14
PARTIAL DERIVATIVES
the points (2, 2, 1) ~nd (  2,  2,  1). The maximum value off on x 2
J( 2, 2,  1)
9 is f(2, 2, 1) = 9, and the minimum is
=  9.
9. f(x, y, z) = xyz, g(x, y, z) = x
zero then x = y
x2
+ y2 + z 2 =
2
+ 2y 2 '+ 3z2 = 6. V f = >.Vg
=:.
(yz, xz, x y)
= .>. {2x, 4y, Gz). If any ofx, y, or z is
= z = 0 which contradicts x 2 + 2y2 + 3z2 = 6. Then.>.= (yz) /(,2x)
= 2y 2 and z 2
= ~y 2 . Thus x 2
+ 2y2 + 3z 2 =
6 implies 6y 2
= 6 or y
= '(xz) / (4y) = (xy)/(6z ) or
= ±1. Then the possible points are
/j), (.,/2, ±1, /j), (.,/2, ± 1, /j), ( .,/2,±1, /j). The maximum valu~ off on the ellipsoid is J:i, occurring when all coordinates are positive or exactly two are negative and the minimum is 7a occurring when 1 or 3 of ( .,/2, ±1,
the coordinates are negative. 11. f(x ,y,z) = x 2 +y2 +z2 , g(x,y, z) Case 1: lf x
f. 0, y f.
0 and z
= x~ +y4 + z 4 = 1
f. 0, then V f
Vf
=?
= >.Vg implies.>. = l /(2x2 )
= 1/(2y2 )
(4>.x 3 , 4>.y3 ,4>.z3 ).
= l /(2z 2 ) or x 2 = y2 = z 2 and
_!_ L) (±L L _ _!_) (±L _ _!_ __!_) VS' V3 ' VS' VJ ' V3 ' VJ' W' V3
3x 4 = 1 or x = ±  1 giving the points (±L L L) (±L _
VJ' VS' V3 ' · VJ'
V3
= (2x,2y,2z)~ >.Vg =
all with an !value of v'3. Case 2: If one of the variables equals zero and the other two are not zero, then the squares of the two nonzero coordinates are
equal with common value ~ and corresponding f value of .,/2. Case 3: If exactly two of the variables are zero, then the third variable has value ± 1 with the corresponding f value ofl. Thus
on x 4
+ y'1 + z4 =
13. f(x, y, z, t)
=
X
1, the maximum value off is
y'3 and the minimum value is 1.
1 y 1 Z 1 t, g(x, y, z, t) = x 2 1 y 2 1 z 2 1
). = 1/(2x) = 1/(2y) = 1/(2z) = 1/(2t) and x
e=
and p.Vh = (0, 2p.y, 2p.z). Then 1 Thus x
+y +z
+z =
t)
1, h(x,y, z ) = y 2
+ z2 =
(2>.x, 2).y, 2).z , 2>.t), SO
+ z 2 = 4 =:.
Vf
~ =
= (1,2,0), .>.Vg =
 p.z or y = 1/ (2p.), z
4 implies p. =±~.Then the possible points are
(>.,>., >.)
= 1/ (2p.).
(1, ±.J2, =FV2)
(1, .J2, .J2) = 1 + 2 .J2 and the minimum value is f (1,  .,/2, .J2) = 1  2 .J2.
17. f(x, y, z ) = yz + xy, g(x, y , z ) = xy = 1, h(x, y, z) = y 2
+ z2 =
1
=:. V f = (y, x + z, y), >.Vg = (.>.y, >.x, 0),
p.Vh = (0, 2p.y, 2p.z). Then y = >.y implies>. = 1 [y f. 0 since g(x, y, z) = 1], x p. = z/(2y) = y/(2y) or y2 = z 2 , and so y 2
©
=
= 2 and the minimum value is
= .>., 2 = .>. + 2p.y and 0 = >. + 2~tz so p.y =
= 1 implies x = 1 and y 2
and the 'mfJ.Ximum value is f
(1, 1, 1, 1)
= y ~ z = t. But x 2 1 y 2 1 z 2 1 t2 = 1, so the possible points are
(±~ , ±~, ±~, ±~). Thus the maximum value off is f(~, ~~ ~~
15. f(x,y, z) = x + 2y, g(x,y,z) = x +y
1 =>
+ z2 =
+z =
.>.x + 2p.y andy= 2p.z. Thus
1 implies y = ±72, z = ±72. Then xy
= 1 implies x =
2012 Ccng.agc Learning. All Rights Rescr\'cd. M:1y not be sc:mncd, copic:c.l, or duplicotcd, or posted to u publicly occcssiblc website, in wll(>Je or in po1rt.
±.J2 and
SECTION 14.8
the possible points are ( ±/2, ± 7:i ,
y2
0
231
72), (±V2, ±~,  );) . Hence the maximum off subject to the constraints is
f(±J2,±~,±jz) = ~ andtheminimumisf(±J2,±~, 'f72) Note: Since xy
LAGRANGE MULTIPLIERS
=
~·
= 1 is one of the constraints we could have solved the problem by solving f (y, z) = yz + 1 subject to
+ z2 = 1.
19. f( x, y)
= x 2 + y 2 + 4x 
4y. For the interior of the region, we find the critical points: fx
= 2x + 4, f u = 2y 
4, so the
onlycritical point is (  2, 2) (which is inside the region) and f (  2, 2) = 8. For the boundary, we use Lagrange multipliers.
g(x, y)
= x 2 + y2 = 9, so "iJf = >."ilg
=;.
Adding the two equations gives 2x + 2y = 2>.x
x
+y = 0
x2
+ y2
=;.
y = x or>.  1 = 0 2
=>
= 9 implies 2y = 9
y
=;.
(2x + 4, 2y  4)
= (2>.x, 2>.y). Thus 2x + 4 =
+ 2>.y =:.
+y =
x
>.(x
+ y)
=:.
(x
2>.x and 2y  4 = 2>.y.
+ y) (>. 1) =
0, so
>. = 1. But >. = 1 leads to a contradition in 2x + 4 = 2>.x, soy =  x and
= ±12. We have f ( ~· 72) = 9 + 12/2 ~ 25.97 and
r( 12, ~) = 9 12/2 ~  7.97, so the maximum value off on the disk x
2
~ y2
:s: 9 is f
( 72· 12)
= 9 + 12/2
and the minimum is f(  2, 2) = 8. 21. f(x , y) = e xy . For the interior ofthe region, we find the critical points: f x =  yexu, / 11
= xe"'Y , so the only
critical point is (0, 0), and f(O, 0) = 1. For the boundary, we use Lagrange multipliers. g(x, y) = x 2
+ 4y2
= 1
=;.
>. "iJ g = (2>.x, 8>.y), so setting "iJ f = >."iJ g we get ye"'11 = 2>.x and xe"'11 = 8>.y. The first of these gives e "' 11 = 2>.xjy, and then .the second gives  x(  2>.xjy) = 8>.y constraint x 2
+ 4y2
f ( ±~ , ±;~)
= 1 gives
X=
= >.(4x3
x4

x3

=.0
±72 and y=±~ . Now t(±~, 'f 2~) = e 1
1 4
~ 1.284 and
= e 1/ 4 ~ 0.779. The former arc the maxima on the region and the latter are the min.ima.
23. (a) f(x , y) = x , g(x,y) = y 2
1
=> x 2 = 4y2 • Solving this last equation with.the
+ x4

x3
== 0
=;.
"ilf = (1, 0) = >.Vg = >.(4x3

3x2,2y). Then
3x 2 ) (1) and 0 = 2>.y (2). We have >. i= 0 from (1), so (2) gives y = 0. Then, from the constraint equation, x 3 (3;
=;.

1)
=0
=;.
x
= 0 or x = 1. But x = 0 contradicts (1), so the only possible extreme value
subject to the constraint is /(1, 0) = 1. (The question remains whether this is indeed the minimum of f.) (b) The constraint is y 2
+ x4 
x3 = 0
¢}
y2 = x3

x 4 . The left side is nonnegative, so we must have x 3

x 4 :2: 0
which is true only for 0 :S: x :S: 1. Therefore the minimum possible value for f( x, y) =x is 0 which occurs for x = y = 0. However,>."iJ g(O, 0)
= >. (0 0, 0) =
(0 , 0) and "iJ f(O, 0) = (1, 0), so "iJ f(O, 0)
(c) Here "iJ g(O, 0) = 0 but the method of Lagrange multipliers requires that "iJg
©
#
# >."iJ g(O, 0) for all values of>..
0 everywhere on the constraint curve.
20 12 Ccngagc Lcruning. All Rights Reserved. May not be scanned, copicc.l, or duplicated, or posted to a publicly accessible website. in whole or in p:m.
232
D
CHAPTER 14
PARTIAL DERIVATIVES
'*
25. P(L,K)=bLo.Kl o., g(L,K)=mL+nK=p
'VP=(abLo: 1 K 1  o.,(1a)bL'""JC'""), >..'Vg=(>..m, >..n).
Then ab(K/ L) o. = >..m and (1 a)b(L/ K)"' =>..nand mL + nK = p, so ab(K/ L?"'/m = (1  a)b(L/ K) 0 /n or 1
naj[m(1 a)]= (L/ K)"'(L/ K) 1 "'. or L = Kno/[m{1 a)]. Substituting into mL + nK = p gives K = (1 a)pjn and L = apjm for the maximum production. 27. Let the sides of the rectangle be x andy. Then f(x, y) = xy, g(x, y) = 2x + 2y = p
>.. 'Vg = (2>.., 2>..). Then A= ~y = ~x i~plies x
2
that g(x, y, z) = x + y + z
+(z + 3)
= 1.
2
= .j(x 2)2 + y 2 + (z +
d=
J(~ 
2) ~· + ( ~ )2 + (
3) 2 , so we seek to minimize
subject to the constraint that (x, y, z) lies on the plane x + y + z = 1, that is,
=*
Then 'V f =A 'Vg
(2(x 2), 2y, 2(z + 3)) = (A, .A, .A), sox= (A+ 4)/2,
y = >./2, z = (>..  6)/2. Substituting into the constraint equation gives
>.. = ~ . sox = ~. y = ~.and z =
'V f(x, y) = (y, x)·,
= y and the rectangle with maximum area is a square with side length iP·
29. The distance from {2,0,  3) to a point (x,y,z) on the plane is d d = f(x, y, z) = (x 2) 2 + y 2
'*
.A~ 4 +
i
+ A;
6
= 1
'*
3A 2 = 2
'*
i· This must correspond to a minimum, so the shortest distance is
i + 3) 2 = ..fi = ~
31. Let f(x, y , z) = d 2 = (x 4) 2 + (y 2) 2 + z 2 • Then we want to minimize f subject to the constraint
g (x,y,z) = x 2 +y2
z 2 = 0. 'Vf = )..'\Jg

'*
(2 (x ; 4) ,2(y 2) ,2z) = (2>..x,2>..y, 2Xz), sox 4 = J..x,
'*
y  2 = .Ay, and z = J..z. From the last equation we have z + >..z = 0 But from the constraint equation we have z two equations. So ).. =  1 and x  4 = >..x
'*
4 + 1  z2 = 0
z
=0
'*
'*
x 2 + y2 = 0
'*
x = 2, y  2 = >..y
z (1 +.A) = 0, so either z = 0 or>..= 1. x = y = 0 which is not possible from the first
'*
y = 1, and x·2 + y 2

z2 = 0
'*
= ±v'S. This must correspond to a minimum, so the points on the cone closest to (4, 2, 0)
are (2, 1, ±v'S). ' 33. f(x,y,z) = xyz,g(x,y,z) = x +y + z = 100 implies x
= y =z
=
1
=* 'Vf = (yz,xz, xy) = )..'\Jg =(>.., >..,>..). Then>.. = yz = xz = xy
~0 .
35. If the dimensions are 2x, 2y, and 2z, then maximize f(x, y, z)
= (2x)(2y)(2z)
g(x, y, z) := x 2 + y 2 + z 2 = r 2 (x > 0, y > 0, z > 0). Then 'V f = .A'V g 8yz
= 2Xx, Bxz = 2
and xy = xz
2
'*
= 8xyz subject to
(8yz , Bxz, 8xy) = .A (2x, 2y, 2z) =*
41JZ 4xz 4xy . . 2 2>.y, and 8xy = 2.Xz , so>.=  · = = .This gives x 2 z = y z X Y Z
'*
z 2 = y2 , so x 2 = y 2 = z 2
'*
equation gives 3x2 = r 2
. f ( ~, ~, ~) = 8 ( ~) (
'*
x = r/../3 = y = z.
7J) (~) = 3 ~ r
3
'*
2
2
x = y (since z
'
x = y = z, and substituting into the constraint
Thus the largest volume of such a box is .
'
© 2012 Cengage Learning. All Rights Reserved. May nor be scnruu.xJ. copied, or duplicatC"d. or postctl to u publicly ucccssiblc website. in whole or in purt.
t= D)
SECTION 14.8 37. f(x ,y,z)
= xyz,
g(x, y ,z)
Then >. = yz = ~xz
= x +2y + 3z = 6 =>
\lf
= {yz,xz ,xy) =
= ~xy implies x = 2y, z = %Y· But 2y + 2y + 2y =
LAGRANGE MULTIPLIERS
D
233
>..Vg = (>.. , 2>., 3>.).
6 soy = 1, x = 2, z = ~ and the volume
is V = ~39. f( x, y, z) = xyz, g(x, y ,.z) = 4(x + y + z) = c 4>.
= yz = xz =
\7 f = (yz, xz, x y), >.Vg = (4>., 4>., 4>.). Thus
=>
xy or x = y = z = fie are the dimensions giving the maximum volume.
41 . If the dimensions of the box are given by x, y, and z, then we need to fi nd the maximum value of f(x, y, z ) = xyz
> 0) subject to the constraint L
[x, y, z
=
..jx2 + y 2 + z 2 or g(x, y, z ) = x 2 + y 2 + z 2 = £ 2 .
\7 f = )..\lg
yz xz (yz , xz, xy ) = >. (2x, 2y, 2z ), so yz = 2>.x => >. =  , xz = 2>.y => >. =  , and xy = 2.Xz 2X
Thus). = yz = xz .=> ~
x2
~
= y2
[since z =f. 0)
2Y ·
(L/ V3)
3
2x
= ).
+ 2J.LX (1),
+ y~  z = 2y
= >. +
From (l) and (2), 2(x  y)
0. \lf
=>
x
2
= £ 2 / 3 · =>
= x2 +
V3 =
y 2 + z 2 subject to the two constraints g(x, y , z)
= 2>. f.t
(3), x + y + 2z = 2 (4), and x 2
= 2~(x  y), so if x =f. y, ~ = 1. Putting this in (3) give,s 2z
+~
= 0 or z = ~. Then (4) and (5) become x
last equation cannot be true, so this case gives no solution. So we must have x 
x = L/
y = z and the
=
x + y + 2z = 2
= (~x, 2y, 2z}, >.\lg = (>. , >., 2>.) and ~\lh = (2J.Lx_, 2~y, ~). Thus we need
2~y (2), 2z
~ = 1 into (l) says>. = 0. Hence z
2x 2
2Z
2X
= £ 3 / (3 V3).
43. We need to fi nd the extreme values of f(x , y, z) and h(x, y, z) = x 2
=>
=> x = y and >.= y z = x y. => x = z [since y =f. 0].
Substituting into the constraint equation gives x 2 + x 2 + x 2 = L 2 maximum vo lume is
=>
z = 0 which imply z = 1  x and z = 2x2 . Thus 2x 2 = 1  x or 2x2
+ y2  z =
0 (5).
= 2>. 1 or >. = z +~. but p~1tting
+ y 3 = 0 and x 2 + y 2 + ~
= y. Then (4) and (5) become 2x + +x
 1
= (2x 
~ 0. The ·
2z
= 2 and
1)(x + 1) = 0 so x= ~ or
x =  1. The two points to check are (t, ~ . ~)and (  1,  1, 2): /(~. ~. ~) = ~and f(1, 1, 2) = 6. Thus(~,~. ~) is the point on the ellipse nearest the origin and (  1,  1, 2) is the one farthest from the origin. 45. f( x, y , z) = yex::, g(x, y, z)
= 9x2 + 4y 2 + 36z2 =
(yex::, ex  z,  yex z) = >.(18x, 8y, 72z)
36, h(x, y , z)
= xy + y z =
1.
\7 f = >.Vg + JL\lh
=>
+ i'(V, x + z, y), so yexz = 18>.x + J.l.V, ex z = 8>.y + ~(x + z ),
 yex :: = 72>.z + ~y. 9x 2 + 4y 2 + 36z 2 .= 36, x y + yz = 1. Using a CAS to solve these 5 equations s imultaneously for x, y, z , >.,
and ~
(in Maple, use the all valu es command), we get 4 realyalued solutions: X~
0.222444,
y ~· 2.157012,
z
~
 0.686049,
>.
~
 0.200401,
~ ~
2.108584
X~
 1.951921,
y
~
 0.545867,
z
~
0.119973,
).
~
0.003141,
~ ~
 0.076238
X ~
0.155142,
y
~
0.904622,
z
~
0.950293,
>.
~
 0.012447,
f.' ~
0.489938
X~
1.138731,
y
~
1.768057, .
z
~
0.573138,
>.
~
0.317141,
f.' ~
1.862675
Subst ituting these values into f gives /(0.222444, 2.157012,  0.686049) ~ 5.3506,
® 20 12 Ceng3gc (.earning.. All Righrs Re~r\'\,.d. May not be scanned, copied, or duplicated, or posted to u publicly acct"Ssibh.: website, in whole or in 11.1r1 .
234
0
CHAPTER 14 PARTIAL DERIVATIVES
f( 1.951921,  0.545867, 0.119973) !(1.138731, 1.768057, 0.573138)
~
~
0.0688, !(0.155142, 0.904622, 0.950293)
~
0.4084,
9.7938. Thus the maximum is approximately 9.7938, and the minimum is
approximately 5.3506.
47. (a) We wish to maximize j(x1, x2, ... , xn) g(x1,x2, .. . , Xn)
\1 f
=
ytx1x2 · · · Xn subject to
= X1 +x2 + ·· · +xn = c and X.;> 0.
= ( ~ (X!X2 · .. Xn).; l (X2 .. · Xn) , ~(X1X2 .. · Xn)'~l (X!X3 · .. Xn),
and >. V g
= (>., >.,
... ,
... , >.),so we need to solve the system of equations 1/n 1/n
x1
1 ( XtX2 · • • Xn ).l.1( ;;: " XtX3 • · • Xn)
This implies n>.x1 But X !
~(XtX2 · .. Xn) !,l (xl · .. Xn1) ~
= n>.x2 =
· · · = n>.x... Note >.
+ X2 + · ·· + Xn = c
have an extreme value is
=>
(~ , ~, n
n
zero (but not equal) as we like,
f
nx1
... ,
= c =>
~). n
x2
1/n 1 /n
= A
X1
X2
1/n
, = nAx 1
1/n
\ = nAX2
· · · Xn · · · Xn
=I 0, otherwise we can't have all x.; > 0. Thus X1 X1
= x2 = · · · = Xn .
= ~n = X2 = X3 = · · · = Xn  Then the only point where f
Since we can choose yalues for (Xt, x2, ... , Xn) that make
f
can
as close to
has no minimum value. Thus the maximum value is
t(;,;, ....;) = \};; ..... ;=;. (b) From part (a), ~ is the maximum value of f. Thus j(x1; x2, . .. , x n )
n
x1
+ x2 + · ·· + Xn
maximum value when Xt
14
,,,,~·
= c, so ytx1X2 · · · Xn
:::;
X1
+ X2 + · · •+ Xn . These two means are equal when f n
~' but this can occur only at the point(;,*'
= X2 = X3 =

= ytx1x2 · · · Xn :::; ~But n .
.
attams 1ts
. .. , * ) we found in part (a). So the means are equal only
c
· · · = Xn = . n
Review
CONCEPT CHECK 1. (a) A function f of two variables is a rule that assigns to each ordered pair (x , y) of real numbers in its domain a unique real number denoted by f(x, y). (b) One way to visualize a function of two variables is by graphing it, resulting in the surface z = f(x, y). Another method for visualizing a function of two variables is a contour map. The contour map consists of level curves ofthe function which are horizontal traces of the graph of the function projected onto the x yplane. Also, we can use an arrow diagram such as Figure I in Section 14.1.
©
2012 Cengage Le'ilming. All Rights Reserved. May nor be scnnncd. copied. or duplicntc:d, or postr..'tl ton publicly accessible wch~itc, in whole or in part.
CHAPTER 14 REVIEW
2. A function
f
0
235
of three variables is a rule that assigns to each ordered triple (x, y, z) in its domain a unique real number
f(x, y , z). We can visualize a function of three variables by examining its level surfaces f(x, y, z) = k, where k is a constant. 3.
f(x , y) = L means the values of f(x, y) approach the nwnber Las the point (x, y) approaches the point (a, b)
lim (:z:,11)+{a ,b)
along any path that is within the domain of f. We can show that a limit at a point does not exist by finding two different paths approaching the point along which f( x, y) has different lim its. 4. (a) See Definition 14.2.4.
(b) Iff is continuous on JR2 , its graph will appear as a surface without holes or breaks. 5. (a) See (2) and (3) in Section 14.3.
(b) See "Interpretations of Partial Derivatives" on page 927 [ET 903]. (c) To find f
., regard 1J as a constant and diffe~entiate f(x, y) with respect to x. To find / y, regard x as a constant and
differentiate f( x, y) with respect toy. 6. See the statement of Clairaut's Theorem on page 931 [ET 907]. 7. (a) See (2) in Section 14.4.
(b) See (19) and the preceding discussion in Section 14.6. 8. See (3) and (4) and the accompanying discussion in Section 14.4. We can interpret the linearization off at (a, b) geometrically
as the linear function whose graph is the tangent plane tothe graph off at (a, b) . Thus it is the linear function which best approximates f near (a, b).
9. (a) See Definition 14.4.7. (b) Use Theorem 14.4.8. 10. See (10) and the associated discussion in Section 14.4. 11. See (2) and (3) in Section 14.5. 12. See (7) and the preceding discussion in Section 14.5. 13. (a) See Definition 14.6 .2. We can interpret it as the rate of change off at ( Xo, yo) in the direction of u . Geometrically, if P is
the point (xo, yo , f(xo, Yo)) on the graph off and C is the curve of intersection of the graph off with the vertical plane that passes through P in the direction u , the directional derivative off at (xo, Yo) in the direction ofu is the ·slope of the tangent line to Cat P. (See Figure 5 in Section 14.6.) (b) See Theorem 14.6.3. 14. (a) See (8) and ( 13) in Section 14.6. (b) Du f(x , y)
©
= "\7 f (x, y) · u or Du f(x, y, z) = "\7 f(x, y, z) · u
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236
0
CHAPTER 14 PARTIAL DERIVATIVES
(c) The gradient vector of a fu nction points in the direction of maximum rate of increase of the function. On a graph of the function, the gradient points in the direction of steepest ascent. 15. (a) f has a local maximum at {a, b) if f(x, y)
:5 f(a, b) when {x, y) is near {a, b).
(b) f has an absolute maximum at {a, b) if f(x, y)
(c) f has a local minimum at (a, b) if f(x, y) (d)
~
:5 f(a , b) tor all points (x, y) in the domain of f .
f (a, b) when (x, y) is near (a, b).
f has an absolute minimum at (a, b) if f(x , y)
f(a, b) for aUpoints (x, y) in the domain of f.
~
(e) f has a saddle point at (a, b) if f(a, b) is a local maximum in one direction but a local minimum in another. 16. (a) By Theorem 14.7.2, iff has a local maximum at (a, b) and the firstorder partial derivatives of j exist there, then / :r(a, b) = 0 and / 11 (a, b)= 0. (b) A critical point off is a point (a, b) such that [.(a, b)
= 0 and j 11 (a, b) = 0 or one of these partial derivatives does
not exist. 17. See (3) in Section 14.7. 18. (a) See Figure II and the accompanying discussion in Section 14.7.
(b) See Theorem 14.7.8. (c) See the procedure outlined in (9) in Section 14.7. 19. See the discussion beginning on page 981 [ET 957]; see "Two Constraints" on page 985 [ET 961].
TRUEFALSE QUIZ 1. True. f v(a, b)
= lim
we get f 11 (a, b)
h~o
= lim
u~ b
f(a, b +h)  f(a , b) from Equation 14.3.3. Let h = ·y b. Ash~ 0, y h
+
b. Then by substituting, ,
f(a , y)f(a,b) b . y
5. False. See Example 14.2.3. 7. Tru~. If f has a local minimum and f is differentiable at (a, b) then by Theorem 14.7.2, f x(a, b) V f(a,b)
= 0 and fu (a, b) = 0, so
= (f,(a,b),fv(a,b)) = (0,0} = 0 .
9. False. V f(x , y)
= (0, 1/ y).
11 . True. V f = (cosx, cosy), so IV / I = v'cos 2 x
Du f(x, y)
= Vf
·u
+ cos2 y . But lcosBI :5 1, so IV!I :5 .J?.. Now
= IVf l l.u lcos{:}, but u is a unit vector, so IDu f (x, y) l :5 .J2 · 1 · 1 = ./2.
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l CHAPTER 14 REVIEW
0
237
EXERCISES ',
+ y + 1) is defined only when x + y + 1 > 0 ~ y >  x  1, so the domain off is {(x, y) I y >  x  1}, all those points above the line y = x  1.
''
1. ln(x
3. z = f(x , y) = 1  y 2 , a parabolic cylinder
5. The level curves are
k
::2:::
y
''
''
'
J 4x2 + y2 = k or 4x 2 + y 2 =
k 2,
0, a fam ily of e ll ipses. y
X
7.
9.
y
2
I
(©J (©J (©J (©J
0
I
h o
eva1uate
~
2
~
4
T( 8 , 4) ; T ( 6 , )
=
86
;
2
80
± 2 and
= 3,
4 4 4 72 80 T( , ) ~ T(B , ) = ~ = 4. Averaging these values, we estimate Tx(6, 4) to 4
2
  2

~e approximately
+ h~  T ( 6, 4) , which we can approximate with h =
) _ T(6, 6}  T(6 , 4) _ 75  80 _ _
'4 
2x y 2(1)(1) 2 + 2y 2 = 12 + 2 (1) 2 = 3 .
~  T (B, 4), so we can approximateT:c (6, 4) by considering h =
h 0
.Lv
lim (z,v )  (1 ,1) X 2
4
3.'5 ° C / m . Similarly, Tv (6, 4} = lim T(B ,
,.., (B
th I" . e 1m1t:
X
using the values g iven in the table: T x(6, 4)
. T.,(6, 4)
is a rational function, so it is continuous on its d omain.
Since f is defined at (1, 1), we use d irect substitution to
2
11 . (a) T.,(6 , 4) = lim T(B + h,
f
,.,., ( ) _ T(6 , 2 )  T(6, 4) _ 87 80 _ 2 · 5 ' .L v 6 , 4 _  ~
2
±2: 
3 .. 5
A . th veragmg ese
values, we estimate Tv(6, 4) to be approximately  3.0° C / m.
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238
CHAPTER 14 PARTIAL DERIVATIVES
0
(b) Here u
= ( 12• )?.). so by Equation 14.6.9, Du T(6,4) = VT(6,4) · u = T, (6,4) ~ + Tv(6,4) ~·Using our
estimates from part (a), we have Du T(6, 4)
l':;j
(3.5) ~
+ (3.0) ~ =
~
0.35. This means that as we move
l':;j
through the point (6, 4) in the direction of u , the temperature increases at a rate of approximately 0.35°C/m.
. Alternatively, we can use Definition 14.6.2: Du T( 6, 4)
=
lim
r(6+h~,4+h~)T(6,4)
ho
which ·we can estimate with h = ± 2 ,;2. Then Du T(6, 4)
h
~ T( 8 , 6 ) ~T( 6 , 4)
80  ~O 2v2
=
2 2
D u T(6 , 4)
~
T(4, 2)  T(6,4) tn  2v2
7480 =~ = 2v2
=
,
0,
3 . ( ) tn' Averagmg these values, we have D u T 6, 4 v2
~
3
27z
_ !_ [T. ( )]  lim T,('x , y +h)  T:r. (x,y) T. ( ) _lim T., (6,4 +h)  T.,(6, 4) () C T.:V!/ ( X 1 y )  a x X1y h , SO xy 6 1 4 h . y ho , ho estimate with h
~
0
I
1.1 C m.
. h
W 1liC We Can
= ±2. We have Tx (6, 4) ~ 3.5 from part (a), but we will also need values for T.,(6 , 6) and T,(6, 2). If we
use h = ±2 and the values given in the table, we have
T,(6 , 6 ) l':;j T(8, 6)  T(6 , 6) 2
=
80  75 2
= 2.5 , T, (6 , 6) ~ T(4, 6) T (6, 6) =
68  75 = 3 .5 . 2
2
Averaging these values, we estimate T.,(6, 6) ~ 3.0. Similarly,
Tx( 6 , 2 ) l':;j T(8, 2) ~ Tx(6, 2)
=
90 ; 87
= l.S, T.,(6 , 2) ~
2
T(4, 2) ~ T(6, 2)
=
74 ~ 87
= 6 .5 .
Averaging these values, we estimate T, (6, 2) ~ 4.0. Finally, we esti~ate T., 11 (6, 4): T.:vy (6 ! 4)
~
T.,(6, 6) T.,(6, 4) _ 3.0  3.5 _ _ 2
~

2

0 •25 >
T. ( X)J
) ~ T.,(6, 2) T.,(6, 4) _ 4.0  3.5 __ 0 •25 • 61 4 ~ 2 2 
Averaging these values, we have Txy (6, 4) ~ 0.25. 13. f(x , y) = (5y3
fv
+ 2x2y) 8
=>
f .,= 8(5y3
= 8(5y3 + 2x2 yf (15y2 + 2x2) =
17. S(u, v,w)
Sw =
= uarctan(vyfw)
+ 2x 2 y)7
= a rctan(v y'W),
Su = u ·
S,.
2 Vw
2
= 4x3 
x y2
= 32xy(5y 3
2 2 3 (16x + 120y )(5y
u· 1 + (:JW) (v· ~w 1 12 ) =
19. f(x,y)
+ 2x 2·yf(4xy)
=> f x = 12x 2

+ 2x 2 y) 7 ,
1
1
+ (vfo)2 (fo) =
ufo
1 +v2w '
(~: v2 w)
y 2 , f v = 2xy, f.,., = 24x, fv 11
= 2_:~:,
fxv
= fvx = 2y
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CHAPTER 14 REVIEW
21. f( x, y , z) = xky1z'"
=>
fx
= kxklylzm., !11 =
lxky 1 1 z m, fz
= mxky1z""' 1 ,
0
239
f:"x = k(k 1)xk 2 y 1z"",
f vv = l(l 1)xky1 2 zm, f zz = m (m 1)xky1z"' 2 , f xv = f vcx: = klx k 1y 1 1z"', ! ex::= f ::;r; = kmx" 1y 1zm t, f vz = f ::v = lmxk yl1zm1 23.
z = x y + xe11 1"'
=>
25. (a) zx = 6x + 2 => zx (l , 2) = 8 and z 11 =  2y =>
zv(1, 2) = 4, so an equation of the tangent plane is
z  1 = 8(x  1) + 4(y + 2) or z = 8x + 4y + 1. (b) A normal vector to the tangent plane (imd the surface) at (1,  2, 1) is (8, 4,  1). Then parametric equations for the normal ·
Ime there are x = 1 + 8t, y =  2 + 4t, z = 1 
27. (a) Let F(x , y ,z) = x 2 + 2y2

· d
.
.
x1 y+ 2 z 1 =   = =t· 8 4
t, an symmetnc equations are  
3z2 • Then F.,= 2x, Fy = 4y, Fz =  6z, so F,.,(2,  1, 1) = 4, F11 (2,  1, 1) =  4,
F: (2,  1, 1) =  6. From Equation 14.6.1 9, an equation of thetangent planeis 4(x  2) 4(y + 1)  6(z  1) = 0 or, equivalently, 2x  2y  3z = 3. · 14620 · equatwns · fior the normaII'1 ~e are X(b) From Equatwns . . , symmetnc 4 29.
2
Y + 1 = =fl· z 1 = =4
(a) Let F (x, y , z ) = x + 2y + 3z sin(x yz ).. Then F,., = 1 yz cos(x yz ), Fv = 2  xz cos(xyz ), F: = 3  x y cos(xyz) , so F:z: (2, 1, 0) = 1, Fy{2,  1, 0) = 2,·F: (2,  1, 0) = 5. From Equation 14.6. 19, an equation of the tangent plane is 1(x  2) + 2(y + 1) + 5(z  0) = 0 or x + 2y + 5z = 0. 1 1 '(b) From Equatwns . 14 .6 .20, symmetriC . equatiOns . <". ' are X 2= y + .or the normaII me  = z or x 2 = y + = z .
.
1
2
5
2
5
Parametric equations are x = 2 + t, y =  1 + 2t, z = St. 31. The hyperboloid is a level surface of the function F (x, y , z) = x 2 + 4y2

z 2 , so a normal vector to the surface at (x 0 ,y0 , zo)
is \1 F( x 0 ,y 0 , zo) = (2xo, 8yo,  2zo). A normal vector for the plane 2x + 2y + z = 5 is (2, 2, 1}. For the planes to be parallel, we need the nonnal vectors to be parallel, so (2xo , 8yo,  2zo} = k (2, 2, 1), or xo = k, yo = t k, and z 0 =  tk. I
But x5 + 4y5  z~ = 4
=>
k2 + i k2

ik2 = 4
=>
k2
= 4 =>
k = ±2. So there are two such points:
(2, ~,  1) and (2,  t , I) .
33.
f (x , y , z) = x 3 .jy 2 + z 2
yx3
zx 3
=> f x(x,'y, z ) = 3x 2 .jy2 + z 2 , f v(x, y, z) = ~· f , (x , Y.> z) = ~ ·
so f(2 , 3, 4) = 8(5) = 40, f x(2 , 3, 4) = 3(4) J25 = 60, f v(2, 3, 4). = ~ =
y2 +z2
2 : ,
and /.:: (2, 3, 4)
=
~ = ~ . Then the
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240
0
CHAPTER 14 PARTIAL DERIVATIVES
linear approximation off at (2, 3, 4) is
f(x ;y,z) ~ /(2,3,4) + J.,(2,3,4)(x ~ 2) + fv(2,3,4)(y  3) + /z(2,3, 4)(z  4) = 40 +60(x 2) + ¥CY  3) +
32 5
(z 4)
= 60x+ 254 y+ 352 z 
120
Then (1.98) 3 J(3.01) 2 + (3.97) 2 = /(1.98,3.01, 3.97) ~ 60(1.98) + ¥(3.01) + ¥(3.97) 120 = 38.656.
du p
35. d
dx au dy au dz 3 2 2 3 = au ,d +;;d +;; d =2xy (1 +6p) + 3x y (per>+er>)+4z (pcosp+sinp) uX p uy p uz p
.
az
37. By the Cham Rule, as
ax az ay = az ax as + ay as. When s = 1 and t =
2,
X
& as
= f,(3, 6)gs(1, 2) +
az at
= f,(3, 6)gt(1, 2) + fv (3, 6) ht(1, 2) = (7)(4) + (8)(10) = 108.
az 39. ax
= 2x!'( X 2 
az Y Bx
az + x {)y
az
az
y 2) '
=1
2
az  y
a z a ax2 = Yax
=
az ay
= (7)(  1) +
2y!'( X 2  y 2)
2
2
(8)( 5)
[w here !'
. . & =  47. Stmiiarly, Bt
= d( x2df
&~
6, so
&~
= ax 8t + ay at, so
y2) ] . Then
2
= 2xyf'(x  y ) + x  2xyf (x  y ) = x.
41 . a = a v+ a 2 X U V X 2
fv (3, 6) h. (1, 2)
= g(l, 2) =;:: 3 andy = h(1, 2) =
I
and
(az). 2y az a (az) 2y az (a z 8 z  y)  y (a z  y a z ) au +xa8v + ~ax av =xaav+Y au ~ + 8v8u~ +~ 8v ~+auavY 2
 y
82 z x 3 8v + Y &u2 2y az
2

2y2 a 2 z X2 &uav
2
2
2
2
2
y 2 8 2z
+ x 4 av2
Thus
.
uv
smcey = xv = 
y
2
ory =
uv.
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CHAPTER 14 REVIEW
1
yf42+(3)2
47. 'V f = (2x y,x2 direction ( 4·,
241
=> 'V f = (2xe 11 ,  x 2 e 11 ), 'V f( 2, 0) = (4, 4). The direction is given by (4,  3), so
45. j(x, y) = x 2 e 11
u =
D
(4,3} = { (4 ,3) and Duf(2,0} ...
= 'Vf{2,0} · u = (4, 4) · t (4, 3) =
t( 16 + 12} = ~.
I'Vf(2, 1)1 = 1(4, ~) I · Thus the maximum rate of change off at (2, 1} is
+ 1/(2JY)),
...
o/ in the
! ).
49. First we draw a line passing through Homestea'd and the eye of the hurricane. We can approximate the directional derivative at Homestead in the direction of the eye of the hurricane by the average rate of change of wind speed between the points where this line intersects the contour lines closest to Homestead. In the direction of the eye of the hurricane, the wind speed changes from 45 to 50 knots. We estimate the distance between these two points to be approximately 8 miles, so the rate of change of wind speed in the direction given is approximately 51 . f(x,y)=x 2  xy+y 2 + 9x 6y+10
j 11 =  x y
+ 2y 6,
= 1, x =
=2=
845 = i = 0.625 knot/ mi.
f, = ·2xy+9,
/ 1111 , f xv =  1. Then f x = 0 and ! 11
4. Thus the only critical point is (  4, 1} and fx:r:( 4, 1)
D( 4, 1) = 3
53. f(x, y)
f xx
=>
50
= 0 imply > 0,
> 0, so f(  4, 1) = 11 is a local minimum.
= 3xy 
=> fx = 3y  2xy  y2 ,
x 2 y xy 2
/ 11
= 3x x 2 
2xy,
= 3 2x 2y. Then f, = 0 implies y(3 2x y) = 0 soy= 0 or y = 3  2x. Substituting into / 11 = 0 implies
f ,, =
 2y, fv·u =  2x, f xv
x(3  x) = 0 or 3x(  1
+ x )' =
0. Hence the critical points are (0, 0), (3, 0},
(0, 3) and (1 , 1). D(O, 0} = D(3, 0} = D(O, 3) = · 9 (0, 3) are saddle points. D(1, 1)
< 0 so (0, 0), (3, 0), and
= 3 > 0 and f xx (1, 1} =  2 < 0, so
!(1, 1) = 1 is a local maximum. 55. First solve inside D. Here f .,
Then f,
=·o implies y =
y = 4 2x into / 11
and when x = 2, y
= 4y2 
2xy2

y3,
f 11 = 8xy 
2x2 y  3xy2 •
= 0 isn't inside D. Substituting = 2 or x = 1, but x = 0 isn't inside D,
0 or y = 4 2x, but y
= 0 implies x = 0, x = 0 but (2, 0} isn't inside D. Thus the only critical point inside
D is (1, 2) and f (l , 2) = 4. Secondly we consider the boundary of D. On L1: f(x, 0) =
f(  y + 6, y)
y
(O, 6)
(6 ,0)
0 and so f = 0 on L 1. On L2 : x = y + 6 and
= y 2 (6 
y)(  2) = 2{6y2

.t
y 3 ) which has critical points
at y = 0 and y = 4. Then !{6, 0) = 0 while !(2, 4) = 64. On La: f(O , y) = 0, so f = 0 on £ 3. Thus on D the absolute maximum off is !{1, 2} = 4 while the absolute minimum is /{2, 4) =  64.
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242
CHAPTER 14 PARTIAL DERIVATIVES
0
57. f(x, y) = x 3

3x + y 4

2y2 2
z 0
l \"''<1l1
1
0 y
From the graphs, it appears that f has a local maximum f( 1 , 0} ~ 2, local minima !(1, ±1} ~  3, and saddle points at (1, ±1} and (1,0}. To find the exact quantities, we calculate fx = 3x2

3 = 0 <=> x = ± 1 and fu = 4y3
y = 0, ±1, giving the critical points estimated above. Also fzx = 6x, fxu = 0, f uy Derivatives Test, D ( 1, 0} = 24 > 0 and f xx( 1, 0) =  6 D (1, ± 1)
= 12y 2 

4y
= 0 <=>
4,so using the Second
<.0 indicating a local maximum / ( 1, 0)
= 2;
= 48 > 0 and / u( 1, ±1} = 6 > 0 indicating local minima / (1, ± 1) =  3; and D ( 1, ± 1) =
 48 and
D(1, 0) =  24, indicating saddle points. 59. f(x,y) = x 2 y, g(x,y)
y = >.. If x = 0 then x 2
= x 2 +y 2 =
+y
2
1 ~
'ilf = (2xy,x2 ) = )..'iJg = (2>.x,2>.y). Then 2xy = 2)..x impliesx = Oor
= 1 gives y = ±1 and we have possible points (0, ±1) where f (0, ± 1) = 0. lfy =).. then
x 2 = 2)..y implies x 2 = 2y 2 and substitution into x 2
+ y2 =
1 gives 3y 2 = 1
~
y = ± 7:J and x =
corresponding possible points are ( ±jf, ±7:J). The absolute maximum is f ( ± jf,
61. f(x , y, z) = xyz, g(x,y,z) = ~2 + y 2 + z 2
= 3.
zero, then x = y = z = 0 which contradicts x 2 y 2 = x 2 , and similarly 2yz2 = 2x2 y
.
~
'ilf
= )..'iJg
~
±jf. The
Ja) = ~ while the absolute
(yz,xz,xy) = >.(2x, 2y,2z). lfany ofx, y, or z is
z 2 = 3. Then ).. = YZ ·=xz = xy ~ 2y 2 z = 2x2 z ~ 2x 2y 2z . Substituting into the constraint equation gives x 2 + x 2 + x 2 = 3
+ y2 +
z2 = x2 •
~
I
x2 = 1 = y 2 = z 2 . Thus the possible points are (1, 1, ± 1), (1, 1, ± 1), (  1, 1, ±1), (  1, 1, ±1). The absolute maximum is / (1, 1, 1) = /(1, 1,  1) = f( 1 , 1, 1) = f(  1,  1, 1) = 1 and the absolute minimum is /(1, 1, 1) = /(1, ....:.1, 1) = f( 1, 1, 1) = f(  1,  1,  1) =  1. 63. f(x ,y,z) = x 2 +y2 +z2 , .g(x,y,z) = xy 2 z 3 = 2
Since xy 2 z 3 = 2, x
~ xz
~ 'ilf = (2x, 2y,2z) = X'ilg = (>.y 2 z 3 , 2>.xyz 3 ,3)..xy2 z 2 ).
=I 0, y =I 0 and z =I 0, so 2x = )..y2 z 3
2 =  2 or y 2 = 3 xy z
f z 2 soy =
±z
q.
(I),
1 = )..xz3 (2),
2 = 3>.xy2 z (3~. Then (2) and (3) imply
2 2 V 3 Similarly (1) and (3) imply y~xz 3 = 3 xyz or 3x = z sox = ± jsz. But
2
?
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CHAPTER 14 REVIEW
D
243
xy2 z 3 = 2 so x and z must have the same sign, that is, x = ~ z. Thus g(x, y, z ) = 2 implies ~z Gz 2 )z 3 = 2 or
z =
±31 / 4 and the possible points are (±3 1 / 4 , 31 / 4 y'2, ±3114 ), (±3 1 / 4 , , 3.. :. 1 ; 4 ._/2, ±3114 ) . However at each of these
points f takes on the same value, 2 J3. But (2, 1, 1) also satisfies g(x, y , z) = 2 and / (2, 1, 1) = 6
> 2 v'3. Thus. f
has an
absolute minimum value of 2 J3 and no absolute maximum subject to the constraint x y 2 z 3 = 2. · . . . f( x, z ) = x 2 +  2 + z 2 . T.I1en Alternate solution: g ( x , y, z ) = xy2 z 3 = 2 .tmp I'tes y., =  2 3 , so mmtmtze 3
xz
2
fx = 2x 
2x3 z 3

X2 Z3
,
f~ =  ;. + XZ
2z, fxx = 2 +
4 X 3 Z'3
,
xz
f ~~ =
24 X Z5
+ 2 and fr~ =
6 X 2 Z4
•
Now f x = 0 implies
2 = 0 or z = 1/x. Substituting into / 11 = 0 implies 6x 3 + 2x 1 = 0 or x .= ~·so the two criti<;al points are
( ± ~, ± yt3).Then D ( ± ~ , ± V'3)
24 3 ) 
= (2 + 4)( 2 +
(
r
7J
> 0 and f xa: ( ± ~, ± V'3)
=
6 > 0, so each point
_2_3 , so the four points closest to the origin are (± 4~ , .:fl,± :y'3), (±+ , ~ , ± V'3). = · v3" V3 V3
is a minimum. Finally, y 2 =
The area of the triangle is ~ ca sinO and the area of the rectangle is be. Thus,
65.
the area of the whole object is f(a , b, c)= ~ca sinO+ be. The perimeter of the object is g( a, b, c) = 2a + 2b + e = P. To simplifY s in 0 in terms of a, b, · and e notice that a 2 sin2
b
c
Thus f(a, b, c)=
o+
4c J4a2 
2
( ~e) = a 2 
=> sino=
..!. v4a2 2a
c2.
. '
c2 +be. (Instead of using 0, we could just have
used the Py1hagorean Theorem.) As a result, by Lagrange's method, we must find a, b, c, and A by solving "il f = A"il g which gives the following equations: ca(4a2

c2 )  1 12 = 2A (1), c = 2A (2),
*(
4a2

c 2 ) 112

i c2 (4a 2
(3),and2a+2b +c = P (4). From(2),A= t candso(l)producesca(4a2  c2 ) ....: 112 =c =>
e2 = a2
4a2

(5),
~  34a
=> · c =
+ b=
.;;a
v'3 a =>
2a + a(l + v'3) + v'3a=P b = ·(2 J3
©
3~ (1 + J3) p
(5). Similarly, since {4a 2
~ 
v;a
= b =>
c 2)

b
112
=> 3a + 2v'3a=P => a =
= 3  6 J3 p and
c=
(2
2
=% (1+ vf3)
c2 )  1 12 + b = A
(4a 2  c2 ) 112 = a
= a and A= tc, (3) gives ~  !:__ 4

4a
+ b = ~ . so from 2
(6). Substituting (5) and (6) into (4) we get:
p 3+ 2 J3
= 2 J3 
=>
3 P and thus
3
V3)P.
2012 Ccngagc learni ng. Atl Ril:l>ts Rcset'·cd. May nol be SC3M<:d, copied. or dup UCill<'d. or posted 10 u pub licly •cccssiblc website, i n whole or in port.
D PROBLEMS PLUS L
1. The areas of the smaller rectangles are A1 = xy, A2 = (L  x)y, A 3 = (L x)(W y),
A4
f(x, y) = Ai +A~+ A~ +A~
= x2 y 2
+ (L 
x?y 2
T w
y
= x(W  y). ForO :::;: x :::;: L,O :::;: y:::;: W, let
1
Wy
+ (L x) 2(W 
= [x2 + (L x?JIY2 + (W 
v?
+ x 2(W y) 2
Lx
X
y)2]
Then we need to find the maximum and minimum values of f(x , y). Here f,(x, y) = [2x  2(L x)Jiy 2
+ (W 
y) 2 ] = 0
· j 11 (x , y) = [x 2 + (L  x) 2JI2y 2(W y)] = 0
+ (W 16(y2 + (W 
f :tx = 4[y2
u?J, f v11 = 4(x2
D =
y) 2 J!x 2
f x:r: = 2W2
=> 4x ' 2L =
0 or x = ~L, and
=> 4y 2W = 0 or y = W / 2. Also
+ (L x?J, and fxv
= (4x 2L)(4y 2W) . Then.
+ (L x) 2 )  (4x 2L) 2 (4y 2W) 2 . Thus when x
> 0. Thus a minimum off occurs at (~L, %W)
=~Landy= ~W, D
> 0 and
and this ·minimum value is f(~L, ~W) = ~L 2 W 2 .
There are no other critical points, so the maximum must occur on the boundary. Now along the width ofthe rectangle let g(y) = j(O, y) = f(L , y) = L 2 [y 2
L 2 [2y  2(W y)) = 0
+ (W Y? ]. 0 :::;: y:::;: W . Then g'(y) =
~
y = ~W.
And g(%) = ~L2 W 2 . Checking the endpoints, we get g(O) = g(W) = L 2 W 2 • Along the length of the rectangle let h (x) = f(x, 0) = f(x , W) = W 2 [x2
+ (L x?]. 0 :::;: x
:::;: L. By symmetry h'(x) = 0 ~
x = ~Land
h( ~L) = %L2 W 2 • At the endpoints we have.h(O) = h(L) = L 2 W 2 • Therefore L 2 W 2 is the maximum value of f . This maximum value off occurs when the "cutting" lines correspond to sides of the rectangle.
3. (a) The area of a trapezoid is ~ h(b1 + b2), where his the height (the distance between the two parallel sides) and b1 , ~ are the Ieng!hs of the bases (the parallel sides). From the figure in the text, we see that h
= x sinO, b1 = w 
2x, and
b2 = w  2x + 2x cos 8. Therefore the crosssectional area of the rain gutter is ~(x, 8) = ~xsin 0 [(w  2x)
+ (w 2x + 2xcos 8)) =
= wx sine  2x2 sin B + x 2 sin B cos
e. 0 < X
(x sinO)(w  2x + xcos8) ~w. 0
::;
< B :::;: i
We look for the critical points of A : oAfax = w sin B  4x sin B + 2x sin B cos B and EJAfEJO = wxcos B 2x2 cosB + x 2 (cos 2 8  sin2 8), so {)A j ax= 0 cos B =
4 x w = 2 w 2x 2x 0
= wxcosB 
=
wx ( 2.
= 2wx ®
(0
<8
:::;:
~ =>
~
sinO (w  4x + 2x cosO) = 0
~
sin 8 > 0). If, in addition, oAf EJB = 0, then
2x2 cosO+ x 2 (2cos2 8  1)
~) ~w 2 
2x
2
(
2
~) + x 2 [ 2 ( 2  ~)
4x 2 + wx + x
2
[
2 
1]
8 4: + ~:  1] = wx + 3x
2
= x(3x  w)
2012 Ccngage Learning. All RighiS Rescn·ed. Mny not be scanned. copied. or duplicated, or postLd lo u publicly accessible websilt.:, in whole or in p3t1.
245
246
CHAPTER 14 PROBLEMS PLUS
0
Since x
> 0, we must have x = ~w, in which case cosO=~. soB =
and A=
f, sinO=~ . k =
'{!w, b1 = ~w,ln = ~w.
{!:w 2 • As in Example 14.7.6, we can argue from the physical nature of this problem that we have found a local
maximum of A. Now checking the boundary of A, let
i w 2 sin 26, 0 < B ::; ~. Clearly g is maximized when
g( B) = A(w /2, B) = ~ w 2 sin B  ~w 2 sin B + iw2 sin Bcos 6 = sin2B = 1 in which case A= iw2 . Also along the line B =
h'(x) = w 4x = 0 *? x
~. let h(x) = A(x, ~)
= i w, and h(~w) = w(~w)
 2(tw)
2
= wx 2x2 , 0 < x < ~w .
=>
= iw 2 . Since kw2 < "If w 2 , we conclude that
the local maximum found earlier was an absolute maximum. (b) If the metal were bent into a semicircular gutter of radius r, we would have w
2
=
~7r.
../3w, 1t• would be better to bend the metal ·mto a gutter WI'th a sem1c1rcu · · Iar crosssection. . >
211'
gy (x, y)
2
2
2
. w S mce
5. Letg(x ,y)
= 1rr and A= ~11'r 2 = ~1r(~)
12
= xf(~) ·
Theng,.(:,y)
= !(~) +xf'(~) ( 
:
2)
= !(~ )  ~ f'(~;) and
= xf' ( ~) (;) = f' ( ~ ). Thus the tange?t plane at ( x 0 , y 0 , zo) on the surface has equation
[t (;~ )  y x0 !' ( ~~ )] x + [r (~~)] y  z 1
0
= 0. But any plane whose equation is of the form ax+ by+ cz
=0
passes through the origin. Thus the origin is the ~ommon point of intersection. 7. Since we are minimizing the area of the ellipse, and the circle lies above the x axis,
y
the ellipse will intersect the circle for only one value ofy. This yvalue must satisfY both the equation of the circle and the equation of the ellipse. Now 2
2
+ Y2 = 1 =>
x
a2
2
= ab2 (b2 y2 ). Substitutingintotheequationofthe
x2
b
2
circle gives
~2 (b2

y
2
)
+y2 
2y
= 0 => 2
In order for there to be orlly o ne solution to this quadratic equation, the discrirninant must be 0, so 4 4a2 b b2
a 2 b2

g(a, b)
+ a 4 = 0. The area of the ellipse is A(a, b) =
= b2

2 2
a b
+ a4
1rab, and we minimize this function subject to the constraint ·
= 0. rrb 2) (1), 2a(2a2  b
.>. = 21rb
2
2 b(l7r~ a = 47ra4
2)
:=:>
2
(2), b a
2

2 2
a b
+ a4 =
2
~a =0
0 (3). Comparing (1) and (2) gives a( ; b_ b2).
= ~ b. Substitute this into (3) to get b = ~
2 2
=> a
= jf.
@) 2012 Ccng"llc Lc""'ing. All Rights Rc>crvcd. Muy not be S
=>
15.1
MULTIPLE INTEGRALS
D
15
Double Integrals over Rectangles
1. (a) The subrectangles are shown in the figure. The surface is the graph of f(x , y) 3
= xy and 6.A = 4, so we estimate
4
2
I: I:
v~
y
i =l j = l
2
f(x;,y1 )6.A
0
= j(2, 2) 6.A + !(2, 4) 6.A + f(4 , 2) 6.A + /(4,4) 6.A + !(6, 2) LU + !(6, 4) 6.A
2
4
6
X
= 4(4) + 8(4) + 8(4) + 16(4) + 12(4) + 24(4) = 288 (b) V ~
3
2
,L: ,L: j (x.:, v1 ) 6.A = /(1, 1) 6.A + f(1, 3) 6.A + !(3, 1) 6.A + !(3, 3) 6.A + !(5, 1) LU + !(5, 3) 6.A
i=lj=l
= 1(4) + 3(4) + 3(4) + 9(4) + 5(4) + 15(4) = 144 3. (a) The subrectangles are shown in the figure. Since 6.A
Jfn.
xe"'11 dA ~
2
2
I: I: i=l j = l
= /(1, ~)
= 1 · ~ = ~ .we estimate y
f (x;1, Yi;) 6.A
I
6.A + /(1, 1) 6.A + /(2, ~) 6.A + /(2, 1) 6.A
I 2
= e 1 1 2 (~ ) +e 1 (~) +2e 1 (~) + 2e 2 ( ~) ~ 0.990
0
.
(b) ffn xe "' 11 dA ~ .
2
2
I: I: f(x;, v1) 6.A
1
1
= !(~, t) 6.A + !(~ , ~) 6.A + f(~, t) 6.A + ! (~. ~) 6.A = ~e l/8a) + ~e318(~) + ~e318(~) + ~e 9/sa) ~ u
I 2
51
5. (a) Each subrectangle and its midpoint are shown in the figure.
f
at each midpoint and estimate 2
2
X
y 4
3
. .
.
.
0
.ffn.f(x , y)dA ~
X
y
i=l j =l
The area of each subrectangle is 6.A = 2, so we evaluate
2
1
.
. .
.
2
2
I: I: f(x.: ,y1 ) 6.A
i = lj = l
= /(1, 2.5) 6.A + /(1, 3.5) AA + /(3, 2.5) AA + /(3, 3.5) AA =  2(2) + (1)(2) + 2(2) + 3(2) = 4
0
2
® 20 12 Ccngage learning. All Ril:l•ts ReserYed. May rn>l be scanned, copiC
4
X
247
248
0
CHAPTER 15 MULTIPLE INTEGRALS y
(b) The subrectangles are shown in the figure.
4
In each subrectangle, the sample point closest to the origin
3
is the lower left comer, and the area of each subrectangle is .6.A
= ~.
2
Thus we estimate
ffnf(x ,y) dA ~
4
4
L:: L::
f(xii,Yii) .6.A
i= l j = l
= f (O, 2) .6.A
0
I
2
3
4
X
.
+ f(O , 2.5) .6.A + f(O , 3)..6.A + f(O, 3.5) .6.A
+ /(1, 2) .6.A + f(1, 2.5) .6.A + /(1, 3) .6.A + /(1, 3.5) .6.A
+ / (2, 2) .6.A + /(2, 2.5) .6.A + /(2, 3) .6.A + /(2, 3.5) .6.A + / (3, 2) .6.A + /(3, 2.5) .6.A + /(3, 3) .6.A + !(3, 3.5) .6.A = 3(~) + <5) G) + < 6) (~) + <4)(~) + < 1) (~) + < 2){ ~) + <3)(~) + <1)G) + 1G) + o(~) + (1){~) + 1 (~) + 2(~) + 2G) ·;~ 1 (~)
+ 3G)
= 8 7. The val~es off (x, y)
= .j52 
x2

y 2 get smaller as we move farther from the origin, so on any of the subrectangles in the
problem, the function will have its largest value at the lower left comer of the subrectangle and its smallest value at the upper right comer, and any otl).er value will lie between these two. So using these subrectangles we have U < V < L . (Note that this is true no matter how R is divided into subrectangles.) 9. (a) With m
= n = 2, we h~ve .6.A = 4. Using the contolir map to estimate the value off at the center of each subrectangle,
we have 2
2
ffn f(x, y) dA ~ L:; L:; J(xi, Yi) .6.A = .6.A[f(1 , 1) + /(1, 3) t /(3, 1) + /(3, 3)]
i'::j
4(27 + 4 + 14 + 17) = 248
i = lj=l
(b)
11. z
/ave= Aln) ffn f(x, y) dA
i'::j
ft(248) = 15.5
= 3 > 0; so we can interpret the integral as the volume of the solid S that lies below the plane z = 3 and above the
rectangle [ 2, 2] x [1, 6]. Sis a rectangular solid, thus ffn 3 dA 13. z
= 4 · 5 · 3 = 60.
= f(x , y) = 4 2y :2: 0 for 0 :::; y :5 1. Thus the integral represents the volume of that
part ofthe rectangular.solid [0, 1] x [0, 1] x (0, 4] which lies below the plane z = 4  2y.
So
ffn (4  2y) dA
= (1)(1)(2) + H 1)(1)(2) = 3 X
© 2012 C.ngnge L
SECTION 15.2
ITERATED INTEGRALS
D
249
15. To calculate the estimates using a programmable calculator; we can use an algorithm ,.
similar to that of Exercise 4.1.9 [ET 5. 1.9). In Maple, we cari define the function
n
estimate
f( x, y) = v'1 + xe
' 1
1.141606
4
1.143191
middlesu m(middles um(f , x= 0 . . 1 ,m),
16
1.143535
y=O .. 1 , m);
64
1.143617
256
1.143637
1024
1.143642
11
(calling it f), load the student package, and then use the
command
to get the estimate with n = m 2 squares of equal size. Mathematica has no special Riemann sum command, but we can define f and then use nested Sum commands to calculate the estimates.
17. If we divideR into mn subrectangles, ffn k dA:::::: m
points,
m
i=1j=l
n
L: L:
i=1j=l
f(xt;, Yii) ~A = k nt
ffnkdA=
15.2
n
= k always and L: L:
But f(xi;. Yi;)
lim
n.
L L:
m ,n +ooi = l j = l
.f _t
<=13=1
~A= area of R n
""
L · L:
i = lj=1
f (xi; . Yi;)
~A for any choice of sample points (xi;, Yi;).
= (b a) (d c). Thus, no matter how we choose the sample
~A= k(b  a)(d  c) and so
f(xi;.Yi;) ~A =
m
lim
rn,n+oo
k
n
L L
i= l j=l
~A=
lim
Tn,n+oo
k(ba) (d c)=k(ba)(d c).
Iterated Integrals
3.
f 14 J02 (6x 2 y 2x) dy dx = f 14 [3x 2 y 2 
5.
J~ J04 y 3 e 2 "' d.yd:c = J02 e 2"' dx .r: y 3 dy
7.
J~ 3 J; 1 2 (y+y 2 cosx) dxdy
2xy]~=~ dx
= f14 (12x2 
4x) dx
=
[4x3

2x2 ] ~
= [te2 "']~ ( h 4 ] : = t(e4 
[as in ExampleS]
= (256 
1) (64  O)
32)  (4  2)
= 32(e 4 ·  1)
= J2 3 [xy+y2 sin xJ::~12 dy = J~3 (~y + y2) dy = [~y2 + h3]~3 = (9411' + 9 
9.
/,4/,2(X+JL 1
1
Y
? ) X
dydx=
(~;  9)]
f01 J01 v(u + v2) 4 dudv
=
J01
[iv(u + v 2 ) 5
= t fo1
1
to [(26 
1
lf ln2 +3ln41 / 2 = ¥ ln 2 ]~=~ dv = t j~ v ((1 + v 2) 5  (0 + v 2) 5 ] dv ~ ln 2 =
1
(v(1 + v2)5 vn] dv
[substitute t
=
2
2 y=
X
= 8ln2 + ~ ln4 11.
= 18
2 4 /,4[x lnlvl+·1 1·y ] 3 ) dx = [ ~x ln 2+~ ln lxlf dx = l (x ln 2+: 2 y=l . 2x 1
= 1+v
2
1)  (1  0)]
=
==:>
Ht. i(l + v2)6 dt
= 222
J:\vl2]~
= 2v dv in the first term]
= to (63 1) = ¥o
€) 2012 Cengnge l..co.ming. All Rights Reserved. May not be scanned, copied, or duplicated, or )"llStcd to o. publicly acccssibh: wcbsilc, in whole or in part.
250
0
CHAPTER 15 MULTIPLE INTEGRALS
13. J~ f 01f r sin2 8 d8 dr = f~ r dr
J; sin2 8 d8
[as in Example 5] =
J
2 0
r dr
J0 t (1 1f
cos 28) d8
[tr 2 )~· ~ [0 ~ sin28)~ = (2  0) · ~ [(1r ~ sin 27r)  (0 ~sinO)]
=
= 2. ~[(7r 0) (0  0)] = 7r
. 15. ffn sin(x  y) dA
= Io1f12 Iorr~
2
= I; 12 [cos(x 
sin(x y) dyd:r:
y)]~~~l 2 dx = Iorrl 2 [cos(x  ~) cosx] dx
1f 12 . = [sin(x~)si?x] = sinOsin~, (sin(~)  sinO] 0
=01(10) = 0
17.ffn X:~ 1 dA = 11
/_33
X::
1 dydx =
11
:r;2:
= ~ (In 2 In 1) . t(27 + 27) =
[t ln(x2 + 1)]: [h3] ~3
1 dx ;_: y2 dy =
91n 2
19. Io1riG j~1rl 3 x sin(x + y) dy dx
= IorriG [xcos(x + y)]~=~13 dx = J; 16 [x cosx xcos(x +f)] dx
= x (sin x =
sin (x +
f) )~ 16 
J0 16 [sin x 1f
sin ( x
+ f)] dx
[by integrating by parts separately for each term]
i [~ 1) (cosx + cos(x + f)]~ 16 =  & [~ + 0 
(1 +
~)]
=
q
1

f2
= J03 (e"' 11J ::~ dy = I:(e 2 il + 1) dy = [~e 2 Y + y]~ ( ~ + 0) = t~6 + ~
21. IInye:z:il dA =I: I: ye"'11 dxdy
= ~ e6 + 3 23. z = f(x, y)
= 4 x 
2y ~ 0 for 0 ~ x ~ 1 and 0 ~ y ~ 1. So the solid
is the region in the first octant which lies below the plane z = 4  x  2y and above [0, 1] x {0, 1].
X
25. The solid lies under the plane 4x + 6y  2z V
= fin(2x + 3y +
1
;)'dA =
= j~ 1 ((19 + 6y) 27. V = I~ 2 f~ 1 (1 f2 = 4 Jo
[
(
13 2 
tx h
1 3 X  fiX 
2

2
)
+ 15 =
0 or z = 2x + 3y +
1/ so
I~ 1 I~ 1 (2x + 3y + Jf) dxdy = j~ 1 [x + 3xy + ¥xJ::~ 1 dy 2
(¥' + 9y) dy =
3y)) dy = I~ 1
dx dy = 4
J; I
1 0
(1 
ix
2

1 2 ] :z; = 1 d 4 Jo f2 ( 11 1 2) d iiY X :z:=O y = i2 iiY y
[Yf y
+ ¥Y2 )~ 1 =
30 ( 21) = 51
~y 2 ) dx dy 11 = 4 [i2Y 
l 3] 2 83 166 vY 0 = 4. 54= 27
29. Here we need the volume of the solid lying under the surface z = x sec2 y and above the rectangle R
= (0, 2] ~ [0,
the xyplane.
V =
I:
Iorr
14
2
x sec y dy dx
= (2 O)(tan f
=
I:
x dx Io"
14
sec2 y dy = [ ~ x 2 ] ~ [tan y] ~
14
 tan 0) = 2(1  0) = 2
© 2012 Congagc Learning. A ll Rights R=:"·cd. May not bo scanned, copkd. or duplicalcd, or poJicd 10 u publicly occcssiblc wcbshc, in who le or in part.
1r /
4] in
l SECTION 15.3
DOUBLE INTEGRALS OVER GENERAL REGIONS
0
251
+ (y 2? and above the plane z = 1 for 1 ~ x ~ 1, 0 ~ y ~ 4. The volume of the solid is the difference in volumes between the solid that lies under z = 2 +x2 + (y 2) 2 over the rectangle
31. The solid lies below the surface z = 2 + x
2
R = (1, 1] x (0, 4) and the solid that lies under z = 1 over R . V
= J; J~ 1 (2 + x 2 + (y :._ 2?J dxdy f04 f~ 1 (1) dxdy = f04 = J; [(2 + ~
+ (y 
[2x + 4x
3
+ x(y 
2?J:: ~ 1 dy
t
1
dx J;dy
2) 2 )  ( 2 ~ (y 2) 2 )] dy [x]~ 1 (y]~
f04 (lf + 2(y 2) 2 ) dy (1  (1)][4 OJ = = [e; + ¥) (o  1nJ 8 = ¥8 = ¥
=
( 134 y
+ t(Y 2?)~ (2)(4)
33. In Maple, we can calculate the integral by defining the integrand as f
and then using the command int ( int ( f, x=O .. 1) , y=O . . 1) ; . In Mathematics, we can use the command
Int egrate[f,{x,0,1}, {y,0,1} ] We find that JJn x 5 y 3 e
x (0, 5]. Thus, A(R) = 2 · 5 = 10 and
35. R is the rectangle (1, 1]
5 1 f' r !( ) dA 1 r 5Jt 2 d d _ 1 rs [ 1 3 ]"' = 1 d _ 1 rs 2 d 1 (1 2] s f,•"• = A(R) Jn x,y =wJo lxy. x Y  wJo 3xv.,=  t Y w Jo3Y Y= to 3Y o = e·
37.
!1
xy 4 dA=
R1+X
1
11 11 1
0
1 1
 xy
1 +X4
dydx =
 1
 x .4 dx 1 +X
11 0
·.
ydy [byEquatJOnS] butj(x) =  x 4 is anqdd 1 +X
1
function so
1
f(x) dx = 0 by (6) in Section 4.5 [ET (7) in Section 5.5]. Thus/" {
xy
ln 1 + x
 1
39. Let f(x, y) = ( x  )
x+y
3
•
4
dA = 0. { y dy = 0.
lo
t and f01 f 01 f(x, y) dx dy = t.
Then a CAS gives J; f0 f(x, y) dydx = 1
To explain the seeming violation of Fubini's Theorem, note that f has an infinite discontinuity at (0, 0) and thus does not satisfy the conditions ofFubini's Theorem. In fact, both iterated integrals involve improper integrals which diverge at their lower limits of integration.
15.3 Double Integralsover General Regions
3.
2
J; J:2 (1 + 2y)dy dx = f01 [y + y
J ~:: 2 dx
= .r; [x + x 2
x2

(x2 ) 2 ) dx
= fo1(x  x4)dx = [tx2  !xs)~ = t i  0+0 = fo
® 2012 Ccngagc LC
252
7.
D
CHAPTER 15 ' MULTIPLE INTEGRALS
IIv y2 dA = I~t I~~2 Y2 dx dy = I~t (xy2J ::~v2 dy = I~l y2 [y  ( y 2)] dy 1 = J11 (2y3 + 2y2)dy = [ly4 + 1y3J = 12 + 13  l2 + 13 = :!.3 2 3  1
9. ffv x dA = ·
f"
~" [xy]Y:'" oin "' dx = f" x sin x dx YJo
ro•ln"' x dy dx =
.lo J t
Jo
= [ x cos x + sin x J~ =
1r cos 1r
+ sin 1r + 0  sin 0 =
[ .
integrate
by parts
wtth v.=x, rlv = sm x dx
]
1r
y
11. (a) At the right we sketch an example of a region D that can be described as lying
betw~en the graphs of two continuous functions of X (a type I region) but not as lying between graphs of two continuous functions ofy (a type IJ region). The regions shown in Figures 6 and 8 in the text are additional examples. 0
X
y
(b) Now we sketch an example of a region D that can be described as lying between the graphs of two continuous functions ofy but not as lying between graphs of two continuous functions of x. The first region shown in Figure 7 is another example.
0
13.
y
X
As a type I region, D lies between the lower boundary y (1, 1)
= 0 and the upper
boundary y = x for 0 ::; x ::; 1, soD= {(x, y) I 0 $ x $ 1, 0 $ y $ x}. lfwe describeD as a type ll region, D lies between the left boundary x
x =l
right boundary x
= 1 for 0 ::; y ::; 1, soD =
= y and the
{( x, y) I 0 ::; y::; 1, y ::; x::; 1}.
X
Thus
IIv xdA =I; IC: xdydx = I 01 [xy] ~ =~ dx = I01 x 2 dx =
.rf'v x dA = f 01 15.
I:
xdxdy =
I
1 0
(tx 2 J : : ~ dy =
y
t J;(1 
y
2
)
!x 3 ]~ = !(1  0) = ~or
dy = t[y  h3 ] ~
= t((1 !)  0] =
~
The curves y = x  2 or x = y + 2 and x = y 2 intersect when y + 2 = y 2
y 2  y 2 =0 ~ (y2)(y+ 1)=0 ~
~
y = 1, y =2,so thepoints .of
inte rsection arc (1 ,  1) and (4, 2) . Tf we de~c ribe D as a type I regi on, the uppe r
boundary curve is y = y =
Thus D = {(x, y)
I 0::; X::; 1,

..fii but the lower boundary curve consists of two parts,
vx for 0 ::; x ::; 1 andy = x 
2 for 1 ::; x ::; 4.
vx::; y::; vx} u {(x, y) 11 ::; X::; 4, X 2::; y::; vx} and
II0 y dA = Io1 I!5z y dy dx + f14 f.,~ y dy dx. If we describe D as a type II region, D is enclosed by the left boundary
.
.
x = y 2 and the right boundary x = y + 2 for 1 ::; y ::; 2, so D
= {(x, y) I 
1 ::; y ::; 2, y 2
::;
x ::; y + 2 } and
.
DOUBLE INTEGRALS OVER GENERAL REGIONS
SECTION 15.3
JJ0
y dA
= f~ 1 J:/ 2 y dx dy.
0
In either case, the resulting iterated integrals are not difficult to evaluate but the region D is
more simply described as a type II region, giving one iterated integral rather than a sum of two, so we evaluate the latter integral:
ffo ydA = f~tf:/2 ydxdy = [jy3+y2
= f~t (xyJ ::~~2 dy = J~1 (y + 2 y2)ydy = f~t (y2 + 2y  ya) dy
h4]~1 =
(! +44) 
(~ +1  i) = ~
19.
0
=
X
21.
3 [§.y 3
!2 !~ (2x 
y
 2
2
y4 ] 1

~
16 ~3
= .§13 
y) dydx
2 /
=
}~ 2 [2x v'4  x 2 : H4  x 2 )
2
[ 2xy
12] !1=~  2Y r.;; dx 2

 2
+ 1 ll 3
y =y4 x
=J~24xv'4x2dx=
+ 2x v'4 x2 + H4 
H4x2)3/2]2
x 2)] dx
= 0
2
[Or, note that 4x v'4  x2 is an odd function, so J~ 2 4x v'4 x2 dx = 0.] y
23.
V=
2 f1 J1a: ( 1 _., 1
10
1 1
=
[ ((1
x
2
x+2y ) dydx=
) 
x(1 x
2
)
f1 [
10
+ (1 
2
y l x yxy + y 2] !1;: _., dx 1
x
2
?)
 ((1 x) x(1 x) + (1 x) 2 ) ] dx
= J01 [(x 4 + x3  3x2 
X
25.
0
y
V=
0
x + 2) (2x2

4x + 2)] dx
3y xy dx dy = J21 [21 x 2 y ] .,"' == 17 3y dy J12J71
X
©
2012 Cengagc Lc=ing. All Rights Reserved. May not be scanned. copied, or duplicated, or posted to o publicly accessible website, in whole or in part.
253
254
27.
0
CHAPTER 15 MULTIPLE INTEGRALS y
f02 f~  ~"' (6 3x 
V =
(0,3)
r2(
= Jo 6y  3xy  y
= J; (6(3
2y)dydx
2] vv= 3 1x = o 2 dx
 ~x) 3x(3  ~x)  (3  ~ x) 2] dx
= J; (~x2  9x + 9) dx = [~x 3  ~x2 + 9x] ~ = 6 0 =
6
29.
31.
y
V
=
111~ 0
{
11[2 ] Y= ~
=
y dy dx
1L
0
1
= Jo
2
1 X  2 dx
2
0
1
1
= 2 [x 
3·c
.
dx
y= O
3 1
Jo = 31
X
= 0 and
From the graph, it appears that the two curves intersect at x
33.
at x
~
1.213. T hus the desired integral is
x x•l JfD X dA ~~ Jorl .213J3x
r L213( = Jo 3X 2 ~
2
X
X
3
dy u.'C .J. 
pla n~
z=
xy ] y =
3
2
"' "'
y = x4.
6) d X= [X 3 
1 4X
4
1 aX
dX
G] O1.213
0.713
= 1  x 2 and y = x 2  1 intersect at (±1, 0) with 1 2x·+ 2y + 10 is above the plane z = 2  x  y, so
35. The two bounding curves y region, the
X
fl. 213 [ Jo
2
x ~x 2
2
1 on ( 1, 1]. Within this

2
•r1_x (2x+2y + 10 ) dydx ,rl V = _1 1 J,. _ 1 Ji~ ., 2 _ ( 2xy) dydx 2 1 1
J
2
( = J•l_1 J1x .,2_ 2x +' 2y + 10 ( 2 x 1
= j •l_1 f,l2:c2 _ 1 (3x + 3y + 8) dy dx =
• y )) dy dx
3 2 Jy = l  :c2 f_1 1 [ 3xy + 2Y + 8y y=z 2 _ dx 1
= J~ 1 [3x(l  x 2 ) + ~(1  x 2 ) 2 + 8(1 x 2 ) 
3x(x 2
= J1 1 (6x"3 
3 .l&x 3
6
= ~  ~
16x2
4 + 6x + 16) dx = [.:!x 2
+ 3 + 16 + ~  ¥  3 + 16 =

1)  ~(x 2  1)2

8(x2

1)] dx
+ 3x2 + 16x] 11
64 3
@) 2012 C
SECTION 15.3
37. The solid lies below the plane z or x
+y +z =
= 1
DOUBLE INTEGRALS OVER GENERAL REGIONS
D
255
x y y
1 and above the region
D={(x,y)IO~x~l,O~y~ 1 x}
in the xyplane. The solid is a tetrahedron. X
,
X
=x3 
39. The two bounding curves y
x andy
= x2 + x intersect at the origin and at x = 2, with x 2 + x > x 3 
x on (0, 2).
· Using a CAS, we find that the volume is V =
121x2 +x o
Z
41 . The two surfaces intersect in the circle x 2
Using a CAS, the volume is '
43.
d d
y
'121:r:2 +:r:(
X=
0
x3 x
!In
(1 x
2
+ y 2= y

2
)
3 4
y
,+. XY2) d y dX=
13,984,7:'!5,616 " 14,549,<>35
1, z = 0 and the region of integration is the disk D : x
dA
=
D
y
X
xa  :z:
. 111~  1
(1  x 2

y
2
)
2+ y
2
~ 1.
= ~.
dydx
2
Vlx2
Because the region of integration is D = {(x,y) I 0 ~ x ~ y, 0 ~ y ~ 1}
we have f 0 J~ f(x, y) dx dy = 1
ffv
= {(x , y) I x
f(x, y) dA =
~ y ~ 1, 0 ~ x ~ 1}
f01
J:
f(x, y) dy dx.
Because the region of integration is
45.
D = {(x,y) I 0 ~ y ~ cosx,O ~ x ~ 1rj2} = {(x,y) I 0 ~ x ~ cos 1 y, O ~ y ~ 1}
we have foTr/2 focou f(x,y)dydx
= ffv
f(x , y)dA
= fol f ocoslv
f( x,y)dx dy.
Because tlie reg_ion of integration is 
47.
= {(x , y) I 0 ~ y ~' Inx, 1 ~ x ~ 2} = {(x ,y)
D
In 2
J
e 11
~ x ~ 2, 0 ~ y ~ ln2}
we have
/, 211" l
49.
11 1
y
o
x
f(x,y).dydx
0
3
3u
ex 2 dxdy
= ~~
f(x,y)dA
= 11n 2[2 f(x,y) dxdy 0
D
= 131:r./3 ex2 dydx = o o
13[
2
ex y
o
cV
]u=x/3dx u=O
(3, 1)
X
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2012 Ccngage Learning. All Rights Rescr\'cd, May not be scanned. cor h.."tl, or dupliCllted, or posted to n pUblicly ncces:~iblc website. in whole or in part.
256
0
CHAPTER 15 MULTIPLE INTEGRALS y
51.
53.
y
111
1
y = sinx or
x= arcsiny
0
\
cosx )1 + cos2 xdxdy
arcsmy
= f 01f
J;
1 ""'
cos x ~:os 2 x dy dx
J;12 COSX Vl + COS 2 X [y]~=~in:c dx f.o7r/2 COS X Vl + COS2 X sin X dx. [Let u = cos x, du ,;, 
= =
X
12
sin X dx, ] dx = du/( sinx)
0
~2 du = Hl+u) 2 3/ 2] =J10 uvl+u 1
=Hv'B1)=H2v'2 1) 55. D = {(x, y) I 0 $ x $ 1,  x
+1$
y $ 1} U {(x, y) I  1 $ x $ 0, x
+1 $
y $ 1}
U{(x,y) I O$x$1,  1 $y$x 1}U {(x,y) l 1 ::;x::; o, 1::;y::; x 1}, alltypel.
fir
x2 dA =
D
11,1 x
1 0
1
=4 {
fo
e 1 116
2
:::;
dydx
+
1011
1
{
l1 "'
 1
2
x dydx
+
x+1
11J•x1 x 0
2
dydx
+
101x1 1
2
2
J,h e(x2+112 ) 2 dA:::; 1 · A(Q) .
2
2 2
+v l dA
1
[by symmetry of the regions and because f(x, y) = x ~ 0]
x 2 dydx
2
< JJQ e  <x
2
x dydx
 1
e<x +t?l $ e0 = 1 s ince et is art increasing fun~tion. We have A(Q) = ~1r (~) =
e 1116 A(Q) :::; 0.1844
2
1x
~ {Ge 1116
:::;
fG, so by Property II,
JJQ e(x2 +112 ) 2 dA:::; {G or we can say
< 0.1964. (We have rounded the lower bound down and the upper bound up to preserve the '
.
inequalities.) 59. The average value of a function f of two variables defined on a rectangle R was
defined in Section 15. I as
!ave
=
AlR) ffn f(x, y)dA. Extending this definition
to general regions D, we have fnvc =
1 A( D)
y
(1,3)
ffv f(x, y)dA.
Here D = {(x, y) I 0$ x $ 1,0$ y $ 3x}, so A(D) = ~(1)(3) = ~and
f .... =
1 A( D)
ffv f(x,y)dA = ~ fo fo x xydydx 1
3
= 1 fl [lxy2] v=3x dx = 1 f1gx3 dx = lx4]1  1 3
Jo
2
v=O
3
Jo
4
o 
4
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X
SECTION 15.3
61 . Since m
DOUBLE INTEGRALS OVER GENERAL REGIONS
0
257
:S j(x, y) :S M, ffD mdA ~ ffD j(x, y) dA ~ ffD M dA by (8) =>
mjf0 ldA ~ fj~ j(x, y)dA ~ lVI .ffD ldA by(?) => mA(D) ~ ffDJ(x,y)dA ~ MA(D) by (10). y
63.
First we' can write JJ0 (x
+ 2) dA =
JJD xdA + JJ0 2dA. But f(x, y ) =xis
an odd function with respect to x [that is, f( x, y) =  f(x , y)] and Dis symmetric with respect to x. Consequently, the volume above D and below the 3
graph off is the same as the voiume below D and above the graph off, so
X
JJD x dA = 0. Also, .f.fD 2 dA = 2 ·A( D) = 2 · ~7r(3) 2 = 97r5ince Dis a half disk of radius 3. Thus JJD(x 65. We can write JJ0 (2x + 3y) dA plane z
= JJ0
2x dA
+ JJD 3y dA.
+ 2) dA = 0 + 97r =
97r.
JJ0 2x dA represents the volume of the solid lying under the
= 2x and above the rectangle D . This solid region is a triangular cylinder with length band whose crosssection is a
triangle with width a and height 2a. (See the first figure.) (a,0,2a)
(O,b, 3b)
z= 3y
y
X
2
Thus its volume is~· a· 2a · b = a b. Similarly, JJ0 3ydA represents the volume of a triangular cylinder with length a, triangular crosssection with width band height 3b, and volume
JJ0 (2x 67.
ff0
(ax3 + In/+ va2

t ·b · 3b · a ~ ~ab • (Sec the second figure.) T~us
+ 3y) dA =
2
a2 b + ~ab2
x2 ) dA = ff0 ax3 dA + JJ0 by3 dA + JJ'a Ja2

3
x 2 dA. Now ax is odd with respect
to x and by 3 is odd witb respect to y , and the region of integration is symmetric with respect to both x and y, so ffD ax 3 dA
JJ'v J a 2 
= ffD'by 3 dA = 0.
x2 dA represents the volume of the solid region under the
graph of z = Ja 2

x 2 and above the rectangle
p, namely a half circular
cylinder with radius a and length 2b (see the figure) whose volume is
t · 1rr2 h = ~7ra2 (2b) = 1ra2 b. Th~s JJD (ax3
+ by3 + Ja2 
x 2 ) dA = 0 + 0 + 1ra2 b = 1ra2 b.
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258
CHAPTER 15 MULTIPLE INTEGRALS
D
15.4 Double Integrals in Polar Coordinates 1. The 'region R is more easily described by polar coordinates: R = { (r, B) I 0 $
r::; 4, 0::; B::;
3 ; }.
Thus JJR f~x, y) dA = J:'~~' 12 f 0 f(r cos fJ, r sinO) r dr dB. 4
3. The region R is more easily described by rectangular coordinates: R = { (x, y) Thus JJRf(x,y) dA = J~ 1 5. The integral
R
J:;t J
2 1
I 1 :=:; x
:=:; 1, 0 :=:; y :=:; ~ x
+ ~}.
J0("'+ 1)/ 2 f(x,y) dydx.
r dr dB represents the area of the region .
y
0=31T 4
= {(r,B) 11::; r::; 2, 7r/4 $ '0::; 37r/4}, the top quarter portion of a
ring (annulus):
tj
J:;
t d()) (f
2 1
r dr dO = ( J:;
2 1
= [B]3'11'/4 [!r2]2 'II' /4
7. The half disk
2 
1
r dr)
=
= 2
{3.,. _ 4
.!!.) . 12 (4 _ 1) = 1!.. i! = 4 2 2
3'11' 4
D can be described in polar coordinates as D = { (r, B) I 0 ::; r ::; 5, 0 ::; B ::; 1r}. Then ffv · x 2 y dA = J0"
9. JJR sin(x
X
J; (r cos B) (r sin B) r dr dB= (J 2
0.,.
[% cos3 OJ~ [ir 5 ]~ =
(!
+ y2 ) dA = JJ:/ 2 J{ sin(r2 ) r dr dB=
12 0.,.
cos2 Osin BdB) ( J~ r 4 dr)
%(11) · 625 =
dO)
(J
3 1
i
12 0
r sin(r2 ) dr)
= [0]~12 [~ cos(r 2 )]~ = (:~) [~(cos9 cos 1)]
11 . f ·r Jn
e
_ ,2 _ u2
=
;f(cos 1  cos9)
dAJ'll'/ 2 f2 r2 d dB J"'/2 dB r2 r2 d '11'/2 Jo e r r  '11'/2 Jo re r 2
2 2 , = [B]"'/ [ler ) = '11'/2 2 0
1r{l)(e4 2

e0 )
= 1!.(1e4 ) 2 y
13. R is the region shown in the figure, and can be described
y=x
by R = {(r,fJ) I 0::; B ::; rr/4, 1$ r::; 2}. Thus
JJR ~ctan(y/x) dA = f0'~~' 14 f 1 arctan(tan B) r dr dO since yjx =tan 0. 2
Also, arctan( tan B) = B for 0 ::; B ::; . .lo f":r/4
f20 '"1 r dr uu 
f'11' / 4"d(} u
Jo
1r
/4, so the integral becomes
f21 r d r 
[!02)"'/4 2 o
[l2 r 2]21 
.,2.
32
~_1_ 2 2  a.t7r ·
15. One loop is given by the region
D = {(r,O) 1rr/6 S 0 $ rr/6, 0 S r $ cos30}, so the area is
11 ! dA
=
7r
/61cos SO
'11'/6
D
.
rdrdO =
0
~.,. /6 [ 1 2] r=coo 30 r '11'/6 2 r=O
16
_! cos2 30 dfJ = 2 f" }_71: 16 2 }0
= {"'
1[
16
_
dO
_! ( 1 +cos 60 ) dO 2
2
6
1 sin 60]"'/ = ;; = 2 0+G 0
®
2012 Ccngnge Learning. All R.icJtts Rcscr"cd. Mil)' not be scnrmcd. copied. or duplicated. or poslt.d to u publicly ncccssiblc websi te , in whole or in part.
SECTION 15.4 DOUBLE INTEGRALS IN POLAR COORDINATES
17. In polar coordinates the circle (x 1}2 + y 2 2
and the circle x + y
2 cos 0 = 1 D
~
2
=1
x2 + y 2
¢}
= 2x
is
= 2r cos 0
r2
~
~ 2 cos 0, 0 ~
0=
7r / 3,
= 2 cos 0,
so the portion of the region in the first quadrant is given by
0 ~ 7r/2}. By symmetry, the total area
O= Tr/3 ,/
is twice the area of D:
2A(D}
1'
259
is r = 1. The curves intersect in the first quadrant when
cos 0 = ~ · .~
= {(r, 0} 11 ~ r
=1
0
r=2oos(J
= 2ffvdA = 2Ja"/3 j l2coao rdrdO = 2fo"/3 ar2)~~~coao dO =
J;/3 (4cos2 0 1) d(} = J;/3 [4 · ~( 1 +cos 20} 1] dO
=
J0" 13 (1 + 2 cos 20} dO= [0 + s~n20] ~;a = f
21. The hyperboloid of two sheets  x 2  y 2 + z2
= 1 intersects the plane z = 2 when x2 
= )1 + x 2 + y 2 and below the plane z = 2 for x
solid region lies above the surface z
II (
V=
+ ~
2
) 1+
x 2 + y 2 ) dA =
2 la " 1av"J ( 2 
2
y 2 + 4 = 1 or x 2
+ y
) 1+
2
:::;
+ y2 =
3. So tlie
3, and its volume is
r2) r dr dO
:r:2 +y2 ~3
= f~" dO
f ov"J (2r rv'f+'Ti) dr
= [0] ~.,.. [r2  t {1 + r 2 ?12 ] :
= 27r (3 i 0 + 1) = ~7r 23. By symmetry, 2 V = 2
Ja 2
//

x2 
y 2 dA = 2 1a "1aa
)a 2  r 2
rdrdO = 2 1az... dO
26. The cone z = Jx 2 + y 2 intersects the sphere x 2 + y 2 + z
V =
loar Ja2  r2 dr
.
~+~~~
/1 (
)1 x
2
= 1 when x 2 + y 2 +
( Jx 2 + y2 )
2 y 2 )x2+y2) dA = la27rlal//2 () 1  r2 
2
= 1 or x 2 +
y2
= ~· ~0
r )rdrdO
.,2 + y2 ~ 1/ 2
= f~1r dO f 011V2 (r JlT2 
r 2 ) dr = [ 0]~7r
[~(1  r 2 ) 3 12
27. The given solid is the region inside the cylinder x 2 + y 2

~r3) :/.12 = 271'( ~) ( ?z  1) = f(2  V2)
= 4 between the surfaces z
= J64  4x2  4y2
and z =  )64 4x2  4y 2 • So V = x2
=
IIs [J
+ 112
64  4x2  4y2  ( :
J 64 
4x2

4
+2
4J~.,. J~ ~ rdrdO = 3 2
= 8Tr(  ~)(12 1
JJ
4y 2 ) ] dA =
162 ' 3 )
=
4J:.,.. dO J: r v'16  r 2 dr = 4
8
; .(64 
+112
2
J 64 
4x2  4y2 dA
~4
[ OJ~". [1(16 r 2 ) 312
J:
24 v'3)
© 2012 Cengage Learning. All Rights Rc"S<m:d. May nol be seanned, copied, or duplicaled, or posl~ 10 a publicly accessible wcbsile, in whole or in part.
260
D
CHAPTER 15 MULTIPLE INTEGRALS
{~ sin(x
3
29. /
3lo ·
2
+ y 2 )dy dx = {"
r sin e) r dr dB
Jo Jo
= fo" dB
J; r sin (r
2
)
dr = [B)~ [~cos (r 2 )J~
= 1r (~) (cos9 1) = ~ (1  cos9) 3
0
3
X
y
31.
f0" 14 fov'2 (r cos B + r sin fJ) r dr dB == f 0" 14 (cos B + sin B) dB f 0v'2 1·2 dr == [sin(} cosB)~/
= (¥ 33. D
4
[i r 3 ]~
 ¥ O+ 1]
. i (2v'2
o)
=¥
= {(r, B) [ 0::; r ::; 1, 0 ::; 8::; 271' }, so 2
2
ffD e<x +1h dA fl
,,.1 d
27rJo re
r
~
= 1~ ... J;
e
2 2 )
r dr dB =
J:" dB f
1 0
4
rer dr
= 271' J01 r er dr. 4
Using a calculator, we estimate
4.5951.
35. The surface of the water in the pool is a circular disk D with radius 20ft. If we placeD on coord inate axes with the origin at the center of D and define f(x, y) to be the depth of the water at (x, y), then the volume ofwater·in the pool is the volume of the solid that lies above D = { ( x, y) I x 2 + y 2
::;
400} and below the graph off (x, y). We can associate north with the
positive ydirection, so we are given that the depth is constant in the x direction and the depth increases linearly in the ydirection from f(O, 20) = 2 to f(O , 20) The slope of this line is 20 ~(_?20 ) ~
= 7. The trace in the yzplane is a line segment from (0, 
t. ~o an equation of the line is z 7 =
ty +~·Thus the volume is given by ffv
independent of x, f(x, y) =
using polar coordinates. Then D = { (r, fJ)
I 0 ::; r
t(y 20)
=?
20, 2) to (0, 20, 7).
ty +
z=
~· Since f(x, y) is
f(x, y) dA, which is most conveniently evaluated
::; 20, 0 ::; B ::; 271'} and substituting x
= r cos B, y = r sin B~he integral
becomes
J;. .1~ tr sin 20
(
(J
+
~) r dr dB = 1~... [i4r 3 sin B + ~r 2 ] ;:: ~ dB =
0
1
0
= ( ~ cos B + 9008] ~"
g•· ( ~
10 0
s in O+ 900) dB
= 180071'
Thus the pool contains 18007r ~ 5655 ft3 of water. 37. As in Exercise 15.3.59,
so A(D)
= 1rb 2 1ra2 1 !ave= A(D)
f,,.•
=
= 1r(b 2 
!1
D
_ 1 7r(b2  a2)
®
A(~J
ffv f(x, y)dA. Here D = {(r, fJ) I a::; r::; b, 0::; ()::; 271'},
a 2 ) and
1 lb 2
1 ~ dA y x 2 + y2
= 7r (b 2 1
a
2)
"
0
. a
1 rdrdB o; r·2
V
= 7r (b 2 1
a2
)
12" ;·b 0
dB
a
dr
[e]2rr (]b
1 ( )(b ) 2(ba) 2 o , rn=11'(b2  a2) 2 7r a = (b+a)( b  a) = a+b
2012 Ccngagc learning. All Rights Reserved . May not be sctmncd, co pied, or duplicated, or posted to n pub licly <Jccessiblc webs ite, in whole or in pan.
SECTION 15.5
APPLICATIONS OF DOUBLE INTEGRALS
0
261
·Y
=
1rr/4 [ r r r cos 8 sinfJdrd8= Jo 4 ?r/4
2
4 ] r=2 .!:.... cos8sin8 d()
3
o
J1
[sin
r = 1
4
2
4
151rr/ . 15 ()] .,.; 15 sm8 cos9d(J =   = = 4 0 4 2 0 16 •
.
2
2
= x and dv =·xex dx. Then du = dx and v =  ~ex , so 2 2 2 2 f "" x 2 e "' dx = lim rt x 2 e"' dx = lim (  l xe "' )t + rt l e "' dx) Jo Jo t+oo 2 Jo 2
41. (a) We integrate by parts with u
t+oo
0
(_.!.tet2) + .!. roo ex 2dx = 0 + .!.
= t+oo lim
2
= 41 f ""
oo
e
2
_.,2
d
Jo
2
2
roo e "' dx Jo
[by !'Hospital's Rule]
2
[since e"' is an even function]
X
[by Exerc ise 40(c)] (b) Let u =
f.0
00
..JX. Then u 2 = x
.../Xe "'dx = lim ·
too
dx
.::?
= 2udu
I; .../Xe"'dx = #
.::?
I
2
lim ,/f.ue" too 0
2I0
00
2u du =
2
u 2 e u du = 2 U.J7T)
[by part(a)]
15.5 Applications of Double Integrals 1. Q = IIv O"(x, y) dA = I~ I: (2x
+ 4y) dydx =I~ [2xy + 2y2 ] ~=~
= I~ (lOx + 50 4x 8) dx = J~ (6x + 42) dx 3.
m
dx
= [3x2 + 42x]~ =
75 + 210 = 285 C
I13 I14 ky 2 dydx = k I13 dx .{14 y 2 dy = . k [xJ~ [h 3 ]~ = k(2)(21) = 42k, 4 4 ~ IIv xp(x,y ) dA = 4 ~k I : I1 kxy2 d ydx = 412 I13 xdx I 1 y 2 dy = f2 [tx 2 ]~ [ky 3 ]~ = 4\(4)(21) = 2, 1 211 85 1 ffD yp(x , Y )dA  42k 1 f 1a .f14kYa dy dx1 f 1sdx Jt r4 ad _ 1 [ ]3 [1 4] 4 m  42 y y  42 X 1 4Y 1  42 ( 2)( 4:;) = 28
= IIv
x =  = Y
p(x, y) dA =
Hence m = 42k, (x, Y) = (2, ~) .
r2f3 x( x+y) dydx = Jo r2 [xy+2Y 1 2]v=3:z: r2 [x ( 3 a x) 5. m= Jo .,12 v=x/ 2 dx =J~ 2
.
x )2  81 x2] dx
(~x 2 + ~) dx = [ ~(tx 3 ) + ~x] ~ = 5,
J02
=
+ 21 ( 3 
r2J·3x( x 2 + xy )dy dx = Jo r2 (x 2y + 2xy 1 2]11=8:z: dx r2 (92x  gX 9 3) d 1l=x/ 2 = Jo x = 92•
lvfv = Jo :r:/ 2 lvfx
= Jof2J3v( ., 12 xy + y 2)dy dx
Hencem
7. m =
r2.[12xy2 + 3Y 1 3)11=3x d r 2 (g  2x 9 )dx = 9 . = Jo 11=, 12 x = Jo
(3 3)
) (lvfy Mx) = 6, (x_, y = m ';:; = 4' 2 .
J~ 1 J~ "'
2
kydydx
= k f~ 1 [~y2 ]~=~"'
2
dx
= ~k f~ 1 (1 
= ~k [x ix3 + %x5 ] ~ 1 = ~k (1  i + k + 1  %+ % )=
x 2 ? dx 8 15
= ~k f~ 1 (1 
2x2
+ x 4 ) dx
k,
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= ~ft.
262
D
Mv
CHAPTER 15 MULTIPLE INTEGRALS
·= .f~ 1 ,{01 :r 2  lk(lx 2 2
Mx
= kf~ 1 axy2 ]~=~:r
2
kxydydx
4 + lx6 ] 1 lx 2 6 1

j 'L j·sin(7rx/L)
o o
2
3 ] v=lx dx (1y 3 v=o
1 = lk s f 1 (1 
x 2 ) 3 dx
= lk { 1 (1'3 .  1
X
.
rL X SID · 2(11'XjL) dx Y dY dx _ .!2 Jo
integrate by parts with
[u
= .!£2 4
l (.!£2 2
4
= x , dv = sin2 (7rx/ L) dx
+ 41f2' L2 
= J 0L J;in("'x f L ) y · y dy dx = f0L ~ sin3 (7Tx/ L ) dx = ~ f 0L [1 
= t( ~)[cos(7Tx/L)  1cos3 (7Tx/L)]~ =  3~ ( 1 +
11 . p(x, y)
_ _
4' (x,y) =
(L /8 2
L/ 4 '
4£/ (911')) L /4
]
= l£2 8 '
L2)
41f2'
.cos2 (7Tx/ L )] sin(7Tx/ L ) dx ::}
~ 1 +
(L 16)
= 2 ' 971'
= k y = kr sin O, m = )~,. 12 J; kr2 sinO dr dB=
du= fsin(7Tx/L))
i) = 9~L.
.
ik )~.,. 12 sin Odtl
= ik [ cosB]~ = ftk, 12
Mv
= J0,.12 J; kr 3 sinB cos BdrdB = i k f 0" 12 sinB cosB dB= ~k [ cos 28] ~12 = lk,
Mx
= J0,.12 J01 kr 3 sin2 B dr dB = ik f 0"'12 sin2 BdB =
Hence (x, y)
x 6 ) dx
2
[substituteu =cos(7Tx/L))
L
+ 3x4 
t sin (7Tx/L) dx =ta x 4~ sin(27rx/L) j~ = ~ L,
[.!x2 + 41f2' L2 cos(211'xj L)]L 4 0
= l£2l 4 2
Hencem =
3x2
t ·x(tx  4r;, sin(27Tx/L)) ] ~  ~ J0L [tx  4~ sin(27Tx/ L)] dx
=
M:z:
= ~ kf~ 1 (x 2x 3 + x 5 )dx
x 2 ) 2 dx
x::::; L.
= J:' J;in(rrxfL) ydydx = g M.y _
= ~kf~ 1 x(1 
(o, ·32k/105) _ (o 4) Bkf l ~ •7 ·
9. Note thatsin(7Tx/ L) ~ 0 for 0 ::::;
m
dx
l2 k (1l + l6  l2 + l2 _ l)  0' 2 . 2 6 

1 = J1 1 Jor1 "'~ ky 2 dydx = kf1
8 k , (x , ) Hencem _ 15 y _
2
lk [B + sin 28] ~
12
= fak .
= (i, ~~) . y
13.
p(x, y)
m
= k jx2 +y2 = kr,
= ffo p(x, y)dA = f 0"' f 12 kr · r dr dB
= k f0"' dB f 12 r 2 dr = k(11') [ ir3 ]~ = ~11'k,
= k [sinO]"' 0
[ .! r 4]2 4 1
= k(O) (.!&) = O 4
.
[this istobe expectedastheregionanddensity function are symmetric about theyaxis]
= ffo yp(x, y)dA = fo,.t
M:r
H ence (x, y ) _
7rrk/ :J

'
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SECTION 15.5 15. Placing the vertex opposite the hypotenuse at
m
= J; foa"' k (x2 + y 2 ) dy dx = k
(0, 0), p(x, y)
foa [ax
2
x

3
APPLICATIONS OF DOUBLE INTEGRALS
D
= k(x 2 + y 2 ). Then
+ 4(a 
x)
3
dx
]
= k[~ax3  tx 4 
4
(a  x) ]~ = ~ ka4 •
1 12
By symmetry,
Hence (x, y) 17. I:r: =
JJ0
=
(~a, ~a).
y 2 p(x, y)dA
= .{~ 1 J; x
1  l4 kf1 (x 8 4x6 +6x4

2
4x2
y 2 · ky dy dx = k j~
1 [h 4J~:~x
9 + l)dx = l4 k [lx 9
7 ix 7
5 + !!x 5
2
2
dx
= tk f~ 1 (1 3
:!x 3.

:t
2 4 )
dx ·
+ x] 1 1 = .M..k 315 '
2
Iy = fio x 2 p(x, y) dA = }~ 1 j~"' kx~ydydx = k I~ 1 [~x2 y 2 J ~:~"' dx = ~k t 1 x2 (1 x 2 ? dx
It 1 <X 2 
= 21 k•
and Io = I, + Iv
2X 4
+X
6)
d X= 21 k [ aX 1 3

2
s X
s
1 1] 1 _ + "'fX  1 
8 k 105 '
= 36~k + t~sk = 38t8sk.
19. As in Exercise 15, we place the vertex opposite the hypotenuse at (0, 0) and the equal sides along the positive axes.
Ix = I~' I;xy2k(x2 +y2)dydx
= k .f'a [lx 2 (a x) 3 + l(a 0 3 5
= k.fo" .fo" "'(x2y2 +y4 )dydx =
x) 5] dx
3 x3 = k [l3 (la 3
2x 4 1a 4
Iv = I; I; x x 2k(x 2 +y2 ) dydx = k J0a.f; "'(x4 +x2 y2)dydx 5  k j 0·a [x 4 (a  x) + l3 x 2 (a  x) 3) dx = k [lax 5
21. I,.
Iv
= IIo y 2p(x, y)dA =

5 + !l.ax 5
2

6 ) lx 6
2x 4 !l.a 4
2
dx
1ka6 l30 (a x) 6 ] a = 180 0 '
y 3 J::~x dx
5 + !!ax 5
= p[ x ]~ [~y3]~ = pb(th3 )
0 ) ) a   1 ka 6 lx 6 0  180 '
= tpbh3 ,
= IIo x 2p(x, y)dA = Io" I~ px2 dxdy =pI: x 2 dx I~' dy = P[ tx3 ]~ (y)~ = ipb3h,
and m = p (area of rectangle) = pbh since the lamina is homogeneous. Hence ¥
I:r: m
= 2
andy = 
3
= .{[0
tpbh = = h3 pbh
x 2 pdA =
f0""12 J;
= {(r, 8) I 0
l
b3 h pbh
m
b2

3
p(r cos 8) 2 r dr d/J
~ r ~ a, 0 ~ 8 ~ ~},so
= p I 0"" 12 cos2 dB
.f0o. r 3 dr
4
ftPa rr,
= p · A(D) = p · trra2 since the lamina is homogeneous. Hence ¥ 2 = y2 ®
l
= ...JL = L _ =
=>
= p(t8 + t sin 28)~ 12 (tr 4 )~ =~Of) (ta4 ) = and m
2
2
23. In polar coordinates, the region is D
! 11
+ h5J~:~x
= k .f0a[x4 y + ~x
0 + l (l a 3 x 3 lx 6. 3 3
Io" I~ py dx dy =pI: dx I~' y dy 2
k.fo"[~x2y3
.l...pa4rr
= ~ 4 pa
1r
a2
= 4
=> x =
2012 Ccngogc Lc:uning. All Rights Re.~rwd. May not be scanned. copied. or duplicated. or posted to a publicly accessible website, in whole or in port.
y = ~2 .
263
264
D
CHAPTER 15 MULTIPLE INTEGRALS
25. The right loop of the curve is given by D = {(r, 0) I 0 ~ r :::; cos 20,  r./4 ~ () ~ 71'/ 4}. Using a CAS, we 2
r:~;4 f oco" 20 r 2 rdrd9 =
2
find m = f fo p(x,y) dA = ff0 (x + y ) dA = 1 · x= 
m
1 y =
i'!r /lr
m
(x, _y )
j "'/4 [
16384J2 2 64 ~Tr/4 1cos20 4 (rcos9)r rdrd8= r cos9dr d8= and 371'  Tr/4 0 1039571'
o
64 xp(x,y)dA= 37r
D
64 ~rr/4 1cos28 64 yp(x, y) dA = . (rsin 9) r 2 rdrd(J = 3 7r rr/4 o 37r
= ( 16384v'2 1039571' '
o)
os20
o
Tr/4
::. Then
j"'/4 Tr/4
cos20
1o
r 4 sin 9dr d0
= 0, so
.
The moments of inertia are
_ .[JD y 2 p ( x, y )dA ' "/4 4 Jo rcos20(r sm . (})2 r 2 r dr dB  J"/4 j ·cos20 r s sm . 2(J d r d(J571' 4 I :r  J Tr/  7r/4 o  384  105,
ly =
JJ0
Io = I,
2 x p(x, y) dA =
J~~;4 J;
28
0 "
2 2 (Hos0) r r dr dO =
J~~~ 4 J 0c"" 20 r 5 cos2 0 drdO = :~ + 1~5 , and
571'
+ I y = 192
27. (a) f(x, y) is a j oint density function, so we know JfR2 j(x, y) dA = 1. Since f(x, Y) .= 0 outside the rectangle [0, 1) x [0, 2], we can say
J JR2 f(x,y)dA = f~oo f~oo f(x,y)dy 'dx .= 2 1 2 = C Jo f x [y + ly ]v= 2 v=;O
Then 2C
= 1 =>
C=
c' Jof 4xdx ~ C (2x 1
2 1 ]
o = 2C
t. = J01 J; tx(1 + y) dy dx
(b) P (X :::; 1, Y ~ 1) = f~oo f~oo f(x, y) dy dx
= Jor l
dx =
f01 J; Cx(1 + y ) dy dx
1 2] Y = 1
2x Y + 2Y 1
[
v=O
d x
· 1 ( 3) dx 1 2] 1 3 = Jor1 2x 2 = 43 [ 2x o = 8 or 0 ·375
(c) P (X+ Y :::; 1) = P ((X , Y) ED) where Dis the triangular region shown in
Y
the figure. Thus
· :s; 1)= If 0 f(x , y)dA = f 01 f 012: 1 P(X+Y 2x(1+y)dydx 1
= Jo t lx [y 2
2 ] v=l:~: + ly 2 11 = o
1 = l41 (x3  4x 2 0
dx
+ 3x) dx =
2 = Jof 1 lx(lx 2 2 3 l4[4 L•I  4L 3
2x + ~) dx 2
D
2] 1 + 3L2()
1
0
X
= 4~ ~ 0. 1042 . 29. (a) f(x, y) ;;::: 0, so f is a joint density function if Jf'it 2 f(x, y) dA
= 1. Here, f(x, y) =
0 outside the fi rst quadrant, so
JJR2 f(x,y)dA= f ooo fo":'0.1e(o.ox+?·211) dydx = 0.1 J~oc f ooo e0.5:re0.2y dy dx = O.l fooo eo.sx dx f ooo e0.2y dy = 0. 1 lim
rt
L+oo Jo
e 0·5 "' dx lim
rt
too Jo
e 0·2 Y dy = O.l. lim ( 2e 0·5 "'] t lim t.+eo
0 t +oo
[ se0·211 ] t0
= 0.1lim [ 2(e 0 · 5 t  1)] lim [ 5(e 0·2 t  1)] = (0.1). (2)(0  1) · (5)(0  1) = 1 t +oo
t +oo
Thus f(x, y) is a j oint density function. . ' © 2012 Ceng;~gc Le:uning. All Rights Rcsc:n·cd. lvfuy not be ocanncd, copied, or duplicated. or JXh1cd 10 a publicly acccssiblc website, in whole or in pan.
SECTION 1505
APPLICATIONS OF DOUBLE INTEGRALS
0
265
(b) (i) No restriction is placed on X, so
Iooo Jt
P(Y ~ 1) = f~oo I 100 f (x,y) dy dx = = o ."1
roo e0 · 5"' dx
Jo
f1
0°1e(OoSx+0. 2 y) dydx
e 0·211 dy = oo1 tim
00
rt e  o.s., dx
t+oo J o
lim
too
It1 e 0 · 211 dv
= 001 lim [2e 0 · 5"']t lim [5e 0·211 ]t = 001 lim [ 2(e 0·5 '  1)] lim [ 5(e0 · 2 ' too
0 too
1
t  oo
t+oo

e 0 · 2 )]
(0.1) 0 ( 2)(0 1) 0 (  5)(0  e 0·2 ) = e 0 · 2 ~ 008187 (ii) P(X $2, Y $ 4) = I~oo I~oo f (x,y) dyd.x =
I 02 J04 Oo1e<0·5"'+0·2 Y) dydx
= 001 1~2 e o.sx dx 1~4 e 0 ·2 ll dy = 0.1 [2e 0 · 5 "'] ~ [. 5e 0 · 211 ]~ = (0.1) · (2)(e= (e 1

1
1)(e 0 · 8
1) · ( 5)(e 0·8 1) 1) = 1 + e J.B  e o.s  e 1 ~ 0.3481

(c) The expected value of X is given by Mt =
IIR2 x f(x, y) dA =
1~00 1~00 :z; [oo1e(o.sx,+o. 2 v)J dydx
= 0.1 ];00 xeO.ux dx ]; 00 e 002!1 dy = 001 lim 0
];t xe 005"' dx
t~oo o
0
lim
rt e 0 ·2 Y dy
t  oo ./o
To evaluate the first inteirat, we integrate by parts with u. = x and dv = e 0·5 "' dx (or we can use Formula 96
I xe 0 · 5 "' dx =
in the Table of integrals):
 2xe 0·5 " ' 
J  2e 0·5 "' do'l =
2xe 0·5 "'

4e 0 · 5 "' =  2(x
+ 2)e 0 ·5 "'0
Thus JJ. 1
= 0°1 tlim [2(x + +oo
2)e 0·5 "'] '· lim [ 5e 0·2 ll] t 0 t +oo
= 001 t lim ( 2) [(t + 2)e...oo = 0.1 (2)(lim t +.
2
t +oo eO.at
0 5 · t 
0
2] lim (  5) [e 0·2 t t+oo
 2)( 5)( 1) = 2

1]
[by !'Hospital's Rule]
The expected value ofY is g iven by J1.2
=
Ifa2 y f(x , y) dA = Io00 fa"" v [oo1e(0.5+0.2 y)]dy dx
= 001 ];0
00
e0 · 5 "' dx {; 00 ye 0 · 211 dy .0
= 001 too. lim {;t eo.sx dx lim rt ye 0 · 2 Y dy o too _J o
To evaluate the second integral, we integrate by parts with u = y and dv the Table oflntegrals) which gives I ye 0 ·211 dy =  5ye 0 · 2ll JJ. 2
= 0.1 too lim [
2e 0 · 0·"] t lim [  5(y 0 t +oo
= e  0·2 !1 dy (or again we can use Formula 96 in
+ .r se 0 ·211 dy =
+ 5)e 0 ·
2
5(y + 5)e 0 ·2 U 0 Then
Y] ' 0
= Oollim [ 2(eo. st _ 1)] lim ( 5[(t + 5)e 0 · 2 t  5]) t+ oo
t oo
=Oo1 ( 2)( 1)o (5)
) lim t+5 .,t  5 = 5 e 0 .... 0
(
t oo
[Py )'Hospital 's Rule]
310 (a) The random variables X and Y are normally distributed with JJ.1 = 45, JJ. 2 = 20, 0'1 = 005, and a 2 = Ool. The individual density functions for X and Y, then, are !J (x) =
h (y) = ®
~ c
001
211'
0.5
~ e <:r<11>)21°·5 and 27r
and Yare independent, the joint density function is the product .
2012 Ccngace Leamin1:.. ;\II Rights Rc~rv~d . May not be scnnnC'd. copi!XI. or duplicated. or rusted ton publicly accessible website. in Y.1l0lc or in part.
266
CHAPTER 15 MULTIPLE INTEGRALS
D
f(x, y) = /I(x)/2(y) =
2 e(x45) /0.s
1
0.5 ..,l2ir 
0.1 ..,f2ir
= lQe 2(x45) 2 so(v2~)2.
"
.
< y < 25) = rso r2s f(x y) dy dx = !.!!. fso f25 e  2(x45)2so(y2o)2 dy dx J 4o J2o . ' .,. . 40 J 2o ·
Then P(40 <X< 50 20 
2 e(y20) /0.02
1
•
Using a CAS or calculator to evaluate the integral, we get P( 40 $ X ~ 50, 20 $ Y $ 25) ~ 0.500. (b) P(4(X 45) 2
+ lOO(Y 20? $
4(x 45) 2 + 100(y 20) 2
2) =· ffv ~ e 2 <"' 45 ) so(v 20 ) dA, where Dis the region enclosed by the ellipse 2
= 2. Solving for y gives y = 20 ±to J2 4(x 45)2, the upper and lower halves of the
ellipse, and these two halves meet where y
x
2
= 20
[since the ellipse is c•~ntered at (45, 20)]
~
4(x 45) 2 = 2 ~
= 4? ±' ~. Thus
!l
2
lQe2(x  45) so(y 20)
D
2
dA =
"
l 45+1/v'2 . lQ . "
451/.,/2
2D+tn/2 4(x45)2
120ftp,/2
Using a CAS or calculator to evaluate the integral, we get P( 4(X  45?
2 2 e2(:.;45) so(y20)
dydi.
4(x45)2
+ 100(Y 
20)2 $ 2) ~ 0.632.
33. (a) If f(P, A) is the probability that an individua l at A will be infected by <m individual at P, and k dA is the number of infected individuals in an element of area dA, then f(P, A)k dAis the number of infectio ns that should result from exposure of the individual at A to infected people in the element of area dA. Integration over D gives the number of infections of the person at A due to all the infected people in D. In rectangular coordinates (with the origin at the city's center), the exposure of a person at A is
E
JL
=
kf(P,A)dA = k
JL fo
[20  d(P, A)] dA
=kj
L(1 foJ(x
xo)2 + (y YoF] dA
(b) If A = (0, 0), then
jL[
E = k
1
=k
1 20
271' 1 10 (1' i
1 0
0
= 27T'k(50
sn
r = 20cos0
Jx2 + y2JdA 0
r) rdrd() = 27T'k[~r 2 
'
=
1 3 10 60 r ] 0
2 0
~ 1T'k ~ 209k
For A at the edge of the city, it is convenient to use a polar coordinate system centered at A. Then the polar equation for the circular boundary of the city becomes 1·
= 20 cos() instead of r = 10, and the distance from A to a point P in the city
is again r (see the figure). So
= k J::~~ 2 (200 cos2 8  4~ 0 cos 3 8) dO = 200k J::~~ 2 [ ~ + ~ cos 20  ~ (1 = 200k [ ~ 8 +
t sin 2() 
2
sin 0) cos OJ dO
~ sin 0 + ~ · ~ sin3 0] =~ 12 = 200k [ :f + 0  ~ + ~ + :f + 0  ~ 2
+ ~)
= 200k(i  ~) ~ 136k
Therefore the risk of infection is much lower ·at the edge of the city than in the middle, so it is better to live at the edge. © 20 12 Cengage Learning. All Rights Reserved. Mny not be scanned, copied, or duplicated, or po!;tcd to 11 pUblicly accessible website, in whole or in part.
SECTION 15.6 SURFACE AREA
0
15.6 Surface Area 1. Here z = f(x , y)
= 2 + 3x + 4y and Dis the rectangle [0 , 5) x
[I, 4), so by Formula 2 the area of the surface is
A(S) = ffD J[f,(x,y)] 2 + [!11 (x,y)]2 + I dA= ffo ../32 +42 +IdA= ../26ffDdA
= ../26 A( D) = ../26 (5)(3) = 3. z = f(x, y)
= 6 3x 
I5 ../26
2y which intersects the xyplane in the line 3x + 2y
= 6, 'soD is the triangular region given by
{(x,y) I0:5 x :5 2,0:5 y :53 ~x}. Thus A(S) = JJD J( 3) 2 5. y 2
+ z2 =
9
+ (2) 2 +IdA= .JI4 ffo dA = JI4 A( D ) = JI4 (~ · 2 · 3) = 3 .JI4
=* z = ~· fx
A(S) = h412
=t { lo lo
= 0, fv = y(9 
y2 ) 
112
=*
Jo2 + [ y(9 y2) 1/2)2 + 1 dy dx = h4 h 2 J9 ~2y2 + I dy dx
2
3
J9 
y2
=3
dydx
t lo
[sin
2 IL] = dx = 3 [(sin 1 (~))x]~ = I 2sin 1(~) ·3 11
1
v=O
7. z = f(x, y) = y 2  x 2 with I ::; x 2 + y 2 ::; 4. Then
A(S)
= Jf D J 1 + 4x2 + 4y2 dA = = (0]~"
9. z
23 2 u2(I + 4r ) 1
= f(x, y) = xy with x 2 + y 2 :5 1, so /ex = A(S) .
= JJD
!L 'lr f2
 .r/2
= /_"/2
acoso
1 o
·
J: = {f(17vT7  5v'5)
y, f v =X =*
1dA
= t" r1 ../r2 +I r dr dO= J;02" 0 Jo
[l(1·2 + 1) 312] r=l dO 3
r=O
I) dO= I(2../2  1) 2
r2
  + 1rdrd0 a2  r2 ar drdO 2 2
Va

r
[ a Ja2...:. r2 ] r=acose dO r=O
 1rj2
1r/2
a( J 1r/2
= /_
f 12 Vf+"4r2 r dr dO = J:.,. dO J12 r ../I + 4r2 dr
x2 +y2
"/2 1 acos6 /
/_
"
a2  x 2  y 2+ IdA
= /_  1r/2 o
=
x2 +
I:'lr H2../2 
=
A(S) =
Jy 2 +
2
.{0
a2 
a2 cos 2 0 a) dO
= 2a2
1 rr/2 0
(I 
Vl  cos
2
0) dO
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267
I
268
0
CHAPTER 15 MULTIPLE INTEGRALS 2
2
2
2
13. z = f(x, y) =ex  v , f.,= 2xex  v , fv
II
A(S)=
2
= 2yex  v
2
. Then
If
v(  2xe xLv2 )2+( 2ye x2 v2 )2+ 1dA =
x2+y2_54
.j4(x2ty2)e 2(x 2 +v 2 lt ldA.
x2+ 11 251
Converting to polar coordinates we have
= J:" .{02 J 4r2e' 2r 2
A(S)
= 27!'
.r; r V 4r e2
15. (a) The midpoints of the four squares are
(i
1
2r 2
i ), (i
1
+ 1 r dr d(J
= ;;11: dl1 .{02 r V 4T 2e 2 ''2 +
1 dr
+ 1 dr ~ 13.9783 using a calculator. %) , ( %, ~)~and (
i
1
~). Here f(x 1 y) = x 2 + y
2
,
so the Midpoint Rule
gives A(S) = JJD .j[fx(X 1 y)F + [fv(x, y)J2 + 1 dA = jjD .j(2x) 2 + (2y) 2 + 1 dA
~ ~ ( J[2(~)J 2 + (2(iW + 1 + V[2(~)J 2 + [2(t~] 2 + 1 +
~(
=
V[2(~)] 2 +
2 [2(i)] + 1 + V[2(%W
+ [2(~)]
2
+ 1)
.ft + 2 JJ + /¥) ~ 1.8279
(b) A CAS estimates the integral to be A(S)
= J[D .)1+ (2x ) 2 + (2y )2 dA =
.1; J
1 0
.)1 + 4x2 + 4y2 dy dx ~ 1.8616.
This agrees with the Midpoint estimate only in the first decimal place. 17. z
= 1 + 2x + 3y t4y2 , so
11
·A(S) =
(~~r + (~~r dA =
1+
1 .f
11 4
4
v 1 +4+(3 + 8y)2dydx =
1
V 14+48y+64y2dydx.
\
Using a CAS, we have
J
4 1
I~ .)14 + 48y + 64y2 dy dx
= ¥ .Jl4 +
~ ln(ll J5 + 3 .Jl4J5)  t~ ln(3 J5 +
v'i4 J5)
45 'P4 orV J.'% + 15 ln 11 J5 + 3 v'70
8
3
16
19: f(x, y)
= 1+
x 2 y2
J5 +. v'70 .
=> f,
= 2xy 2 , fv =
2x 2 y. We use a CAS (with precision reduced to five significant digits, to speed
up the calculation) to estimate the integral A(S)
=
1!1~ Vfi  1
 vhx.2
+ F~ + 1 dy dx
.
'
= 1!1~ ..j4x2 y 4 +  1
4x4 y 2 + 1 dy dx, and find that A(S) ~ 3.3213.
~
·
21. Here z = f(x , y) = ax+ fnJ + c, f.,(x 1 y) =a, / 11 (x, y) = b, so I
A(S) = JJD ,Ja2 + bz + 1 dA
= ,ja2 +
b2 + 1JJD dA
= ,Ja2 +
b2 + 1 A(D).
23. If we project the surface onto the xzplane, then the surface lies "above" the,' disk x 2 We have y = f(x, z)
=x
2
A(S) =
+ z2
~ 25 in the x zplane.
2
+ z and, adapting Formula 2, the area of the surface is J{ ,;2 +z 2 .$25
.j(f,. (x ,z )]2+[fz (x,z)]2+1dA =
JJ
,J4x2 +4z 2 +1dA
,2 +=2 .$25
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SECTION 15.7 TRIPLE INTEGRALS
Converting to polar coordinates x
= r cos fJ; z = r sin fJ we have
15.7 Triple Integrals 1.
3.
v=2 d d · ri r3 3 2 d z d x If,JBr xy~ 2 dV = Jori Jor3 J2_ 1 xyz 2 d y d z dx = Jori Jor3 [21 xy2 z 2] v=1 z x = Jo Jo 2 x z = Jor1 (lxz3] ==3 dx = fl TI xdx = 27 x2 ] 1 = TI 2 ==O Jo 2 4 o 4
2 · x  .y)cJ:J: dydz=Jor2 J;r· 2 [x 2 J02 J;" Joru "(2 =
5.
7.
J; J
2
2
(z
0"
x=yxy] x=O
r2 Jo r= 2 [(y z) 2  (yz)y ] dydz dydz =Jo
z
yz) dydz =

J~2 [yz 2  h 2 z] ~=~
2
dz =
J02 (z4  %z5 ) dz
v=ln:r; d d = !2 r2z ( x z xe  ln x + xe0) dxd z !12Jor2= Jor tn x x e  y dy dxdz = !21 Jor2= [ x e v] v=O 1 Jo 2 = Jt J:= (i + x) dxdz = g [x + %x J::~= dz 2 2 2 3 16 · = f 1 ( 2z + 2z ) dz = [ z + ~ z ] ~ =  4 + 3 + 1  ~ = ~
J0.,.12 J~ J0"' cos(x+y+z)dzdxdy = j~rr/
2
g
[sin(x+y +z)J::~ dxdy
= f0.,.12 J~ (sin(2x + y) = j~,./
2
sin(x + y)J dx dy
[ %cos(2x + y) + cos(x + y)J::~ dy
= f 0,.12 [ ~ cos 3y + cos 2y + ~ cos y  cos y] dy =
9.
JJJE ydV = J;
1. . [ 6 SID
3
1
2
.
fa"' J;::: ydzdydx = J; J; [yz]~==~~ dydx =
 r3 (1y3] y = x dx  r3 1x3 dx  lx4] 3  81  Jo 3  Jo 3 G o 6 11=o
11 1=~ + 4
11. / " ( ( } } E
 2 z
+Z
X
2
1
dV
=
4
1
=
4 1
71
4
f f 11
0
X
11 1
1
(tan (1) tan (o)] dzdy
] .,. /2 0
1
1
1
= G.  2 = S
f 03 f0"' 2y2 dydx
= l2l
4
dx dz dy =
Z
1
.
y + 2 S Ln y  2 sm y
4
[z ·
71
=f
4 1
.! tan  l ~] x = z dz dy Z
Z
x =O
I:(~ o) dz' dy = ~ f 14 [zJ::: dy
= ~ f 14 (4y)dv= ~ [4v  h 2 ]~ · = ~ (168  4+~) = 9; 13. Here E = {(x,y,z) I 0 :::; x:::; 1, 0 :::; y:::; fo, O :::; z :::; 1 +x +y}, so
JJJE 6xydV = f01 fov'X J;+.,+u 6xydzdydx = J~ =
f~ fov'X 6xy( 1 + x +
= t0 (3x2 + ©
1
fov'X [6xyzJ ::~+"'+11 dydx 2
y) dy dx = / 0 (3xy + 3x 2 y2
3x3 + 2x5f2) ·dx
=
+2x~ )~=~ dx
[x3 + ~x·l + :!x7f 2] 1 = 4 7 0
3
65
28
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pan.
0
269
270
D
CHAPTER 15
MULTIPLE INTEGRALS
HereT = {(x, y,z) I 0 $ x $ 1, 0 $ y $ 1  x, O $ z $ 1 x  y}, so
15.
Jffr x2 dV ::::: fot J;a: fot a:y x2 dz dydx
= fot fotx x 2(1 x y) dy dx
= J; fot a:(x2 ~3  x2y ) dy dx = fot [x 2y.= J;
tX2y2J::~a: dx
xay 
[x2 (1  x) x 3 (1  x)  t x 2(1 x?] dx
4 = j '01 (lx 2
x3
2 5 ) dx = [...!. + lx 2 10 x 
l4 x .4
+ l6 x 3 ) 10
1  w  41 + s1 = oo1
The projection of E on the yzplane is the disk y 2 + z2 $ 1. Using polar
17.
coordinates y = r cos 8 and z = r sin 8, we get
f.ffe x dV = ffv [J4~2 +4:2 x d.x] dA = ~ ffv [42

(4y
2
+ 4z2 ) 2 ) dA
= 8 J;""J:(1 r 4 ) 1· drd8 = 8 J~ " dB J0\ r  r 5 ) dr 2
X
_ 8(211") [lr2 _ lr6] I _ 2 6 0

19. The plane 2x
2x + y + 0
+y + z =
=4
=>
16rr 3
4 intersects'the xyplane when 4
y = 4  2x, so
E = {(x, y, z) I 0 $ x $ 2, 0 $ y $ 4 2x, 0 $ z $ 4  2x  y} and V = fo2 J~l 2a: fo42xv dz dy dx = fo2 fo4 2:r. (4  2x  y) dy dx
= Jor2 [4y  2x y  21 Y2] vy ==4.O 2x dx = J; [4(4  2x) 2x(4  2x)  t(4 
=J; (2x
2

8x + 8) dx =
[~x 3  4x2 + 8x]~ ='¥
E
= {(x, y, z) I.  1 $
V
= JJJ ~ dV = .C1 J:2f01 11 _

[1
1
3
1, so
2
x $ 1, x $ y $ 1, 0 $ z $ 1  y} and
[y 21 Y2] v=1 y=x2
2X  . 3 X
y
2x? ] dx
= 1 intersects the xyplane in the line y =
21. The plane y + z
1 = ! 1
X
dz dy dx
dX = }'1 (1  1 2 
+ lOX· 1 5] 1 1  1 = 2 
1
3
= f~ 1 f:.2 (1 X
'.!
y) dy dx
+ 21 X") dX
+ 101 + 2l 
1
J
+ lO1
( 1, 1, 0) _

R
15
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SECTION 15.7 TRIPLE INTEGRALS
D
271
23. (a) The wedge can be described as the region
D = { (x,y,z) I y2 + z 2
::;
1, 0 ::; x::; 1, 0 ::; y ::; x}
= { (x, y , z) I 0 ::; x ::; 1, 0 ::; y ::; x, 0 ::; z ::; } 1  .Y2
}
So the integral expressing the volume of the wedge is
JffodV=fo1 fo"'J/1 v2 dz dydx. . (b) A CA S gives
r
l f"' n /l. vz Jo Jo Jo
dz dy dx
= 4"
l  3·
(Or use Formulas 30 and 87 from the Table of Integrals.) 25. Here
f(x, y, z) = cos(xyz) and~ V
= ~·
t · ~ = ~,so the Midpoint Rule gives
1 1) + J(l4 •4• 1 43) + !(1 3• 4l) + f(l4 • 43• 43) = s1 [!(14 • 4•4 4•4 + !(~ ,
t. t) + !(~ , t. ~) + !(~, ~, t) + ! (~, ~. ~) ]
= ~ [cos l4 +cos ~ +cos 6~ +cos f4: +cos 6~ +cos i4 +cos f4: + cos~~ ) 27. E
~ 0.985
= {(x , y, z) I 0 ::; x::; 1, 0::; z::; 1 x, 0 ::; y::; 2 2z },
the solid
bounde~
z = 1  x, y
by the three coordinate planes and the planes
=2 
2z.
X
y
29.
4
X
~"x' + 4z
1
I
= 4
.t
 I
If D 1 , D2 , D3 are th~ projections of Eon the xy, yz, and xzplanes, then
D 1 = {(x, y) I  2::;
X::;
2, 0 ::; y::; 4  x 2 }
= {(x, y) I 0 ::; y::; 4, I
 v4  y::; .
D 2 = {(y,z) 1o ::; v ::; 4,  ~v4 y::; z::; ~v4 v} = {(y,z) 11 ::; z::; 1,
DJ = { (x,z) I x 2 +4z 2
::;
X::; .J4=Y } o::; v ::; 4  4z 2 }
4} [continued]
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272
0
CHAPTER 15
MULTIPLE INTEGRALS
Therefore
E
= {(x , y , z) 1 2 < x < 2'0< y< 4 ·_ = { (x , y,z) I 0:::; y:::; 4,
= {(x,y,z) I 1 $ = { (x, y, z)
x2 ,
 y'4""='1j:::; x:::;

:z;2 
y<  z <  1 2 yl4 
yf4  y 4z2
 ~ J 4  x2
I 1 5: z 5: 1,
 .J4  4z 2
:::;
:::;
z :::;
t J4 
y}
:z;Z 
x 2  y:::; z ~ ~yf4 x 2
h/4 
:::;
x:::; yf4 
 ty'4""='1j $ z $ ty'4""='1j,  j4 y 4z 2
= {(x, y, z) I  2 5: x 5: 2, = { (x, y, z)
y'4""='1j, 
z 5: 1, 0 5: y 5:4  4z 2 ,
I 0 5: y 5: 4,
12 yl4 

:::;
x~, 0 :::; y :::; 4  :z: 2
x:::; .J4 4z 2 , 0:::; y:::; 4 x 2

y  4z2
 y}
}
x:::; yf4  y 4z2 
}
4z2 } 4z 2 }
Then
Iffsf (x,y,z) dV = !_ 2 ; 0 2
r4x2 JJ4x2yf2
I
f4  4z
=
I  tJo
=
f _2
2
2
~/ /
(
JV4y4z2
(
.J4 _
11
_
2
z 4 2
r;:;
)
,
f x, y, z dx dy dz
J~/2 r4x2 4:2' y4  :z:2/2
r4 f14ii JJ4:z:2y j 2
)
Jo
(
)
x,y,z dzdydx=:Jo 14ii ~ 12 ! x,y,z dzdxdy 2
J V4y4z ( = Jof4 f/4!i/2 14ii/2  J 4 _ _ 4 : 2 f x, y, z) dx dz dy 11
f(x. ,y,z) dydzdx= fl_ 1 J
2
J4 4:;2 r4o: ~Jo
y44z2
4:2 f ( x,y,z ) dydxdz
y
31 .
( 2,4,0; JC
X
(2,4,0)
4
y
2
2
JC
If D1, Dz, and Da are the projections of Eon the xy, yz, and xzplanes, then
= {(x, y) I 2 $ x 5: 2, :z:2 ~ y 5: 4} = {(x, y) I 0 :::; y :::; 4, y'Y $ x:::; vY }, D2 = {(y,z) I 0 :::; y $ ,4, 0 :::; z 5: 2 ~y} = {(y,'z) I 0:::; z 5: 2,0 ~ y:::; 4  2z}, and
D1
Ds = { (x, z) I  2 $ x :::; 2, 0 5: z ~ 2  ~~ 2 }
®
= {(x, z) I 0 :::; i ~ 2,  .J4 2z :::;. x
$ v'4  2z}
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SECTION 15.7 TRIPLE INTEGRALS
E = { (x, y, z) I 2
Therefore
= { (x,y,z)
0
273
~ X~ 2, x 2 ~ y ~ 4, o' ~ z ~ 2 h}
I 0 ~ y ~ 4,
Jv ~ x ~ V"Y. 0 ~ z ~ 2  tv}
JY} = { (x, y, z) I 0 ~ z ~ 2, 0 ~ y ~ 4 2z,  VY ~ x ~ vY} = { (x,y,z)I O ~y~4,0 ~z~ 2ty, vv~ x~
= {(x,y,z)
l2
~ x ~ 2, 0 ~ z ~ 2  tx2 , x 2 ~ y ~ 4  2z}
= {(x, y, z) I 0 ~ z ~ 2,
 v'4  2z
~ x ~ y'4 2z, x 2 ~ y ~ 4  2z }
1JJ~ f(x, y, z ) dV = I~ 2 I:2 I~ 1112 f(x,y,z) dzdydx =
Then
4
I 04 I~ I 02 v 12 f (x, y, z) dz dxdy
2
= 1~ J~v/ 1~f(x, y,z) dxdz dy =I~ Io4  2 " J~f(x,y,z)dxdydz 2 f2= J2 Jo
33.
%2 / 2 !4 2:: z2
J( x , y,z) d y d z d·X= Jo r2 Jv'4=2z f4  "'4  2z z2
2z
. I ( x ,y , z ) dydxdz
y
The diagrams show the projections of Eon the xy, yz , and xzplanes. Therefore ~++ 0 I X
1
Io
I{z 1~l  v f (x, y, z ) dz dydx =I: I/ I:rJ f(x, y, z) dzdx dy = f Iol,zIt f(x, y, z) dx dy dz 1 0
rt rt u1'tJ0 = JoJo 2
=
2
J( x,y,z ) d x d zdy= Jo rt Jo rt../XJ1..;x z f (x,y,z)dydzdx
fl / ·(l  z) Jl z
J o. 0
..;x f ( x,y,z) dydxdz z
y
35.
.
X
I: I:Icf f(x,y, ~)dzdxdy = IIIEf(x , y ,z)dV where E = {(x,y,z) I 0 ~ z ~ y, y ~ x ~ 1, 0 ~ y ~ 1}. If D 1, D2, and D3 are the projections of Eon the xy, yz and xzplanes then
= [(x , y) I 0 ~ y ~ 1,, y ~ x ~ 1} = {(x , y) I 0 ~ x ~ 1, 0 ~ y ~ x}, D2 = [(y,z) I 0 ~ y ~ 1, 0 ~ z ~ y} = {(y,z) I 0 ~ z ~ 1, z ~ y ~ 1}, an~ Dt
D3 = [(x,z) I 0 ~ x ~ 1, 0 ~ z ~ x}
= { (x, z ) I 0 ~ z
~ 1, z ~ x ~ 1}.
[continued]
®
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274
CHAPTER 15
0
MULTIPLE INTEGRALS
Thus we also have E
= {(x, y, z) I 0 ::::; x::::; 1, 0 ::::; y ::::; X, 0 ::::; z::::; y} = {(x, y , z) I 0 ::::; y::::; 1, 0::::; z.::::; y, y ::::; x ::::; 1} = {(x,y,z) I 0 ::::; z::::; 1, z::::; y::::; 1, y::::; x ::::; 1} = {(x,y,z) I 0 ::::; x $ 1,0::::; z ::::; x , z::::; y ::::; x} = {(x, y, z) I 0::::; z::::; 1, z::::; x::::; 1, z::::; y::::; x}.
Then
J; fv f~ f(x,y , z) dzdx_dy 1
= f 01 J; f~ f(x,y ,z) dz dydz = f01 g
{=]; J;
I:
1:
f(x,y, z ) dxdz dy
f(x,y,z) dxdyd~ = J; J; f:'" f(x,y , z) dydzdx
"~
=
f 01 f~ [;"' f(x, y, z) dy dx dz 1
37. The region Cis the solid bounded by a circular cylinder of radius 2 with axis the z axis for 2 ::::; z $ 2. We can write
fff0 (4 + 5x2 yz2) dV = fffc 4dV + fffc 5x2yz 2 dV, but f(x , y,z) = 5x2yz 2 is an odd function with . respect toy. Since Cis symmetrical about the xzplane, we have fffc 5x 2yz 2 dV
39. m
= 0. Thus
= JJJE p(x, y,z) dV = f 01 fofi fol+x+v 2dz dydx = J: J0fi 2(1 + x + y) dydx. = Jorl (2y + 2xy + y2] y=D y=fi dx = Jo rl
1
(2 v'x + 2x3/ 2 + x) dx = [ix3/ 2 + lx2] = so 79 3 2 + ix5/ s 2 0 '
My: = fffE xp(x, y, z ) dV =
1 f0
fofi J~+o:+v 2x dz dy dx = J; fofi 2x(1 + x
+ y) dydx
= Jr1( 2 + x2) dx = 2xs12 + 2xsf o .
1
, = r1 [2 xy + 2x2y + xy2]v=fi dx Jo v=D
Mx~
= JJJE
yp(x,y,z)dV
s
3
1 0
= 179 105
= f 01 fofi J;+x+v 2ydzdydx = J01 f 0fi2y(1 +x+y)dydx
= Jor1 [y2 + xy2 + 1y3] v=fi r1 (x + x2 + 1x3/ 2) dx = 3 v=O dx = Jo 3
. lvlxv
[ixo/ 2 + ix7/ 2 + J.x3]
[lx2 2 + lx3 3 +
1
~x:>/2) = 15 0
11 10
r l Jo r fi Jo r1+:r+v 2z dz d y .dx = J·l J·./i. [z 2] ==D ~=l+:r+v d.y dx = Jo r1 Jo r./i. (1 + x + y )2 dy dx = fffE zp (x , y, z ) dV = Jo 0 0 =
J01 J/"(1 + 2x + 2y + 2xy + x 2 + y2) dydx = J; [y + 2xy + y 2 + xy2 + x 2 y .+ ty 3 J::f dx
r1 ( 'x + 1x3/2 + x _ Jo v"' 3
Thus the mass
.
IS
+ x2 + x5/2)
79
30 and the Center Of mass
.
dx _ ['l3 x3/2 + llx5/ 2 + l2 x2 + lx3 1s · 3 (
IS X,
_ _) Y, Z
M:r= Mxv ) = (My= :;;:;:, :;;:;:, m
Mv: = J; f 0a f 0a [x3 + x(y 2 + z2 )] dxdydz = J; J; [ta4
ru (.4a 1s _ Jo ) H ence (x,y,z ®
1 a Z 2) dZ = + 01a s + 2a
1 6 + 3a 1 6 4a
+ ~a
2
(y 2
1 6 = na = M :z:z =
=
+ 'lx7/ 2) 7
1 0
_ 571

2 10
(358 33 571) 553 ' 79 > 553 ·
+ z2 ) ] dydz M xy by symmetry 0 fE and p(X, y, Z )
) = (112 a, 127 a,' 127a.
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SECTION 15.7 TRIPLE INTEGRALS
43. I,.
= foL foL foL
k(y2
0
275
+ z2)dz.dydx = k foL foL (Ly2 + ~L3) dydx = k foL ~L4 dx = ~kLs.
By symmetry, l x = 111 =I.= ~k£5 •
45. 1:::
(b)
=
2
JJJE(x + y
2
)
p(x, y, z) dV
=
JJ
z2+y2:5a2
(x,y, z) wherex =
~= ~t
1
[J;• k(x
2
+ y 2 ) dz] dA
=
"* f~ 1 !:2f01 11 x Jx2 + y 2dzdyd$, y = "* f
J:2 J;  v z Jx 2
k(x 2 + y 2 )hdA
JJ
:z:2+y2:5a.2
1
1
J:z f 0 v y Jx 2 + y 2 dz dy dx, and
+ y 2 dzdydx.
f Jo r~ Jof"(1 + x + y + z ) d z d y dx 49. (a) m = Jo 1
=
3 ""
11 32 + 24
(b) {X, y, z) = ( m 1 J; fo~ f~ x(1 + x + y + z) dzdydx ,
m 1 f 0
1
f0~ J~ y(1 + x + y + m
.
28
307r
1
z) dzdydx,
f 01 fo~j~ z(1 + x + y +z)dzdydx)
+ 128
457r + 208 )
= ( 9n + 44 ' 45n + 220 ' 135n + 660
(c) 1:::
t
= fo
{~ {"
fo
.
fo (x2 + y2)(1 + x + y + z) dz dy dx =
68 + 1571' 240
51. (a) f(x, y, z) is a joint density function, so we know JJJR 3 f(x, y , z) dV = 1. Here we have
JJJR3 f(x,y ,z) dV = f~oo f~oo }:"00 f(x,y, z) dzdydx = C J: xdx
Then we must have 8G = 1
=> C 
+ Y + Z::; 1) =
and the plane x + y + z
P(X .+ Y
f 02 ydy f 02 zdz = C[~x2 )~ [h2 )~ [~z 2)~ =
r1
d
'
rl
Jo 11 Y Jo z
d z = ii1 [ 21 x 2]1o [ 2Y 1 2] 1 [ 1 2] 1 o 2z o
= 81 ( 21 )3
1 = 1M
P((X, Y, Z) E E) where E is the solid regio·n ·in the first octant bounded by the coordinate planes
= 1. The plane x + y + z = 1 meets the xyplane in the line x + y = 1, so we have
+ Z::; 1) = fff Ef(x, y, z) dV = j~1 j~ x j 01:z:y kxyz dz dy dx
= k J; fot  :z: xy[~z2 J ;:~:z:y dydx = fs fo1 fol z xy(l = 1~
f f
=
fo1 [(x3  2x2
110
1 1 0 0 
"'[(x3
1 f1(  192 Jo X  4X 2
®
8C
18'
= ii1 Jort x d X (c) P(X
= f 02 J: J: Cxyzdzdydx

2x2
+ x)y + (2x2

2x)y2
x  y)2 dydx
+ x y 3 ] dydx
+ xHy2 + (2x2 2x)h3 + x(fy'')J::  :r: dx
+ 6X .3 .
4X 4
+ X 5) d X 
1 ( 1 ) 192 30 
1 57iii)
2012 Ccngagc learning. All Rigllls Rcsco·cd. May not be SCttnncd, copied, or duplicah:d, or posted loa publicly 3Cces.sibJe website, in whole or in pnr1.
276
0
CHAPTER 15 MULTIPLE INTEGRALS
1 L2 L2 L2
La
£ 3·222 = 8 55. (a) The triple integral wi ll attain its max imum when the integrand 1  x 2

2y2

3z 2 is positive in the region E and negative
everywhere e lse. For if E contains some region F where the integrand is negative, the integral could be increased by excluding F from E, and if E fails to contain some part G of the region where the integrand is positive, the integral could be increased by including Gin E. So we require that x 2 e llipsoid x 2
+ 2y2 + 3z2
(b) The maximwn value of x 2 + 2y2
+ 3z 2 =
+ 2y 2 + 3z 2
$ 1. This describes the region bounded by the
= 1.
JjJE (1  x 2 2y2 
3z 2 ) dV occurs when E is the solid region bounded by the ellipsoid
1. The projection of E on the xyplane is the plaf1ar region bounded by the e llipse x 2
+ 2y2 = i, so
and
us ing a CAS.
15.8 Triple Integrals in Cylindrical Coordinates From Equations I, x
.1. (a)
y '= rsinfJ Y ·
= r cos 9 =
= 4sin ~ = 4 · ~ 3
4 cos
i
= 4·
~ = 2,
= 2.J3, z = ~2, so the point is
(2 , 2.J3, 2) in rectangular coordinates ..
I
 2 •I
X
(b)
l (4, 3. 2)
(2.¥. 1)
x
= 2cos(~) =
and z
0, y = 2sin(  ~l)
== 1, so the p o int is (0, 
=  2,
2 , 1) in rectangular coordinates.
X
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SECTION 15.8 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES
3. (a) From Equations 2 we have r 2 = (  1) 2 quadrant of the xyplane, so 8 = (b) r 2 = ( 2) 2
5. Since (;I =
f
v'2; tan{;/=
= 4 r 2 =
~
277
_: 1 = 1 and the point ( 1, 1) is in the second
4'~~' , 1).
3
4; tanB = ~ =  v'3 and the point (  2, 2v'3) is in the second quadrant of the
3. Thus, one set of cylindrical coordinates is ( 4,
but r and z may vary, the surface is a vertical halfplane
halfline y = x, x
1. z
+ 2mr; z =
2 ;
2 so r =
s; + 2mr; z = 1. Thus, one set of cylindrical coordinates is ( v'2,
+ (2J3) 2 = i6 so r =
xyplane, so (} =
+ 12 =
0
i~cluding
2 ; ,
3).
the z axis and intersecting the xyplane in the
0.
4  (x 2
+ y 2 ) or 4 
x2
y2,

SQ
the surface is a circular paraboloid with vertex (0, 0, 4), axis the zaxis, and
opening downward. 9. (a) Substituting x 2 + y 2 = r 2 and x
x
+ y2 + z2 =
=x2 
y becomes
= r cos 8, the equation x 2 
1 becomes r 2

r cos 8 + z 2 = 1 or
z 2 = 1 + r cos 8  r 2 . (b) Substituting x = r cosO andy = r sinO, the equation z
z = (rcos8) 2

2
(rsinB? = r 2 (cos2 8 sin 2 B) or z = r 2 cos 28. 1 0
11.
~
r
~
2 and 0
~
:z
~
1 describe a solid circular cylinder with
radius 2, axis the z axis, and height 1, but  11'/2 ~ 8 ~ 11'/2 restricts the solid to the first and fourth quadrants of the xyplane, so we have a halfcylinder.
13. We can position the cylindrical shell vertically so that its axis coincides with the zaxis and its base lies in the xyplane. If we
use centimeters as the unit of measurement, then cylindrical coordinates conveniently describe the shell as 6
o ~ o ~ 211', o .~ z 15.
~
~
r
~
20. The region of integration is given in cylindrical coordinates by
E
= {(r,B,z) 1
11'/2 ~ 8 ~ 11'/2, 0 ~ r ~ 2, 0 ~ z ~ r 2 } . This
represents the solid region above quadrants I and IV of the xyplane enclosed by the circular cylinder r = 2, bounded above by the circular paraboloid z
= r 2 (z = x 2 + y 2 ), and bounded below by the xyplane (z = 2
/ 2 j '2 rr I 71'/ 2 0 Jo T<
rdz
d d(}
r
=
I" rr/ / 2 f 2 [ ] z= r 2 Jo rz z=O
= I c'lf/2 1f12
2
2
3 dB Jo f r dr
d dB r
0).
'l ' / 2 2 Jof 2 r 3 drd(} = I 71'/
4 2 = toJr. 1271'/2 (lr ] 4 o
= 11'(4 0) = 411'
© 2012 Ccngage luming. All RighiS Rescrvctl. May nol be scannctl, copiod, orduplicalcd. or poslc'
7,
278
CHAPTER 15 MULTIPLE INTEGRALS
D
17. In cylindrical coordinates, E is given by {(r, B, z} I 0 ~ B ~ 21r, 0 ~ r ~ 4,  5 ~ z ~ 4}. So
19. The paraboloid z = 4  x 2

y 2 = 4  r 2 intersects the xyplane in the circle x 2 + y 2 =
4or r 2 = 4
=?
r = 2, so in
I
cylindrical co~rdinates, E is given by { (1·, B, z } 0 ~ (} ~ 1r/ 2, 0 ~ r ~ 2, 0 ~ z ~ 4 r 2 }. Thus ·
JJJE (x + y +
2
Io" 12 I; I04 (rcos{J + r sin(}+ z) r dzdr d(} = I0"12 I; [r2 (cos 0 + sin O) z + Io"/2 J; [(4r2  r 4 )(cos 0 + sin 9) + ~r(4 r· 2 ?] dr d{J
z) dV = =
3 = .Jfo'1r / 2 [(1r 3
= I 0"
12

l r 5 ) (cos(} +sin 0) 5
[~~(cosO + sin 0) +
= lfi(1 
0) +
Jf] dO= [~(sin 0 
¥. ~  ~(0 
~
8
~
dr d{J
r 2 ) 3 ] rr=O = Z d9
 .1.. (4 12
cos B)+ JfB];
12
1) 0 = ~7r + \ 258
21 . In cylindrical coordinates, E is bounded by the cylinder,. = 1, the plane z
E = {(r, 0, z) I 0
2
~7·z2 ]~=~r
r
= 0, and the cone z = 2r. So
27r, 0 ~ r ~ 1, 0 ~ z ~ 2r} and
IIIE x 2 dV = J;'T( f 01 J;r r 2 cos 2 9 r dz dr dO = I02 J01 [r 3 cos2 0 z ]::~·· dr d9 = I;" Io1 2r4 cos2 8 dr dO = I 02'T( [%r 5 cos2 B]~=~ dO= %I02" cos2 0 d(} = ~I;"~ (1 +cos 28) dO,;, t [B + t sin 28] ~1f = 2; 1f
23. In cylindrical coordinates, E is bounded below by the cone z
con~ and the sphere ·intersect when 2r2 = 2
r
=?
= r and above by the sphere r 2 +
z2
= 1, so E = {(r, 8, z) I 0 ~ 0 ~ 21r, 0 ~ r
= 2 or z = v'2 
~ 1, r ~ z ~ v'2  r 2 }
and the volume is
IIIE dV = J;" I01 .fr~ rdzdrdO =I;" I; [rzJ~::P drdO = I J (r~ r 1 2 2 3 2 3 = J0 dO Io (rv'2  r 2  r dr = 21r [~(2 r ) 1  ~r J: 2 0"
2
1f
= 27r ( ~) (1 + 1 2312 ) = 25. (a) The paraboloids intersect when x 2 + y 2 is D .E
=
{(x,y)
I x2
= {(r, B, z) I r 2
V =
g"I; fr
+ y2
::;
3x2

3y2
36 3 2 2 
r
2
)
drd(}
~7r
(V2 
x2 + y2
=?
1)
= 9, so the region of integration
9}. Then, in cylindrical coordinates,
~ z ~ 36  3r2 , 0 ~ r ~ 3, 0 ~ (} ~
r dz dr d(} =
(b) For constant density K , m = Nfyz
= 36 
1 0
)
 j7r (2 2V2) =
J(V
J;"I; (36r 
27r}
4r3 ) dr dB =
and
J;" [18r2 
r 4]
r 2 . The
::~dB= J02 "
81 dO= 16271".
= 1627rJ( from part (a). Since the region is homogeneous and symmetric,
= Mxz = 0 and
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SECTION 15.8 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES
M xv =
I~1f I: I:26  3r
2
= J( I~1f I: r [~ z2 ) ;:~~3 ,. dr d() 3 4 5 3 r ) dr d() = if I~:r d8 I 0 (8r  216r + 1296r) dr 2
 K (27r) [.!!.r6  216 r4
2
6
279
(zK) r dz dr d()
I: r((36 3r2 ) 2 
=if I 02 "
0
4
+ 1296r2] 30 = 11"!((2430) .= 243011"J( ~
11 z Mxz M:xv ) MThus (x, __ y, z ) = ( , ,  . m m m
2430rrI<) = (O , O, ~ = (0 , 0, 15).
+ 4y2 intersects the plane z = a when a = 4x2 + 4y2 or x 2 + y 2 = ia. So, in cylindrical 2 { (r, 8, z) I 0 $ r $ h!ci, 0 $ 8 $ 21r, 4r $ z $a}. Thus
27. The paraboloid z = 4x2 coordinates, E =
12"1Va/2 1a
K r dz dr d8 =
m =
0
0
 }(1211" [lar 
2
0
12"1Va/2(ar 
2 
0
r=,fi/2
d() 
r 4] r=O
T.(
.1·
4r 3 ) dr d8
0
12" ...!.. a
.
16
0
Since the region is homogeneous and symmetric, lvfvz = Mxz
Hence (x, y, z ) =
J(
4r2
2 }(
l a 1r 8
2 d() 
= 0 and
(0, 0, ~a).
29. The region of integration is the region above the cone z
=
Jx 2
+ y 2 , or z = r, and.below the plane z = 2. Also, we have
2 :::; y $ 2 with  /4  y 2 $ x $ )4  y 2 which describes a circle of radius 2 in the xyplane centered at (0, 0). Thus,
1 ~! 2
1~
2
..jz2+v2
 2
xzdzdxdy=
r1f {
2
Jo Jo
1 2
2
(rcos8)zrdzd;d0 = {
2 1f
{
1.
2
Jo Jo ,.
r
r 2 (cos O)z dzdrd()
•
= ~ Io2 I~ r 2 (cos 0) (4 r 2 )
=
J~1f I~ r 2 (cos 8) [~ z 2 ] ::~ dr d()
=
~ I~" cos 8 d() I~ (4r2  r 4 ) dr = ~ [sin 8]~" [~r 3
Tf

ir
5
]
dr d8
~= 0
31. (a) The mountain comprises a solid conical region C. The work don~ in lifting a small volume of material !:. V with density
g(P) to a height h(P) above sea level is h(P)g(P) t:.V. Summing over the whole mountainwe get W =
IIIc h(P)g(P) dV.
(b) Here C is a solid right circular cone with radius R
= 62,000 ft, height H =
12,400 ft,
and density g(P) = 200 Ibj ft3 at all points Pin C. We use cylindrical coordinates:
W = I~" IoH IoR(l z/ H) Z • 200r dr dz d8 = 211" foH 200z ar2 J ~::(l z/ H) dz = 4007r
1fl
2
~
z
z
2
= 2001rR2 [2 2
= ¥1rR H
2

( 1
·~
r
dz = 2001rR
2
1fl
( z :
2
z  ]H= 2001rR2 (H +3H 4H 2 2 2z
3
4

2
= ¥11"(62,000) (12,"400? ~ 3.1 x 10
19
3
~ + ~2 ) d7 2

0
2
2
2H 3

2) + H 4
!:._ = H z = 1 _ .=_ R
H
ft lb
© 2012 Ceng.!Jge Le:.ming. All Rigl11s Reserved. May not be scanned, copied. or duplicated. or posted to u publiCly accessible website, in whole or in part.
H
280
CHAPTER 15 MULTIPLE INTEGRALS
0
15.9 Triple Integrals in Spherical Coordinates 1. (a)
If 1
I I I
I '1r
y
= p sin rP sin 9 =
z
= pcos¢' =
I I
I
= psin¢cos9 =
From Equat!ons 1, x
(6.¥.f) 7i / 6 : I I I
6 sin ~ sin ~
6cos 2J
6 sin 2J cos i
=6·~·~ =
!.
== 6 · ~ · :1/ = ~, and
= 6 · :1/ = 3VJ, so the point is(~ . ~ ,3../3) in
I
I
rectangular coordinates. y
X
(b)
x
= 3 sin 3_; cos ~ =
y = 3 sin
3 ;
z= 3cos
3 ;
sin ~
3· ~ ·0
= 0,
=3 ·~ · 1=
¥, and
= 3 (~) = ·~¥. so thepointis
(o.¥ ,  ¥ ) in
X
·rectangular coordinates.
. 3. (a) From Equations 1 and 2, p = cos 9 = p s:
rP
.
.jx2 + y2 + z2
= 2 sin~71'/ 2)
(b) p = . v'1 + 1 + 2
= 0
z = 2, cos¢ = P =
X  1 cos 9 =   = ::::::::7 • psin ¢ 2sin(311'/4)
= ../02 + (2)2 + 02 = 3
=> 9 = ;  J2  2
::::}
 1
p= '02 = 0
=>
¢'
71'
= 2'' and
3 ( 2, ; ,
i).
371' 4 , and 371' 4
1
,..:::::    => 9 = 2 (J2/2) 
z
< 0). Thus spherical coordinates are
[since y
=
2, cos¢ =
J2
[since y
> 0). Thus spherical coordinates
37!') .
are ( 2, 371' , 4 4 5. Since ¢ =
x2
~,
+ (y 
9. (a) x
the surface is the top half of the right circular cone with vertex at the origin and axis the positive z axis.
~) 2 + z 2 = ~ . Therefore, the surface is a sphere of radius ~ centered at ( 0, ~, 0) .
= psin¢>cos8, y
= psin ¢>sinO, and z = pcos ¢,so the equation z 2 = x 2
2
(p cos¢) = (p sin cos 8)
2
+ (p sin ¢> sin 8) 2 or p2 cos2 ¢J =
+ y 2 becomes
p 2 sin2 ¢'. If p ;f 0, this becomes co~ 2 = sin2
rp. (p = 0
corresponds to the origin which is included in the surface.) There are many equivalent equations in spherical coordinates, such as tan 2 ¢ = 1, 2 cos 2 ¢' = 1, cos 2¢ = 0, or even rP (b) x 2
+ z2
= 9
<*
(psin cos 8)
p2 (sin2 ¢cos2 8 + cos2 ¢)
2
+ (pcos ¢') 2
= ~ . ¢' = :~4,..
= 9 <*
p2 sin2 ¢>cos2 9 + p2 cos2 ¢> = 9 or
= 9.
® 20 12 Ccngugc Lcamjng. All !Ugllls RcscrYcd. May not be scanned, copied, or duplit11ted, or posted to a publicly accc:ssible ""bsite, jn whole or in part.
SECTION 15.9 TRIPLE INTEGRALS IN SPHERICAL COORDINATES
0
281
11. 2 :S p :S 4 represents the solid region between and including the spheres of radii 2 and 4, centered at the origin. portion on or above the cone ¢>
0
:S ¢> :S
f
restricts the solid to that
= f, and 0 :S 8 :S 1r further restricts the
solid to that portion on or to the right of the xzplane.
13. p :::; 1 represents the solid sphere of radius I centered at the origin. 3 ;
15. z
:S ¢> :S 1r restricts the solid to that portion on or below the cone ¢> = ~;.
2': Jx2 + y 2 because the solid lies above the cone.
2z 2 2:_ x 2 cos¢>
p
2':
+ y 2 + z2 =
p
2
:::?
~, or equivalently 0
z
2
= p 2 cos2 ¢> 2':
:S ¢> :S
*.
Squaring both sides ofthis inequality gives z 2
~p2
:::?
2
cos ¢>
2': x 2 + y 2 ·
:::?
2': ~· The cone opens upward so that the inequality is
In spherical coordinates the sphere z
= x 2 +. y 2 + z 2
is p cos lp
= p2
:::?
= cos¢. 0 :S p :S cos¢> because the solid lies below the sphere. The solid can therefore be described as the region in
spherical coordinates satisfYing 0
:S p :S cos ¢>, 0 :S ¢> :S
*.
The region of integration is given in spherical coordinates by
17.
E = {(p, 8, ¢>) I 0 :S p :S 3, 0 :S 8 :S 7r/ 2, 0 :S ¢> :S 1r/ 6}. This represents the solid region in the fi rst octant bounded above by the sphere p
= 3 and below by the cone
¢> = 1T/6.
J0" 16 J0" 12 .1; p2 sin ¢> dp dB d4> = ./~"16 s in 4> d¢> f0" 12 dB J: = [cos¢>] ~
X
=
16
p2 dp
( 8]~ 12 [~Pa]~
(1 ~) G)(9) 9; (2 v'3) =
19. The solid E is most conveniently described if we use cylindrical coordinates:
E
= {(r,B,z) I 0 :s e :s i. O:s r :s 3,0 :s z :s 2}. Then
.1; ./~ f(r cos f)_ , rsinB, z) 1' dz d1·d8.
JJJE f(x , y, z) dV = f0" 12
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Resc~rvcd.
May not be: scnnnc:d. copied. or duplictuctl, or poS1ec.l 10 u publicly ucccssible wcbsilc~ in whole or in p3rl.
282
0
CHAPTER 15 MULTIPLE INTEGRALS
21. In spherical coordinates, B is represented by {(p, 0, 4>) I0 ~ p ~ 5, 0 ~ 0 ~ 27r, 0 ~ cf> ~ 1r}. Thus
JJJ8 (x 2 + y 2 +
z2) 2 dV = J; f 02 "' J:(p 2 ) 2 p2 sin cf>dpdO dl/> = Io"' sin 1/>dl/> J~"' dB I~ p6 dp = [ coscf>]~ [0]~" [tp 7 ]~ = (2)(27r)(1s.~211) = 312tJ07r ~ 140,249.7
23. In spherical coordinates, E is represented by {(p, 0, ¢) 12 ~ p ~ 3, 0 ~ 0 ~ 27r, 0 ~ cf> ~ 1f} and
IIIs(:c 2 + y 2 ) dV
= ]~"' g•r I 23 (p2 s in2 4>) p2 sin cf>dpdO dcf> = Io"sin3 cf>dcf> I~"' dO I: p4 dp
I 0"'(1 cos2 1/>) sinl/>dl/> [ 0]~,. [ip5 )~ = [coscf>+ ~ cos3 4>]~ (27r) : i(243 = (1  ~ + 1  U (27r) Cil) = 16~:" =
I
25. In spherical coordinates, E is represented by { (p, 0, 4>) 0 ~ p ~ 1, 0 ~ 8 ~ 2
2
JJJE xe"' + 112 +z dV = .=
J0"/2 J0Tr12 I:(psin cf>cosO)eP fo"12 H1 
cos 2¢>) d¢
2
IoTr12 cos(} d(} ( ~ p2eP
2
[integrate by parts with u = a¢
p2 sin
%• and
= p3 , so the integral becomes 11
sin ¢ '[~lJ :~ cos,P dBd¢ =
= i(4096)J0"
= 20348
5
[~ sin2 B]~ 12 [ip5 ] :
(Jorr/ 4. (1  cos2 1/>) sin 1/> dl/>)
( ~)
12
(211')
i J0" 1 2 f02 " sin c/> (4096 cos6 ¢) dOd¢
2 0 cos ¢sin ifJdifJ J 0 " dB =
20 48
3
( ~cos 7 ¢]~
12
[ B ]~"
= 40i~rr
43. In ·cylindrical coordinates, the equation of the cylinder is r = 3, 0
~
z
~
10.
The hemisphere is the upper part of the sphere radius 3, center (0, 0, 10), equation 1·
2 + (z  10) 2 = 32, z ~ 10. In Maple, we can use Ute coords=cylindrical option
in a regular plot3d command. In Mathematica, we can use Parametri c Plo t 3D.
45. If E is the solid enclosed by the surface p
E = { (p, 8, ¢) I 0 ~ p ~ 1 +
V(E)
= fJ{s
dV
= 1 + t sin 60 sin 5¢, it can be described in spherical coordinates as
t sin 68 sin 5¢, 0 ~ 8 ~ 211', 0 ~ ¢ ~ ,1f}. Its volume is given by
= fo" fo2 rr J~l+(sin60sin5r/>)/u P2 sin cf>dpd(Jd¢ = 1~~"
47. (a) From the diagram, z = r cot ¢ 0 to z = ~. r to r= asin ¢ (or use a 2  r 2 = r 2 cot2 ¢ ). Thus 0
V
[using a CAS).
=0
0
= Jor 2" Jora sin
rdzdrdB
0
z = rcot r/> 0 y
=
2:r [
(a2  a2 sin2 c/Jo)3/2  a3 sin3 ¢o cot¢o + a3]
;, ~7ra 3 [1
 (cos3 c/>0 + sin2 ¢ 0 cos 1/>0 )]
= lrra3(1  cos ¢ 0 )
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SECTION 15.1 0 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
0
285
(b) The wedge in question is the shaded area rotated from B = 81 to B = 82. Letting
= volume of the region bounded by the sphere of radius P;
V.i
and the cone with angle ¢; (B = 81 to 82) and letting V be the volume of the wedge, we have
y
V = (1122 1l21) (V12  Vn) · = 1(82  B1)[pg(1 cos¢2 )

p~(l  cos¢1 )  p~(1 cos¢2 )
= 1(82 81) [(P~ p~)(1 cos ¢ 2 ) Or: Show that V =
021P2 s in
1
p 1 s in 4> 1
01
(P~

pi) (1 cos ¢ 1)]
+ p~(1 cos¢1 ) ]
= i(B2  (Jt)[(p~
pi) (cos ¢ 1 
cos¢2 )]
1
r dz dr dB.
r cot 4> 2
(c) By the Mean Value Theorem with f(p) = p3 there exists some p with p 1 ~ p ~ p2 such that
f(p 2 )

f(p 1 ) =
!' (p)(p2 
p1 ) or p~  p~ = 3Ji D.p. Similarly there exists¢ with ¢ 1 ~ ~ ~ ¢ 2
such that cos ¢ 2  cos ¢ 1 ·= (sin¢) D.¢. Substituting into the result from (b) gives
D. V
15.10 1. x
= (p2 D.p)(B2 B1)(sin ¢)
Change of Variables in Multiple Integrals
= 5u v, y = u + 3v.
1) TheJacobianis a(x ~( ,y) v
3. x
= er sinB,
8(x,y)
 
8(r, B) 
5. x
D.¢ = p2 sin¢> D.pD.¢ D.B.
= ufv,
y
u,v
:;=
= l8xf8u
8x f 8v I = 5 8y f 8u oyfav 11
ll 3
=5(3) (1)(1) =16.
er cos B.
18xf8r 8xj8fJ I ~ e''sinB ercosBI · · r . = e•·er sin2 B ere'· cos2 B = sin2 fJ cos2 ()or cos 2() ayj8r ayjaB er cos() e smfJ . . y = vfw, z = wfu. oxfou 8xfav 8x/8w a(x,y,z) ayfau ayf8v 8yf8w a(u,v,w)azjou 8z/8v 8z/8w 2
1/ v
 u /v 2
0
0
1/w
v jw 2
 wfu2
0
1/u
I
2
= .!_11/w  vjw ! ( u) 0 vjw ! 0 I 0 1/w I v 0 1/u   V 2 .:..wfu 2 1/ u + wfu2 0 =.!.
1 1 (...!_  o) + ~ (o ~) + o= uvw     = o u w uvw v2
v uw
2
1. The transformation maps the boundary of S to the boundary of the imageR, so we first look at side 8 1 in the uvplane. S 1 is
described by v = 0, 0.::; u.::; 3, sox = 2u + 3v
@ 20 12 Ccng:~ge Learning. All Right5 Resencd.
= 2u andy =
u v = u. Eliminating u, we have.x = 2y, 0 _::; x
MBy not be sccumed. copied. or duplicated.
or posted 10 a publicly acccssiblo wcb~itc, in whole or in prut.
~
6. 8 2 is
286
0
CHAPTER 15 MULTIPLE INTEGRALS
= 3 y =? x = 6 + 3(3 y) = 15 3y, 6 :::; x :::; 12. S3 is the line segment v = 2, 0 :::; u :::; 3, sox = 2u + 6 andy = u 2, giving u = y + 2 =? x = 2y + 10, 6 $ X $ 12. Finally, S4 is the segment U = 0, 0 $ V $ 2, SO X = 3v and y = V :=} X = 3y, 0 $ X $ 6. The image of
the line segment u
:::; 2, sox= 6 + 3v andy = 3  v. Then v
= 3, 0:::; v
set S is the region R shown in the xyplane, a parallelogram bounded by these four segments. IJ
y
sl
(0, 2)
s.
s
0
s,
(3,2)

(6, 3)
T
sl (3, 0)
(12,1) 0
II
X
(6, 2)
9.
= v, 0$ u ~ 1, soy= v = 'U and X = u 2 = y2 • Since 0 ~ u ~ 1, the image is the portion of the 2 2 parabola X = y ' 0 :::; y :::; 1. s2 is the segment v = 1, 0 :::; u ~ 1, thus y = v = 1 and X = u ' so 0 :::; X :::; 1. The image is the l.ine segment y = 1, 0 $ X $ 1. 83 is the segment U = 0, 0 $ V $ 1, SO X = u 2 = 0 and y = V :=} 0 $ y $ 1. The
sl
is the line segment u
image is the segment x = 0, 0 :::; y :::; 1. Thus, the image of S is the region R in the first quadrant bounded by the parabola
x = y 2 , theyaxis, and the line y = 1. y
T
u
0
X
= 2x 1, y = 2x + 1 and the parallel lines y = 1 x , y = 3  x. The first pair of equations can be written as y  2x = 1, y  2x = 1. If we let u = y  2x then these lines are mapped to the ' vertical lines u =  1, u = 1 in the uvplane. Similarly, the second pair of equations can be written as x + y = 1, x + y = 3, and setting v = x + y maps these lines to the horizontal lines v = 1, v = 3 in the uvplane. Boundary curves are mapped to boundary curves under a !ransformation, so here the equations u = y 2x, v = x + y define a transformation r  1 that maps R in the xyplane to the square S enclosed by the lines u = 1, u = 1, v = 1, v = 3 in the uvplane. To find the transformation T that maps S to R we solve u = y  2x, v = x + y for x, y: Subtracting the fi_rst equation from the second gives v  u = 3x =? x = ~ (v  u) and adding twice the second equation to the first gives u + 2v = 3y =? y = Hu + 2v). Thus one possible transformation T (there are many) is given by x = Hv u),y = ~('u + 2v).
11. R is a parallelogram enclosed by the parallel lines y
3
v=3
s
u '= 1
u= 1
T
+1
 I
0
v 1 u
1
y=2t1
'',f' :
•
X
SECTION 15.10
CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
0
287
13. R is a portion of an annular region (see the figure) that is easily described in polar coordinates as
R = {(r , 8) 11 ::; r ::; ../2, 0 ::; (} ::; 7T/ 2}. If we converted a double integral over R to polar coordinates the resulting region of integration is a rectangle (in the rOplane), so we can create a transformation T here by letting u play the role of rand v the
= u sin v and T
role of 0. Thus T is defined by x = u cos v , y
maps the rectangle S
= {(u, v) I 1 ::; u ::; ../2, 0 ::; v. ::; 7T / 2}
in the uvplane to R in the xyplane. v
y
E.
2

T
s
0
15 B(x, y) = 1
· 8(u, v)
2 1
1 2
1
= 3 and x 
II
3y
= (2u + v) 
X
3(u + 2v ) = u 5v. To find the regionS in the uvplane that ·
corresponds toR we first find the corresponding boundary under the given transform ation. The line through {0, 0) and (2, 1) is
y = ~x which is the image of u
+ 2v =
image of (2u + v) + (u + 2v) = 3
u
+ 2v =
2(2u + v)
jj~ (x  3y) dA
. 8( u, v)
2 0 1
0 3
=>
v = 0; the line through (2, 1) and (1, 2) is x
= j 01 j 01 u ( u 
I= 6, x
2
(u 
u
2
5v) 131 dv du
+ ~{1  u?)
= 3 f 01 du =
= 4u2 and the planar ellipse 9x2
.
JJ
ffn x 2 dA =
+y =
3 wh ich is the
=> u + v = 1; the line through (0, 0) and (1, 2) is y = 2x which is the image of
=> u = 0. Thus S is the triangle 0 ::; v ::; 1  u, 0
=  3 j; 17 B(x, y) =
~(2u + v)
(4u2 )(6) dudv =
u2 +v2~ 1
~
u ::; 1 in the uvplane and
[uv + ~v 2 J::~u du .
3[~u 2  iu3

~(1 u?]~
=
3(~  ~ + ~)
=  3
+ 4y 2 ::; 36 is the image ~fthe disk u 2 + v 2 < 1. Thus 
f 02 "' f 01 (24r2 cos2 8)rdrd8 =
24J;.,.. cos 2 8d8 J01 r 3
dr
4
= 24[~ x +~sin 2x]~"' ar ] ~ = 24(7r)(i) = 67!' 2 1 19. 8(x, y) = , /v  ufv
o
8(u, v)
1
'
=
.!., xy = v
u, y =xis the image of the parabola v 2 = u, y
== 3x is the image of the parabola
v 2 = 3u, and the hyperbolas x y = 1, xy =a are the images of the lines u = 1 and u = 3 respectively. Thus
jL 21 . (a)
11: 3
x y dA =
:ix, z~ = y, u,v,w
1 (In v3u 3
u ( ; ) dv du =
u
a 0 0 0 b 0 = abc and since
u= ~. v= a
ln
vfu) du = J: u In J3 du = 4ln J3 = 2 ln 3.
J!.b' w =
~the solid enclosed by the ell ipsoid is the image of the c
0 0 c
JJJEdV =
JJJ u2+v2+w2
abcdudvdw
=
(abc)(volumeof tlie ball) = ~1rabc
:S 1
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288
0
CHAPTER 15 MULTIPLE INTEGRALS
i
2
2
2
(b) If we approximate the surface. of the earth by the ellipsoid ;., + + ; _= 1, then we can estimate 6 82 6 782 6 562 the volume of the earth by finding the volume of the solid E enclosed by the ellipsoid. From part (a), this is
fffE dV =
~11"(6378)(6378)(6356) ~ 1.083
X
(c) The moment ofintertia about the z axis is I%=
1012 km3 •
JJJE(x2 + y 2) p(x, y, z) dV, where E is the solid enclosed by
I
xz yz zz 18(x y z) 2 + b2 + 2 = 1. As in part (a), we use the transformation x = au, y = bv, z = cw, so ( ' ' ) = abc and a . c 8 u,v,w I, =
JJJE (x 2+ y 2) kdV =
2 2+ b2 ·u2 )(abc) dudv dw
f.(.[
k(a u
u2 + v2+w2
2
= abck f 0" f 0 "
~
1
J01 ( a 2p2 s in2 ¢ cos2 () + b2 p2 sin2 ¢ sin2 8) p 2 sin ¢ dp d8 drjJ
1 = abck [a2 f0,.. f~,.. f0 (p2sin2 rP cos2 B) p2sin 1/> dp d() dlj> + b2 = . a 3 bck
f0" f~,.. f01(p2 sin2 rjJ sin2 B) p 2 sin rP dp dB dr!>]
J0" sin3 rjJ drjJ f02" cos2 BdB J; p4 dp + ab3 ck fo.,. sin3 rjJdrjJ f~" sin2 BdB J; p4 dp
= a3 bck [ ~ cos3 ¢cos I/>] ~ [~B +
i sin2B]~"
= a3 bck (t) (1r) ( t) + ab3 ck (1)
(1r) (i)
23. Letting u = x  2y and v = 3x  y, we have x
[iP5 ] ~ + ab3 ck [t cos3 rjJ cosr/J]~ [~B i sin 28] ~" [tP 5 ]~
= rt7r (a 2 + b2) abck 8 f(v 3u). Then ((x, y)) = 8 u,v
= f.(2v u) and y = 0
and R is the image of the rectangle enclosed by the lines u = 0, u
0
= 4, v =
~ 3/5 1/5
215 1
1/5
= .!. 5
1, and v = 8. Thus
25. Letting u = y x, v = y + x, we nave y = !(u + v), x = Hv  u). Then
8(xy) 1.:...1/2 1/ 21 1 (u: v) = i = 2 and R is the 8 1 2 112
image of the trapezoidal region with vertices (  1, 1), ( 2, 2), {2, 2), and {1, 1). Thus
!h
R
11"
I
2
y x cos  dA=
Y +X
'1
112[
11
cos u   du dv = 2 2 v V
u]u=v
v sinV
1
u=
v
112
dv = 
2
.
27. Letu =x+y andv=x+y.Thenu+v = 2y => v=Hu+v) andu  v=2x 1 8(x ( 'y) ) = 11/2 1/ 2! =.Now[u[= [x+y[< [x[+[y[<1 2 1/ 2 1/2 . 8 u, v
[v [ = [x + y[ ~ [x [ + [y[ ~ 1 =>
2v sin{1) dv = ~sin 1
1
=> X=~(uv).
=>  l
 1 ~ v ~ 1. R is the image ofthe square
y 1
region with vertices {1, 1), (1, 1), (  1, 1), and (1, 1). So ffn e"+" dA
= ~ f~ 1 f~ 1 e" dudv =
He"t
1 [v
t
1
=e
e
1
X
.
® 2012 Crngogc Leoming. All Rights Reserved. May not be: scnnned, copied, or dupliea~<:d. or posted loa publicly accessible wcbsi~e, in \\i10lc or in part.
CHAPTER 15 REVIEW
0
289
15 Review CONCEPT CHECK m
1. (a) A double Riemann sum off is
n
,L ,L f (xi;, Yi;) b.A, where b. A is the area of each subrectangle and (xi; , Yi;) is a i=lj = l
S3J!lple point in each subrectangle. If j(x, y) ;::: 0, this sum represents an approximation to the volume of the solid that lies above the.rectangle R and be low the graph off.
(b) ffn. f(x, y) dA
= m,n+oo lim f f= f(xi;. Yi;) b.A i = 1 j =1
(c) If f(x, y) ;::: 0, ffn f(x, y) dA represents the volume of the solid that lies above the rectangle Rand below the surface
z = f(x, y). Iff takes on both positive and negative values, ffn j(x, y) dA is the difference of the volume above R but below the surface z
= f(x , y) and the volume below R but above the surface z =
f(x, y).
(d) We usually evaluate ffn. j(x, y) dA as an iterated integral according to Fubini's Theorem (see Theorem 15.2.4). (e) The Midpoint Rule for Double Integrals says that we approximate the double integral Riemann sum
JJR f(x, y) dA by the double
f f: f(xi, fi;) b.A where the sample points (:x,, fi;) are the centers of the subrectangles.
i=lj= l
(f) fnvc = A
~R)
IIn f(x, y) dA where A (R) is the area of R.
2. (a) See ( I) and (2) and the accompanying discussion in Sectio n 15.3. (b) See (3) and the accompanying discussion in Section 15.3. (c) See (5) and the preceding discussion in Section 15.3. (d) See (6)  (1 I) in Section 15.3 . 3. We may want to change from
re~tangular
to polar coord ina tes in a double integral if the region R of integration is more easi ly
described in polar coordinates. To accomplish this, we use ffn f(x, y) dA =
J: J: f (rcosB, rsin 9) r dr d(} where R is
given by 0 ::::; a ::::; r ::::; b, a ::::; B ::::; {3. 4. (a) m
= JJD
(b) M ,
p(x, y) dA
= ffv yp(x, y) dA, M 11 = ffv xp(x, y) dA
(c) The center of mass is
(d) I ,
= ffv
5. (a) P(a ::::;
(x, y)
where
2
y p(x, y) dA I v
= ffv
X::::; b, c:=; Y::::; d)=
m
and y= lvfx .
m
x p(x, y) dA, I o = ffv (x2
1:J:
(b) f(x , y) ;::: 0 and JJ~~ f(x, y) dA · (c) The expected value of X is JJ.
x == M" 2
+ y 2 )p(x, y) dA
j(x, y)dydx
= 1.
1= ffR2 xf(x, y) dA; the expected value ofY is p..z = JJT:/.2yf(x, y) dA.
@) 2012 Ccngage Learning. All Rights Rcscn·ed. May not be scanned. coried. or duplicatt.."tt. or posted to a publicly accessible wcbsilc. in whole or in pnrt.
290
CHAPTER 15 MULTIPLE INTEGRALS
0
6. A(S)
= JJ0 J
[f.,(x, y))2 + [f11 (x, y))2
(b) We usually evaluate
+ 1 dA
f[JB f(x, y , z) dV as an iterated integral according to Fubini's Theorem for Triple Integrals
(see Theorem 15.7.4). (c) See the paragraph following Example 15.7. 1. (d) See (5) and (6) and the accompanying discussion in Section 15.7. (e) See (10) and the accompanying discussion in Section 15.7.
(f) See (11) and the preceding discussion in Section 15.7. 8. (a) m
= JJJE p(x, y, z) dV
71 = JJJE xp(x, y, z) dV,
Mxz =
. (b) M z
· (   ) J (c) The cent er of mass ts x, y, z w 1ere x
JJJEyp(x, y, z) dV,
Mxy =
JJJEzp(x, y, z) dV.
Myz _y = Mxz, an d _z = . lvfx11 = , m m m
9. (a) See Formula 15.8.4 and the accompanying discussion.
(b) See Formula 15.9.3 and the accompanying discussion. (c) We may want to change from rectangular to cylindrical or spherical coordinates in a triple integral if the region E of integration is more easily described in cylindrical or spherical coordinates or if the triple integral is easier to evaluate using cylindrical or spherical coordinates.
10
I
I
a (x, y) ax;au ax;av ax ay ax By · (a) a (u, v) = ay;au ayjav = au av  av au (b) See (9) and the accompanying discussion in Section 15.10. (c) See (13) and the accompanying discussion in Section 15.10.
TRUEFALSE QUIZ
1': This is true by Fubini's T heorem. 3. True by Equation 15.2.5.
5. True. ~y Equation 15.2.5 we can write J~ this becomes
7. True:
I; f(x) dx .{
JJ0 J 4 
1 0
x2

f(x) dx
f01f(x) f(y) dy dx = I~ f (x) dx
I~ f(y) dy. But J~ f(y) dy
= J~ f(x) dx so
= [J; f(x) dxf.
+ y 2 + z 2 = 4 and above the xyplane = ~ (the volume ofthe sphere x 2 + y 2 + z 2 = 4) = ~ · t1r(2) 3 = 136 1r
y 2 dA = the volume under the surface x 2
® 2012 Cc:nguge Learning. All Rights Re served. May not be scc1.nncd copied. or duplicutcd. or posted to a publicly accessible website. in whole or in p.1rt. 1
CHAPTER 15 REVIEW
0
9. The volume enclosed by the cone z = ...jx2 + y2 and the plane z = 2 is, in cylindrical coordinates,
V = J~" J~
J: r dz dr dO i= f
2 0"
J~
J: dz dr dO, so the assertion is false. EXERCISES
1. As shown in the_contour map, we divideR into 9 equally sized subsquares, each with area b.A
ffn f(x, y) dA by a Riemann sum with m 3
= n = 3 and the sample points the upper right comers of each square, so
3
L: L:
ffnf(x,y)dA>:::J
= 1. Then we approximate
f(x i, yj)b.A
i =lj=l
= b.A [!(1, 1) + /(1, 2) + /(1, 3) + !(2, 1) + !(2, 2) + f(2, 3) + !(3, 1) + !(3, 2) + !(3, 3)] Using the contour lines to estimate the function values, we have
ffnf(x,y) dA 3.
7.
>:::!
1(2.7 + 4.7 + 8.0 + 4.7 + 6.7 + 10.0 + 6.7 + 8.6 + 11.9] ~ 64.0
f 12 J~ (y + 2xeY) dx dy = J? [xy + x 2 e 11] ::~ dy = f 12(2y + 4eY) dy = [y2 + 4eY] ~ = 4 + 4e 2  1  4e = 4e2  4e + 3
J; J; J0~ ysinx .d zdy dx = f J; [(ysinx)zJ::F dydx·= f f 0"
.
0"
1 0
y
...jl y 2 sinxdydx
= f 0" [ ~ (1 y2 ) 312. sin x)·u :l dx = f 0" ~ sin x dx = ~ cosx]~ = ~ y~
9. The regi?n R is more easily d~scribed by polar coordinates: R
= {(r, 0) I 2 s; r s; 4, 0 s; 8. s; 71"}. Thus
ffn f(x, y) dA = fo"f f (rcosO, r sinO) rdrdO. 4 2
. w hose area .IS g.tven . by Jo fTr / 2 r•in 20 d dO . The reg1on Jo r r IS
11.
·{ ( r, 0) I 0
s; 0 s; %, 0 s; r s; sill 20}, which is th~ region contained in the
loop in the first quadrant of the fourleaved rose r =sin 28 .
./~
1
13.
J: cos(y
2
)
dy dx =
J; g cos(y
2
)
dx dy
= f 01 cos(y2 ) [ x ] ::~ dy = J01 y cos(y2 ) dy X
®
2012 Cengogc Le:uning. All Rights Reserved. May not be scanned, copied. orduplic:ued, or posted to • publicly accessible website, in whole or in pan.
291
l
292
CHAPTER 15 MULTIPLE INTEGRALS
0
!1.r
y
17.
D
y 1 + x2 dA =
111r..rx o
= l2 0
o
1
,
.
1 : x2 dy dx =
1+ X
  dx 0 1 x2
1 o 1 + x2 [
[! In(1 + x
=
2
·I
2
1
))
fj ,0 ydA =
f 2J,By~
Jo
o = !4 In 2
ydxdy
112
= fo2 y[x] ::~~v2 dy = J;
3)
= f02 (8y 
/L (xz +
y
2y dy
y2)3/ 2 dA
=
[4y2
y(8
~ y2 y2) dy
tY 4 ]~

=8
= 17r/313 (r2)3/2r drd0 = 11r/3 dO 13r4dr = [0)~/3 [ir:;)~ 1f
35
811f
=35=5 23.
fffe xy dV
= .1: j~x J;+v xy dz dy dx = .{03 .fox xy [ z J::~+v dy dx =J; ./~"' xy(x + y) dy dx = ]~3 J;(x 2 y + xy 2 ) dydx = .r; [tx2 y 2 + ~x1lJ;;:~ 6 / '3 = 6, 0
25. Iff v 2 z 2 d\l=f E
1
 1
X
4
dX
[GX 1 5] 3 0
=
J~ ]'1 v ~ 0
= lls Jof " r4 ( 2 29 . V _ Jro2 JJ X
COS
sin3 () dO
= J; (tx 4 + tx4 )
dx
= 281 = .40 '5 2
2  z
 1 · Vl y2
1·2 )rdrd() = J02 Tr
40) [ 6r 1 6 r = 1 d()  ii1 1' 8 ] r=O
= lll5 [
2
rv'~~~ y 2 z 2 (1v2 
y2 z 2 dxdzdv=t
= J;1r J; (r2 cos2 O)(r2 sin2 O)(l = Jor2 rr S1 (1 
dx
1 [0 = TIJ2 
3 cos 0 + i3 cos OJ "o .
d ._ + 4y2) d y d X _ 10.2 [X 2 y + 34 y 3J u=4 u= l , X 
Jd ;t sin
2
2(J(r5
. 49] 2,41 SID 0
=

2Tr 192
z 2 ) dz dv
r 7 ) drdO
=
" 96
= !11 15 r2 ( 3 2 X
Jo
+ 8 4) d:r.. _
176
31: V
= Jof 2 j o'11 .lor<2  ul/ 2
dzdxdy
y..rx
h )y:o
X
19.
21 .
11
= .Jorz.lorv (1ly) dxdy 2
y
X
<£) 2012 C
dx
CHAPTER 15 REVIEW
33. Using the wedge above the plane z for m
293
= 0 and below the plane z = mx and noting that we have the same volume for .m < 0 as
> 0 (so use m > 0), we have
V = 2 foa/3 35. (a) m
D
J/a29y2 mx.dxdy = 2 foal~ tm(a2 
= fo1 fo1  v2 ydxdy = J~1 (y 
y3) dy
2
c
2
= Jor l Jor1 
= m[a2y  3y3)~/3 = m(ia3 
ia3) =
~ma3 .
= ~ ~ = ~
rl rl v ri 2Y 1 ( (b) Mv = Jo Jo xy dx dY = .Jo 1  Y2)2 dY
M,
9y2) dy
v y 2 dx dy = r1 y 2  y 4) d.y 10
= u1 ( 1 
'y
2)3) o1 = 12' 1
1 s ) = 152 . Hence cx, ) y = (_ 3 , 15 .
2 r1 r1v ( 3 s) dy=u, 1 (c)lx=JoJo y 3 dxdy= J·1 0 y  y r1 rl ly = Jo Jo
v
2
r1 1 ( _ 24 1 (1 1 yx 2 d x d Y _ Jo 3Y 1  Y2)3 dY Y2)4] 1o _ 24'
_ I I _ 1 =2 _ .!.L_g _ 1 .lo  "'+ v 8• Y  1/ 4  3
=?
1 d= 2 _ Y  73• an x 
= _
1/24 _ 1/ 4 
•= !!
o
;r;2+y2 ~ 11 2
2
=
+ y 2, 0 $
z $ h. Then V = ~ 1ra2 h is the
= !vfx:: = 0; and
1a (a r 2ar 2
2
0
o·
ooo
·
Hence the centroid is (x,y,z)
(b) I ,
1 = ?s·
11•(11/a),jx2+y2 z dz dA = 12,.1a1(hfa)(ar) rzdzdrd/J = 1r 1a.r h2 (a r) ; dr
M;r;y
= 1rh a2
x
=?
37. (a) The equation of the cone with the suggested orientation is (h  z) = ~ .jx 2 volume of one frustum of a cone; by symmetry M 11::
=
1
6
2
4
4
2a
2 2
4
(a 1rh a + r 3) dr = 1rh 2  2a3 + a4 ) = a2 12
= (0,0, ~h).
2"1"1(/•/.a)(ar) r. 3 dz dr d() = 27r 1" h (ar3  r 4) dr = 27rh (a5  a5) = 7ra4h a a 4 5 10 1 o o 0
0
39. Let D represent the given triangle; then D can be described as the area enclosed by the x  and yaxes and the line y = 2  2x,
or equivalently D
z
= {(x , y) I 0 $ x $
1; 0 $ y $ 2  2x }. We wantto find the surface area of the part of the graph of
= x + y that lies over D, so using Equation 15.6.3 we have 2
A(S)= =
Jfv
1+
.r; .../2 + 4x
(~;y + (~~Y dA = Jfv v 1+(2x)2+(1)2dA= 2
[y J~:~
2
;r;
dx
11 1
2

2 ;r;
)2 + 4x2dydx
= J;(2  2x) .../2 + 4x2 dx = · J01 2 J2 + 4x2 dx JC: 2x J2 +
Using Formula 21 in the Table oflntegrals with a
= J2. u. = 2x, and du =
4x2 dx
2 dx, we have
J 2 J2 + 4x2 dx = x J2 + 4x 2 + ln(2x + J2 + 4x 2 ). lfwe substitute u = 2 + 4x2 in the second integral, then du = Bx dx and J 2x ,J2 + 4x2 dx = ~ J .JU du = i · tu 312 = i (2 + 4x 2 ? 12 . Thus A(S) =
=
[xJ2 +
4x 2 + 1n(2x +
J6 + ln (2 + JB) 
= ln( J2 + v'3)
J2 + 4x2) 
.
1
t{2 + 4x2 ) 312 Jo
~(6) 3 1 2  ln J2 +
4 = ln 2t# + {1
+ 4 ~ 1.6176
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294
0
CHAPTER 15 MULTIPLE INTEGRALS
41.
f
rJ
3
2
9  x
1o 1J9x2
(x3 + xy2 )dydx
= {
3
{~
1o 1~
x(x 2 +y2 )dydx
= J::~~ 2 f~ (r cos O)(r =
J::~~2 cos 0 dO
= (sino]
J; r
2
4
)
r dr dO
dr
:::/2 (ir5]~ = 2 ·. i (243) =
4 6 :
= 97.2
43. From the graph, it appears that 1  x2 = e"' at x ~  0.71 and at
x
= 0, with 1 
> e"' on (0.71, 0). So the desired integral is
x2
JJ"n y2dA ~ f~o.n J.1., x2 y2 dydx = l3 Jo [(1  x2)3  e3"'] dx 0. 71 =o.25
45. {a) f(x , y) is a joint density function, so we know that
JJR2f( x, y) dA =
1. Since f(x , y)
= 0 outside the rectangle
(0, B) x (0, 2], we can say
JJR2 f (x,y) dA = f~oo f~oo f(x ,y) dydx = J: J: C(x + y) dydx = C .r; [xy + h2J ~:~ dx = C J03(2x + 2) dx = C[x2+ 2x)~ = 15C Then 15C
=1
(c) P(X + Y ~ 1)
C
:::?
= ft.
= P ((X, Y) E D) where Dis the triangular regiol} shown in
y
the figure. Thus
P(X +Y ~ 1) =
JJ0 f(x,y) dA = f 01J;x ft (x+y)dydx
=· ..1.. rl 15 J o = 
fs f; 1
[
1 xy + 12Y 2]u= u=O "' dx
[x(1 
f1(1
30 Jo
.z
47.

x) + ~ ( 1  x?] dx
0
I
X
.2)dX  30 1 [ 1 3]1 1 X 3X 0 = 45
X
( , zdzdydx= ) J.l0 .r.1=Jv"ii  v"ii ! (x,y,z) d xdydz J_l 1 j'l.,2 j 0'1 IJ fx,y 0
X
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CHAPTER 15 REVIEW
49. Since u = x y and v = x
8(x Thus ' y) = 8(u,v)
51 . Let U
=y 
I~~::~~
+ y, x
I1/2 1/2 ·1/2 1/2 1= 12 and flRx+y xy dA = 1/ .o u (1)  du dv =  £4 dv = In 2 v 2 , v · 4
and V
= y +X SO X =
I= I ~:~ .~~~~I
with vertices (u, v)
=
y :... U
= (V 
X)  U
=:>
_ 2
X
2
= ~ (V 
U)
and y =
V 
~ ( V  U)
= ~ (V + u).
1~ (~)  ~(~)I = '~I = ~.· R is the image under this transformation .o f the square J
= (0, 0), (2, 0), (0, 2), and ( 2, 2). So
This result could have been anticipated by symmetry, since the integrand is an odd function of y and R is symmetric about the xaxis. 53. For each r such that Dr lies within the domain, A(Dr)
exists (xr, Yr) in D,. such that f (xr, Yr) =
~
7TT
so lim
ro+
~ 1fT
295
= t(u + v) andy= ~(v u).
2
X
0
= 71'1'2 , and by the Mean Value Theorem for Double Integrals there
j'lv,. { f(x, y) dA. But lim (xr, Yr ) = (a, b), ro+
j'{ f (x, y) dA = lim f(xr, y,.) = f(a, b) by the continuity of f. } Dr
r+O+
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D PROBLEMS PLUS 1.
y
Let R ,=
5
u;=
1
Ri, where
R , = { (x , y ) I x+ y
4
~
i + 2, x +y < i+3, 1 ~ x ~ 3, 2 ~ y ~ 5}.
5
3
I:
ffn [x+y) dA =
2
5
f fn .[x+y)dA= L: [x+ylffn . dA , since
i =1
x+y=5 0
2
3
[x + y]
i= l
1.
= constant = ~+
'
2 fo r (x, y) E R; . Therefore
Jfn[x + y) dA = E?=1 (i + 2) [A(R;)]
X
= 3A{R I) + 4A(R2) + 5A(R3) + 6A(&) + 7A(R s) = 3(4) + 4(~) + 5{2) + 6(~) + 7 ( ~) = 30
rb
I
t [J.t cos(t)dt
I
3. f ,,.• = b _ ala f (x )dx = I  0 Jo =
J; J:cos(t
2
)
1
dt dx = ,{0
?
J; cos(t
2
)
dx dt
]
dx x=t
[changing the ordeqlf intcsrntion]
0
5. Since
lxvl <
1 1, except at (1, 1), the formula for the sum of a geometric series gives I 
f
=
 ~
.
n~
(xy)", so •.
"' 1 1 "' 00
=
1
00
=
L.., n+1 ' n+1
n=O
= 1!! +
L.., (n+11)2 n= O
= "L..,n'""= 1 ~1
1 + 32" 1 + ··· 21
1 
7. (a) Since ixyz i < 1 except at (I , I , I), the formula for the sum of a geometric se;ies gives 
1 X 1JZ
111111 o
11lll1I:
1  d:r;dydz=  _xyz o .o o
o o 1
(x yztdxdy dz
00
n =O
= I: 00
[
n=O 00
.
(xy z tdx dydz
o
0
00
[
,[01 y"' dy]
1
= ...~o (n + 1)3
1
[ Jf o1
1
z" dz ]
= I:  11 · n =O
1
= 13 + 23 + 33
.
n
1
o o o 1
xyz
dx dy dz
=
111·11·1I: ( 00
o
o o
xyz)" dx dy dz
n=O
+
n
1
+1
·  1
n
+1
1
00
+ . .. = n~1 n 3
(b) Since 1xyzl < 1, except at (1, 1, I ), the formula for the sum of a geometric series gives
111111 +
n=O
11111·1
n=O 0
00
= I: .f01 x" dx ]
E(xy z )n, so
=
= 2: 00
n=O
1 1 +xyz
=
f= ( x yz)" , so
n= O
11.1111( o
o
o
xyz)" dx dy dz
·
= n=o f ( 1)". (1~1 x" dx] [J; y"dy] [J; z" dz] = n=O f (1) "n+11 n +1 1 . n +1 1 00 ( 1)" 1 = n~o (n + 1)3 = 13

00 1 1 (1)"  1 23 + :P  . . . = n~o n3
© 2012 Ccngagc: Lc:.1ming. All Rights Resc!rv.:d. May not be scanned, copied. or duplicated, or posted to a publicly oc~ssiblc website. in whole or in parl
[continued]
297
298
D
CHAPTER 15 PROBLEMS PLUS
To evaluate this sum, we first write out a few tenns: s = 1  ; 3 a1
.
+ ;3

;
3
+
;~
 ;3
~ 0.8998. Notice that
= 7\ < 0.003. By the Alternating Series Estimation Theorem from Section 11.5, we have .Is s61~ a1 < 0.003.
This error of0.003 :viii not affectlhe second decimal place, so we have s .~ 0.90.
.
9. (a)x=rcosO, y =rsiDO, z=z.
I ou ou . (} S.lffil. 1ary {)(} =ax TSID
82u ao2
a2 u '
au. 8u ox au ay au az au au. . Then or= axar + ayar + ozor =ox cosO+ oysmOand
(} d + 8u oy rcos an 2
8 2 u.
.
= 8x2 r 2 SID2 0 + 8y2 r 2 cos2 (} 82 u !:i:2 vr'
lau
8 u 2 . au 8u . 2 oy ax r SID 0 cos (}  OX r cos (}  {}y r SID 0. So
1 8 2u _ 8 2 u a2 u 82 u . 2 a 2 v. . au cosO 2 <>o2 + !i'2 = !i"2 cos 0 + !i'2 sm 0 + 2 .<~.. <> cos 0 sm 0 + ;:; 
+ r ar + 2r
u
uz
ux
uy
82u . 2
+ ux. ~sm
vyuX
82u
2
82v.
. (}
O + ~cos 0  2~a sm uy uy X
8ucos9 8usin9  ox  r   oy  r 
82u
uX
cos
r
8u sinO
+ ;:;uy r
(}
82u
+ 8z 2
82 1~
82 u
=  ++ {)z2 . 8x2 {)y2 (b) x = psinrf>cosO, y = psin ¢>sin0, z ·= pcos¢>. Then
au
au~
8u8v
&&
&.
& ..
au .
op = ox op + oy op + oz ap = OX sm¢> cosO+ {}y sm¢> s m9 + {)z cos rf>, and
. . 1 I 8u 8u ~.. 8u ~.. . (} ou S1m1ar y Br/> = &xp.cos'l'cos 9 + oypc~'l'sm  azpsinrj>,and
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CHAPTER 15 PROBLEMS PLUS
au
au
And afJ =  ax
.
A. •
psm.I' Slll
f)
299
. A. () h'l + au ay pSID'I'COS , W I e a2u
a2u afJ
a2u ·2 2 a 2 p2 sin ¢ sin ()
·
= 2 ya a p2 sin2 if> cos() sin 0 + X
2
D
X
2
. 2 A. au psm'l' . A. cos 0 'l'cos2 0 a + aa yu2 p2 sm X
au p sm . 'l' A. • 0 a sm y
Therefore
2
2
2
2 au cot¢> au 1 8 u 1 a u aap2u +  + p2a¢  + + ~;:opap p2 aq? p2 sin2 if> aB2
= ::~ [(sin2 ¢> cos 2 B) + (cos2 ¢> cos2 B) + sin2 B) + ~~ [(sin2 ¢> s in 2 0) + (cos2 ¢> sin2 0) + cos2 OJ + ~~ + au
[cos2 ¢> .+ sin2 ¢)
[ 2 sin if> cos()+ cos ¢> c~ 0  sin ¢> cos 0  cos 0] 2
2
ax
2
psm¢> 2 2 [ 2sin ¢> sinB + cos ¢> s~n() sin ¢> sin() sinO ] ay psm¢>
+ au
2
But 2 sin2 ¢>cos 0 + cos2 if> cos 0  sin2 ¢>cos 0  cos 0 = (sin2 if>+ cos 2 if>  1) cos 0 = 0 and simi larly the coefficient of
aujay is 0. Also sin2 ¢co?>2 0 + cos2 if>cos2 0 + sin2 e = cos 2 fJ (sin2 ¢ + cos2 ¢) + s in2 () coefficient of
a2uj8y
11. Io"' I~ Io" f(t)dtdz dy
2
= 1, and simi larly the
is 1. So Laplace's Equation in spherical coordinates is as stated.
= IIIE f(t)dV , where.
y
· E = {(t, z, y) I 0 ~ t ~ z, 0 ~ z ~ y, 0 ~ y ~ x}.
t =z
If we let. D be the projection of Eon the ytplane then D = {(y, t)
I0 S t S
x, t S y S x}. And we see from the diagram
that E = { (t, z , y) I t ~ z ~ y, t ~ y ~ x, 0 ~ t ~ x }. So }~"' I~ I; f(t) dt dz dy = Io"' I /' .ftv f (t ) dz dy dt = Io"'
= J;' [(h = Io"' =
©
2

[ft"' (y t) f(t) dy] dt
ty)f(t) J~:; dt
[~x2  tx +
tt
2
]
f(t) dt
2
= Io"'
[~x  t x ~t
= Io"'
ax
2

2
+ t 2 ]f(t) dt
2tx + t 2 )f(t) dt
~ I;(x  t) 2 f(t) dt
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300
0
CHAPTER 15 PROBLEMS PLUS
13. The volume is V
= Jff R dV where R is the solid region given.
From Exercise 15.10.2 l(a), the transformation x =au,
= cw maps the unit ball u 2 + v 2 + w 2 :::; 1 to the solid ellipsoid x2 y2 z2 i o(x y z) 2 + b + 2 :::; with o( ' ' ) = abc. The same transformation maps the
y = bv, z
a
2
plane u
+v +w =
c
u,v,w
X
1 to 
a
. R. + yb + cz . = 1. T.hus tl1e regton 111 xyzspace
corresponds to the regionS in uvwspace consisting of the smaller piece of the . unit ball cut offby the plane u + v
+w
= 1, a "cap of a sphere" (see the figure).
We will need to compute the volume of S, but first consider the general case where a horizontal plane slices the upper portion of a sphere of radius r to produce a cap of height h. We use spherical coordinates. From tile figure, a line through tile origin at angle ¢ from tile zaxis intersects the pl_ane when cos¢
= (r  h)/a
=:>
a = (r·  h)/ cos¢, and the line passes through the outer rim of the cap when
= cos 1 ((r h)/r). Thus the cap
a = r =:> cos¢ = (r. h)/r =:> ¢
I (r h)/ cos¢:::; p:::; r, 0:::; (}:::; 21r, 0:::; ¢:::; cos 1 ((r h)/ r)} and its volume is
is described by { (p, 8, ¢)
v= =
1
f27r rcos 
Jo
.J o
f27r
f COS  l
Jo
((rh)/r)
((rh)/r)
Jo
112" 1cos
= 3 _ 1

3
0
0
2
[l
sin¢] P=•·
3 3P
((rh)/r) [ 3
r sin¢ 
· A._
r cos 'I'
1( _ 2
r
sinA.d dd.d(} "~'
P
'I'
d¢ d(}
p = (•·h)/cos¢
1
f27r [  3
.I 0
rr
J(rh)/ cos 4> P
h)3
cos
(r _
h? ] ¢ sin rjJ d¢ dO
cos3
 2 ,;.] ¢=cos~ ¢ =O
1
((r h)/•·) d()
.
(rh) h) +r3+ 2 r h) 3] d()  r   2(rh) 3(r r
r7f' [r3 = 31 ./o
2
1
1(
(This volume can also be computed by treating the cap as a solid of revo lution and using the single variable disk method; see Exercise 5.2.49 [ET 6.2.49].) To determine the height h ofthe cap cut from tl1e unit ball by the plane u
+v +w =
1, note that the line u
= ·u = w passes through the origin with
direction vector (1, 1, 1) which is perpendicular to the plane. Therefore this line coincides with a radius of tile sphere that passes through the center of the cap and
h is measured along this line. The line intersects the plane at ( t, sphere at (
t, t) and the
plane u + u + w = I
'7a, '7a, '7a). (See the figure.)
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CHAPTER 15 PROBLEMS PLUS
The distance between these points is h
V=
IlL
=
V jg  t Y= v'3 ( jg  t) = 3(
dV= 2
Ills 1:(~:~:~~ ~
= abc· 1rh (r  th) = abc· 1r ( 1 
= abC7r ( ~ 
©
dV = abc
1
Ills
73. Thus the volume of R is dV = abcV(S)
7J Y[1  i _(1 73)]
"7a) (j + aTa) = abC7r ( ~ 
~) ~ 0.482abc
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0
301
16
VECTOR CALCULUS
0
16.1 Vector Fields 1. F (x, y) = 0.3 i  0.4j All vectors in this field are identical, w ith length 0.5 and parallel to (3, :4) .
y
3. F (x,y) =  ~ i + (y  x) j T he length of the vector  ~ i
vt. +
+ (y 
x) j is
(y  x) 2 . Vectors along the line y
=X
are
horizontal with length ~.
5 F( )  y i +x j . x,y  Jx2 +y2 The length of the vector
~ is 1. x2 + y2
7. F (x, y , z) = k All vectors in this fie ld are parallel to the z axis and have length 1.
9. F(x,y,z)
=xk
At each point (x, y, z), F (x, y, z) is a vector of length For x
,,
lxl.
> 0, all point in the direction of the positive z axis,
while for x
< 0, a ll are in the direction of the negative
z axis. In each plane x = k, all the vectors are identical.
X
]j
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303
304
0
CHAPTER 16
VECTOR CALCULUS
11. F(x, y) ·= {x, y) corresponds to graph IV. In the first quadrant all the vectors have positive xcomponents and negative
ycomponents, in the second quadrant all vectors have negative x  and ycomponents, in the third quadrant all vectors have negative x components and positive ycomponents, and in the fourth quadrant all vectors have ~ositive x  and ycompdnents. In addition, the vectors get shorter as we approach the origin. 13. F (x, y) = (y, y + 2) corresponds to graph I. As in Exercise 12, aJI vectors in quadrants I and II have positive x~components while all vectors in quadrants III and IV have negative xcompo nents.Vectors along the line y
= 2 are horizontal, and the
vectors are independent of x (vectors along horizontal lines are identical).
15. F (x, y, z) = i
+ 2j +
17. F (x, y, z)' =·xi+ y j
3 k corresponds to graph IV, since all vectors have identical length and direction.'
+ 3 k corresponds to graph III; the projection of each vector onto the xyplane is x i + y j , which points
away from the o rigin, and the vectors point generally upward because their zcomponents are all 3.
19.
4.5'
1~I ~
I  4.5 1
I I
I
l
I
I
'
I
\
.
'
'
.. I
'
.
.. '
.'
I
I
'
' \
'I
I
The vector field seems to have very short vectors near the line y = 2x.
I
I
For F (x, y) = (0,
I
o'} we must have y 2 
2xy = 0 and 3xy 6x 2 = 0.
I
I 4.5 I
The first equation holds if y = 0 or y = 2x, and the second holds if
I
I
x = 0 or y = 2x. So both equations hold [and thus F (x, y) = 0] along
t
the line y = 2x.
I ~ \
\
4.5
21.
f(x, y)
= xe"'Y
=>
'i7 f(x , y) = f:,(x, y) i
+ jy (x, y) j
= (xe'"Y · y + e'"Y) i + (xe"'Y. x) j = (xy + l )e"'Y i + x 2 e'"!l j
23. 'ilf(x,y, z) =f:z:(x,y,z)i+fy(x,y,z) j +f::(x,y, z ) k =..) x i +. J y j +..) z ~k x 2 +y2 +z2 x 2 +y2 +z2 x 2 +y2 +z 25. f(x,y)
= x2 
y => 'il f(x,y)
The length ofV' f(x, y) is ..)4x2
= 2xi 
j.
+ 1. When x =I= 0, the vectors point away
from theyaxis in a slightly downward direction with length that increases as the distance from the y axis increases.
27.Wegraph\lf(x, y)=
2x 2 22 l +x+ y
i+
4y jalongwith 2 2y2 1 +x+
a contour map off. The graph shows that the gradient vectors arc perpendicular to the level curves. Also, the gradient vectors point in the direction in which f is increasing and are longer where the level curves are closer together.
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SECTION 16.2
29. f ( x , y)
= x 2 + y 2 =>
V f( x, y )
LINE INTEGRALS
0
305
= 2x i + 2y j . Thus, each vector V f (x, y ) lias the same direction and twice the length of
the position vector of the point (x , y), so the vectors all point directly away from the origin and their lengths increase as we move away from the origin. Hence, V f is graph III.
31 . f (x, y) = (x + y) 2
=> V f (x , y) = 2(x + y) i + 2(x + y) j. The x andycomponents of each vector are equal, so all
vectors are parallel to the line y
= x . The vectors are 0 along the line y = :x and their l~ngth increases as the distance from
this line increases. Thus, V f is graph II.
33. At t = 3 the particle is at (2, 1) so its velocity is V (2, 1)
=
(4, 3). After 0.01 units of time, the particle's change in
location should be approximately 0.01 V(2, 1) = 0.01 (4, 3)
= (0.04, 0.0~) . so the particle should be approximately at the
point (2.04, 1.03) .
35. (a) We sketch the vector field F (x; y )
=xi
y j along with
several approximate flow lines. The flow lines appear to be hyperbolas with shape similar to the graph of
y =
±11x, so we might guess that the flow lines have
equations y
= C / x.
= y(t) are parametric equations of a Aow line, then the velocity vector of the flow line at the point (x , y) is x' (t) i + y' (t) j . Since the velocity vectors coincide with the vectors in the vector field, we have x' (t) i + y' (t) j = x i  y j => dx Nt = x, dy I dt =  y . To solve these differential equations, we kno~ dxldt = x => dx/ x = dt => In lxl = t +. C => x = ±~t + c = Aet for some cons~nt A, and dy I dt =  y => dy / y = dt => In Ivi =  t + I< => y = ±e t + K = B e  t for some constant B. Therefore x y = A et B~t = AB = constant. If the flow line passes through (1, 1) then (1) (1) = constant= 1 => x y = 1 => y = 1l x, x > 0.
(b) If x = x(t) andy
16.2 Line Integrals 1. x = t 3 andy
= t, 0 :::; t :::; 2, so by Formula 3
.Ia y ds ~ 1\ 3
3 (
= ..l. . 1 (9t4 :ill 3
3. Parametric equations for C are x
fc x y 4 ds =
~~ Y+ ( ~~ Ydt =
1 2
3
2
3 2 + 1) 3 / 2) O = ..l. G4 (145 1 
= 4 cos t , y = 4 sin t, 
~
2
t y'(3t )2
+ (1)2 dt
=
1) or ..l. (145 v'I4554
:::; t :::;
~.
1\a
yf9t 4 + 1 dt
1)
Then
J:~~ 2 (4 cos t) (4sin t) 4 y'( 4sin t )2 + (4cos t)2 dt = J:~~ 2 4 5 cost sin4 t yf16(sin2 t + cos2 t ) dt
= 45
2 Jrr/ (sin4 t cos t )(4) dt =  rr/ 2
2
(4) 6 [l5 sinGt) Tr/ =  Tr/ 2
2 · 40 5
= 1638.4
® 2012 Ccns:~ge Lc.:uning. All Rights Rcscn·cd. May not be scnnncd, copied, or duplicated, or posKd to a publicly accessible wcbsile, in whole or in p..1n.
308
VECTOR CALCULUS
CHAPTER 16
0
5. If we choose x as the parameter, parametric equations for Care x = x, y = v'xfor 1:::; x:::; 4 and
fc ( x2y3 v'x) dy =
v'x] 2 ~ dx =
J;' [xz. (Vx)3 
= 12 [14 x ·1
4J14 (x3 1) dx

x] 41 !2 (64 
4  14 + 1)  ill 8
X
= x, y =
=> dy = ~ dx, 0 :::;
y
7.
(2, 1)
On cl:
~X
X :::;
2.
On Cz: x = x, y = 3  :v => dy = dx, 2 :::; x:::; 3. Then
f 0 (x + 2y) dx + x 2 dy = fc 1 (x + 2y} dx + x 2 dy +.J02 (x + 2y} dx + x 2 dy =J: [x+2(~x)+x 2 (~)]dx+J: [x+2(3x}+x2 (1}]dx =
9. x
= 2sint,
y = t, z
fc xyz ds =
= 2cos t,
f 02
0:::; t:::;
(2x + ~x 2 ) dx +
7T.
J23
(6  x x
2
)
dx
Then by Formula 9,
J0 (2 sin t)(t)(  2 cos th/(!!fd + {!fltf + C¥tf dt = J; 4t sin t cost .J(2 cos t)2 + 11)2 + (2 sin t) 2 dt = J0 7T
7T
= 2 y'5 j~'ff t sin2tdt = 2 y'5 [~ t cos 2t
 2tsin 2t
+'i sin2t] ~
.J4(cos 2 t + sin2t) + 1dt
integrate by parts with ] [u t , dv s in 2t dt
=
=
= 2¥'5(~  0) = J57T 11 . Parametric equations for Care x = t, y = 2t, z = 3t, 0 :::; t :::; l. Then
15. Parametric equations for Care x
=1+
3t, y = t, z = 2t, 0 :::; t :5. 1. Then
fc z2 dx + x 2 dy + y 2 dz = J;(2t? · 3 dt + (1 + 3t}~ dt + t 2 · 2 dt
= [¥tl + 3t2 + t]~ 17. (a) Along the line x
=
= J; (23t2 + 6t +
1) dt
¥ + 3+ 1= ¥
=  3, the vectors ofF have positive ycomponents, so since the path goes upward, the integrand F ·Tis
always positive. Therefore f.e l F · dr = f.e 1 F · T ds is positive. .
© 2012 Cengoge L
SECTION 16.2
LINE INTEGRALS
0
307
(b) All of the (nonzero) field vectors along the circle with radius 3 are pointed in the clockwise direction, that is, opposite the
direction to the path. So F · Tis negative, and therefore
fc2 F · dr = fc2 F · T ds is negative.
19. r(t) = llt 4 i + t 3 j,, so F (r (t)) = ( llt4 )(t3 ) i + 3(t3 ) 2 j = llt 7 i + 3t6 j and r' (t) =· 44t3 i + 3t2 j . Then
21 .
fc F · dr =
J; F(r(t)) · r '(t)dt = J;(ne · 44e + 3t6 · 3t2 )dt =
fc F · dr =
J; (sin t\ cos(t
2
= J;(3t2 sint3

), t
4
) ·
=
i
10
+ 9t8 )dt = [44t 11 +t 9 )~ = 45.
(3t2 ,  2t, 1) dt
2tcose + t 4 ) dt = [ cost3
fc ~ ·d~ = i
J; (484t
2
F (r (t )) · r '(t)dt = 2 [e
2
t t
2 
2te 
2 t

i
sint + tt 5 )~ =~cos 1 sinl 2
2
2
2
[ett et :rsm(et
sin ( e 
2 t )
)
. (
 '2 te t
2 ) ]
dt
J dt ::::J 1.9633
25. x = t 2 , y = t 3 , z = t 4 so by Formula 9,
fc xsin(y + z) ds = =
27. We graph F ( x, y)
= (x 
J;(e) sin(t3 + t 4 )vf(2t)2 + (3t2)2 + {4t3 )2 dt f 05 t 2 sin(t3 + t 4 ) J4t2 + 9t 4 + 16t6 dt ::::: 15.0074
y) i + x y j and the curve C. We see that most of\he vectors starting on C poi~t in roughly the same
direction as C, so for these portions of C the tangential compon~nt F · T is positive. Although some vectors in the third quadrant which start on C point in roughly the opposite direction, and hence give negative tangential components, it seems reasonable that the effect of these portions of C is outweighed by the positive tangential components. Thus, we would expect
fc F · dr = fc F · T ds to be positive. To verifY, we evaluate
fc F · dr: The curve C can be represented by r (t) = 2 cost i + 2 sin t j , 0 $ t
so F (r(t)) = (2 cost 2 sin t) i
+ 4 cos tsin t j and r ' (t) =
$
3 ;,
 2 sin t i + 2 cos t j . Then
fc F · dr = ./;"12 F (r (t)) · r ' (t) dt =
J;"12 [2 sin t(2 cost 
=
3 4J0 ,. 12 (sin2
2 sin t) + 2 cos t(4cos t sin t )) dt
t  s in tcost + 2sin tcos 2 t) dt
[using a CAS)
@) 2012 Ccngogc Lcruning. All Rights Reserved. May nol be 5Cl!Mcd. copied, or duplicnled, or pos lcd lo • publicly accessible wcbsilc, in whole or in part.
0
308
CHAPTER 16
(b) r (O)
= 0,
VECTOR CALCULUS
F(r(O))
(l
1 ) _ r ( 72 
2'
= (e 1 , 0);
1 ) 272' ,
2.1
F(r(ll ) /
F(.r (72
r(1) = (1, 1), F (r(1))
1 )) _
1 )·  /\ e1/2 ' 4,72 '
= (1, 1).
In order to generate the graph with Maple, we usc the 1 ine command in 0 ~....:.....:....c!...__
the plot t ools package to define each of the vectors. For example,
_ _+i 2. 1
0.2
vl : =line ( [0 , 0),. [exp(1), 0 ] ):
generates the vector from the vector field at the point (0, 0) (but without an arrowhead) and gives it the name vl. To show everything on the same screen, we use the display command. In Mathematica, we use List Plot (with the Plot Jo ined 
31. x
= et cos4t,
> True option) to generate the vectors, and then Show to show everything on the same screen.
y = et sin4t, z = et, 0 ~ t :::; 2rr .
Then:= e 1 (sin4t}(4)  etcos4t =  e 1 (4sin4t+cos4t),
dy dt
= e 1 (cos4t)(4) et sin4t =
e t(  4cos4t + sin4t), and dd.z =  e L, so
t
(~~ y ~; y ~: y + (
= v(e 1 )2((4sin 4t + cos4t)2 + (  4cos4t + sin4t)2 + 1]
+ (
= et J16(sin 24t + cos2 4t) + sin2 4_t + cos2 4t + 1 = 3 ..f2 et
J~ x 3 1/ z ds
Therefore
= J0211" (et cos4t) 3 (et sin 4t) 2 (e 1) (3 ..f2 et) dt  Jo r 2 .r3.,J2e 7tcos3 4t sin2 4tdt
33. We use the parametrization x = 2 cost, y = 2 sin t,  ~ :::; t :::;
ds =
172·704 5,632,706
..f2(1  e 14.r)
'i. Then
2
(',~n + (~~) dt = v( 2sin t) 2 + (2cost) 2 dt = 2dt, som = J~ kds = 2kJ::~~2 dt ~ 2k(7r),
2
Hence (x,·il) = · (~, 0). 35. (a) x = ..!._ { xp(x , y , z) ds, y = ..!._ ; · yp(x , y, z) ds, z = ..!._ { zp(x, y, z) ds where m =
m}c
(b) m =
x= z=
m
c
m }c
fc kds = k J~.r V4sin2 t + 4cos2 t + 9 dt
= k
vTIJ;"" dt =
1 1 211" iM 1 2.r 2k JI3 2kVlJ sintdt = 0, y = 27rk 13 0 27rk 13 0 2
M { .r (k Vi3) (3t) dt =
1
27rk
13
/o
3 (2rr2 ) 27r
./~ p(x , y, z) ds.
2rrkJI3,
v'i3 costdt =
= 3rr. Hence (x, y, z) =
0,
(0, 0, 37r).
® 2012 CcngJgc LcJ.ming. All Rights Rcsr.:rvcd. Mny not l>c scumH:~I. copic(l, oi· duplicntcd. or posted to 11 publicly ncclsslblc wcbsilc. in who le or in r:1rt.
SECTION 16.2
LINE INTEGRALS
D
309
37. From Example 3, p(x, y) = k(1 y), x = cost, y = sint, and ds = dt, 0 :S t :S 1r ::::}
1:. = Icy2 p(x,y) ds = IoT< sin2 t [k(1 sint)] dt = ki0" (sin2 t  sin3 t) dt = ~k I 0.,. (1 cos2t) dt k Io"(1 cos2 t)sintdt
Let u [
= cost, du =  sin t dt ] in the second integral
= k[~ + I1 1 (1u2 )du] =k (~  ~)
lv = Icx 2 p(x, y)ds = ki0.,. cos2 t(1 = k(~
sint)dt = ~ I 0.,.(1 +cos2t)dt
kio" cos2 tsintdt
i), using t?e same substitution as above. ·
39. W = I c F · dr = I;.,. (t sint,3  cost)· (1  cost,sint) dt
=
I;.,. (t 
t cost  sin t + sin t cos t + 3 sin t  sin t cos t) dt
=I;,. (t tcost + 2sin t) dt = [~t 2  (tsin t + cost)  2 cos t)~,..
integrate by .parts ] [ in the second term
41. r (t) = (2t, t, 1  t), 0 :::; t :::; 1. 1
W = Ic F · dr = Io (2t t 2 , t (1 1
= Io (4t 2t2 + t 1 + 2t  t 2 43. (a) r (t) =
at2 i +
3
bt j
::::}

t? ,1 1+
t (2t) 2 ) · (2,1,  1) dt
t + 4t2 ) dt =I; (t2 + 8t  2) dt = [lt3 + 4t2  2t)~ = ~
v (t) = r' (t) = 2at i + 3bt2 j
::::}
a(t) = v' (t) = 2a i + 6btj, and force is mass times
acceleration: F(t) = ma(t) = 2mai + 6mbtj. (b) W = f c F · dr =
J01 (2mai + 6mbtj ) · (2at i + 3bt2 j) dt = I 01 (4ma2 t +
18mb2 t 3 ) dt
= [2ma2 t 2 + ~mb2 t 4 )~ = 2ma 2 + ~mb2 45. Let F = 185 k. To parametrize the staircase, let x = 20 cost, y = 20 sin t, z
W = f c F · dr =I:.,. (0, 0, 185) · ( 20 sin t, 20 cost, ~) dt = (185) ~
=
~~ t =
J:.,.. dt =
1! t,
0 :::; t :::; 61r ::::}
(185)(90) :::::: 1.67 x 104 ftlb
47. (a) r (t) = (cost, sin t}, 0 :S t:::; 271", and let F =(a, b). Then W = I c F · dr = I ;" (a, b) · ( sin t,cost) dt = I~'lf(asint + bcost) dt = [a cost + bsin t)~.,.
=a +O  a+ O=O (b) Yes. F (x, y)
W
= kx =
(kx, ky) and
= I c F · d r =I;,. (k cost, k s~ t} · (sin t , cost} dt = J;" ( k sin t cost+ k sin t cost) dt = I~.,.. 0 dt = 0.
49. Let r(t) = (x(t), y(t), z(t)) and v = (v1, v2, va). Then
I c v · dr
= I:
(v1, v2, va) · (x' (t), y' (t) , z' (t)) dt
= I:
[v1
x' (t} + v2 y' (t)
+ va z' (t)] dt
= [v1 x(t) + v2 y(t) + va z(t)) : = [v1 x (?) + V2 y(b) + va z(b)J = v 1 [x(b) x(a)J + v2 [y(b) y(a)] + va [z (b)  z(a)] =: (v1, v2, va) · (x(b) x(a), y(b)  y(a) , z(b) z(a)) =
[v1 x(a)
+ V2 y(a) + vs z(a)]
(v1, v2, va) · [(x(b) , y(b) , z(b)} (x(a), y(a), z (a))] = v · (r(b) r (a)]
@ 201 2 C
310
CHAPTER 16
0
VECTOR CALCULUS
51. The work done in moving the object is fc F · dr 7 segments of equal length t:.s
= fc F · T ds. We can approximate this integral by dividing C.into
= 2 and approximating F
· T , that is, the tangential component of force, at a point (xi , yi) on
each segment. Since C is composed of straight line segments, F · T is the scalar projection of each force vector onto C. If we choose (xi, yi) to be the point on the segment closest to the origin, then the work done is 7
fc F
E
· T d.s ~
(F (xi, yi) · T(xi, yi)] t:.s
= (2 + 2 +
i=l
2 + 2+ 1 + 1 + 1)(2)
.
= 22. Thus, we estimate the work done to
.
be approximately 22 J.
16.3 The Fundamental Theorem for Line Integrals 1. C appears to be a smooth curve, and since \Jf is continuous, we know f is differentiable. Then Theorem 2 says t hat the value of fc \Jf
· dr
is simply the difference of the values off at the terminal and initial points of C. From the graph, this is
50 10 = 40. 3. 8(2x 3y) j8y
= 3 =
8( 3x + 4y 8)j8x and the domain of F is IR2 which is open and simplyconnected, so by
Theorem 6 F is conservative. Thus, there exists a function f such that \J f = F , that is, f:r:(x, y)
= 3x + 4y 
f 11 (x, y)
8. But f,.(x, y)
equation with respect to y gives f 11 ( x, y)
g(y) = 2y2

= 2x 
3y implies f(x, y) = x 2
=  3x + g' (y). Thus  3x +
By+ [(where I< is a constant. Hence f(x, y) = x 2


= 2x 
3y and
3xy + g(y) and differentiating both sides of this 3:~.:
4y  8 =
3xy + 2y2

+ g' (y) so g' (y ) =
4y  8 and
8y +I< is a potential function for F .
5. 8(e"' cosy)j8y :=  e"' sin y, 8(e"' siny)f8x = e" sin y. Since these are not equal, F is not conservative. 7. o(ye"' + sin y)j8y =ex+ cosy= o(e"' + X cosy)f8x and the domain ofF is IR2 • Hence F is conservative so there exists a function
f v(x, y)
= e"' +
f
such that \J f =F. Then fx (x, y)
xcos y
= ye" +sin y implies f( x, y) =
+ g'(y). But f 11 (x, y) = e"' +
ye"
+ x sin y + g(y) and
x cosy so g(y) = K and f(x, y) = ye"
+ xsin y +
K is a potential
function for F. 9. 8(lny + 2xy 3 )j8y = 1/y + 6xy 2 = 8(3x 2 y 2 + xfy)j8x and the domain ofF is {(x, y) I y> 0} which is open and simply
connected. Hence F is conservative so there exists a function f such that \J f = F . Then f:z:(x, y) =In y + 2xy3 implies
f(x,y) g(y)
=
= xlny + x 2 y3 + g(y) and ! 11 (x , y) = [(and f( x, y)
=
xlny
+x
2
y
3
xfy + 3x2 y 2 + g'(y). But f 11 (x,y) = 3x 2 y 2
+ xjy so g'(y) =
0
=}
+ [(is a potentia l function for F .
11. (a) F has continuous firstorder partial derivatives and
8 8 2xy = 2x = X (x2 ) on JR 2 , which. is open and simplyconnected. 8 8y
Thus, F is conservative by Theorem 6. Then we know that the line integral ofF is independent of path; in particular, the value of fc F · dr depends only on the endpoints of C. Sin.c e all three curves have the same initial and terminal points, J~ F · dr will have the same value for each curve.
®
2012 Cengagc Lc:uning. All Rights Reserved. Mny not be scanned, copied. or duplicatcdt or posted to a publicly occcssiblc ~vcbsitc, jo whole or in pan.
SECTION 16.3
THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS
(b) We first find a potential function f, so that \l f = F. We know ! :r:(x, y)
0
311
= 2xy and j y(x, y) = x 2 • Integrating
fx(x , y) with respect to x, we have f(x , y) = x 2 y + g(y). Differentiating both sides with respect toy gives
· J11 (x, y)
= x 2 + g'(y), so we must have x 2 + g'(y) =
x2
=?
g'(y)
=0
=> g(y) = K, a constant.
Thus f(x, y) = x 2 y + K. All three curves start at (1, 2) and end at (3, 2), so by Theorem 2,
fe F · dr = !(3, 2)  !(1, 2) = 18  2 = 16 for each curve. 2
13. (a) f r(x , y) = xy 2 implies J(x, y) = ~x y
g(y) = K, a constant. We can take K
2
+ g(y) and fy (x, y) =
= 0, so f( x, y) =
x 2 y + g'(y). But /y(x , y) = x 2 y so g'(y) = 0 =>
~x 2 y 2 .
(b) The initial point of C is r (O) = (0, 1) and the terminal point is r(1)
= (2, _1), so
fe F · dr = /(2, 1)  f(O, 1) = 2  0 = 2.
= xyz + g(y, z) and so jy(x, y , z) = xz + g11 (y, z). But jy(x, y, z) = = h(z) . Thus f( x, y, z) = xyz + h(z) and f:(x , y, z) = X'!f + h'(z). But
15. (a) f x(x , y, z ) = yz implies f(x , y, z)
gy(y, z ) = 0 => g(y, z) f:(x, y, z)
= xy + 2z, so h'(z) =
(b) fe F · dr
= !(4,6,3)  !(1, 0, 
2z =>
h(z ) = z2 + K.. Hence f(x, y, z) = xyz + z 2 (taking J( = 0).
2) = 8 1  4 = 77.
17. (a) f:r:(x, y, z) = yze"'= implies f(x, y, z) = ye"'=
gy{y,z) = 0
=?
+ g(y, z) and so fv(x, y, z) =
+ gy(y, z) . But jy(x, y, z ) = e"'= so.
e"'=
g(y,z) = h(z). Thus f( x,y,z)·= ye"'= + h(z) and f z (x , y,z) = xyex=
f:.:(x,y,z) = xyex=, so h'(z) = 0 (b) r (O) = (1, 1, 0), r(2)
xz so
=?
+ h'(z). But
h(z ) = K. Hence f(x,y ,z) = ye"'= (taking!(= 0).
=(5, 3, 0) so fe F · dr = /(5, 3, 0)  /(1 ,  1, 0) = 3e
0
+ e0 =
4.
19. The functions 2xeY and 2y  x 2 e v have continuous firstorder derivatives on JR2 and
~
Ulj
(2xev) = 2xe Y =
Theor~m
~
vx
(2y x 2 e:v), so F(x, y) = 2xeY i + (2y x 2 eY) j is a conservative vector field by . .
6 and hence the line integral is independent of path. Thus a potential function f exists, and fx(x , y) = 2xev
implies f(x, y) = x 2 e 11 + g(y) and fv(x, y)
g'(y)
=
g(y)
2y =?
= y 2 + K.
.fe 2xev dx + (2y  x 2 e v) dy
=  x 2 eY + g'(y). But f v(x, y) =
We can take K = 0, so f(x,y)
= /(2, 1) 
f(l , 0) = 4e 1
2y x 2 ev so
= x 2 e 11 + y 2 • Then
+ 1  1 = 4/e.
21. IfF is conservative, then fe F · dr is independent of path. This means that the work done along all piecewisesmooth curves
that have the described initial and terminal points is the same. Your reply: It doesn't matter which curve is chosen. 23. F (x, y) = 2y3 12 i
+ 3x Jyj , W
=
.fe F · dr. Since 8(2y3 12 )/8y = 3 y'Y = 8(3x Jy )/ ax, there exists a function f
suchthat\lf=F.Infact,fx(x, y)='2y 3 12
=> f(x,y)=2xy 312 +g(y) =>
/y (x, y) = 3x y'Y so g'(y) = 0 or g(y) = K. We can take I< = 0 W = fe F · dr
= j(2, 4) :.... j(I,·l) = 2(2)(8) 2(1) .=
=?
fv(x,y)=3xy 1 12 +g'(y). But
f(x, y) = 2xy3 12 • Thus
30.
© 2012 Ccngogc Leoming. All Rights Resen'Od. May nol b< scanned, copied. or duplicalcd. or posl
312
0
CHAPTER 16
VECTOR CALCULUS
25. We know that if the vector field (call it F) is conservative, then around any closed path C, fc F · dr = 0. But take G to be a circle centered at the origin, oriented counterclockwise. All of the field vectors that start on G are roughly in the direction of motion along C, so the integral around G will be positive. Therefore the field is not conservative.
27.
r1
..
'
I
I
I
'
1
t
2'11'
I
'
'
\
I
t 1
r
UI
I
\
I
'
,' I
.
'
I
I
I
I
I
paths, the number and size of the field vectors pointing in directions similar to that of the path seem to be roughly the same as the number and size of the
r 1
I
' '
'
From the graph, it appears that F is conservative, since around all closed
\:!l
. ' ' , 1r it '
'
I I
I I
I \
f r r ' ' ' ' I
I I l I
'
'
I
'
''
I
:
I
I
I
I
vectors pointing in the opposite direction. To check, we calculate . .
° (~
o (sin y) = cos y = ay 0X
~ ~
+ x cos y). Thus F is conservative, by
Theorem 6. 29. Since F is conservative, there exists a function
f such that F = \1 f, that is, P
= /z, Q = ! 11 , and R =
Q, and R have continuous first order partial derivatives, Clairaut's Theorem says that 8P / EJy = fx 11
f= · Since P ,
= fvx = EJQf EJx,
EJP/8z = fxz = /zx = 8 R /.8x, and 8Q/8z = fyz = f zv = 8 Rf8y. 31. D = {(x , y) I 0
y
< y < 3} consists of those points between, but not
3
on, the horizontal lines y = 0 andy = 3.
 
(a) Since D does not include any of its boundary points, it is open. More formally, at any point in D there is a disk centered at that point that lies entirely in D.
0
·.: X
'(b) Any two points chosen in D can always be joined by a path that lies entirely in D , so D is com1ected. (D consists of just one "piece.") (c) Dis connected and it has no holes, so it's simplyconnected. (Every simple closed curve in D encloses only points that are in D.) 33. D
= {(x, y) II ::; x
2
+ y2
::;
4, y
y
2: 0} is the semiannular region
in the upper halfplane between circles centered at the origin of radii l and 2 (including all boundary points). (a) D includes boundary points, so it is not open. [Note that at any boundary point, (1, 0) for instance, any disk centered there cannot lie entirely in D .] (b) .The region consists of one piece, so it's connected. (c) D is connected and has no holes, so it's simplyconnected. y 8P y2  x2 x 8Q y 2  x2 8P 35. (a)P=  x 2 +y2 , .A..= and Q = = .Thus VIJ (x2 + y2) 2 x2 +y2' OX (x2 +y2 )2 8y
(b) G 1:
X
= cost, y = sin t, 0 ::; t ::; ;rr' c 2:
X
= cost, y = sin t, t = 21!' to t =
1l'.
8Q =,.. UX
Then
{ F ·dr = ("'( sin t)(s~n t)+.(~ost)(cost)dt= ("' dt = 7rand { F· dr = ("' dt=  1!' }0 cos t + sm t ./0 lc2 J 21r
Jc,
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SECTION 16.4
GREEN'S THEOREM
3
Since these aren' t equal, the line integral ofF isn't independent of path. (Or notice that J~ F · dr = 2
Cs is the circle x + y
2
f 02
1r
dt
= 1, and apply the contrapositive of Theorem 3.) This doesn't contradict Theorem 6, since the
Green's Theorem
1. (a) Parametric equations for C are x = 2 cos t, y
= 2 s in t,
0 :::; t :::; 27r. Then
J;" [(2 cost 2sin t )( 2 sin t) +
f c(x  y) dx + (x + y) dy =
2
= J;"(4sin t+ 4 cos 2 t)dt
(2 cos t+ 2 sin t)(2 cost)] dt
= J~" 4dt = 4t]~"' =
81r
(b) Note that Cas given in part (a) is a positively oriented, smooth, simple closed curve. Then by Green's Theorem,
f c(x  y)dx + (x + y) dy= ffv [: ., (x+y)  : 11 (x  y)] dA = ffv [1 (1)] dA = 2ffvdA = 2A(D) 3. (a)
y
Ct: x = t
(1,2)
= 27r(2) 2 = 81r => dx = dt, y
C2: x = 1 =>
dx =Ddt, y
T hus
c1
(I, OJ
= t =>
dy
= dt,
D:::; t $ 2.
=> dx = dt, y=22t => dy =  2dt, D$ t$ 1.
C3: x= 1  t
o
=> dy = Ddt, D :::; t :::; 1.
= D
X
fcxydx + x 2 y 3 dy
=
xydx+x 2 y3 dy
f C1 + C2 + C3
2 3
(b) f c xydx +x y dy = ffv
= f 01 Ddt + f 02 t 3 dt +
J; [(1 t)(2  2t) 2(1 t?(2 
= D+ [~t 4 ]~ +
t? +
[:x (x y 2
= Jort [21 xy4 5.
) 
~(1 t) 6 ]~
= 4
2t) 3 ] dt
¥=~
/J~ (xy)] dA = f01 J02 ~(2xy 3  x)dydx
y=2z d rt (8xs  2X2) dX xy] y=O X= Jo
= 4f
 32 = 3'2
The region D enclosed by Cis given by {(x, y) I D :::; x:::; 2, x :::; y $ 2x }, so
y 4
3
[t(1 
(2,4)
f c xy2 dx + 2x2 ydy
= ffv ~
[ :, (2xiy)  : 11 (xy2 )] dA
I: I:"' (4xy  2xy) dy dx
= Jr2 [xlY2] y= y=2x d~ o x "" =
7. fc
(v + efi) dx+ (2x +cosy ) dy = Jfv [:x(2x + cosy 2
2
= fol Jyfi (2 
) 
1) dx dy =
:v·
2
Jf o
2
4 3x 3 dx = !!.x ] = 12 4 o
(v + efi)] dA
fol (yl/2 
y2) dy
=~
@ 2012 Ccngage Learning.. AU Rights Reserved. May nor be sc:a.nncd, copied, or dupliCDtctl, or posted to n publicly .:JCC("SSiblc website. in whole or in JUrt.
313
= 21r where .
domain of F , which is IR 2 except the origin, isn' t simplyconnected.
16.4
D
314
CHAPTER 16
0 3
VECTOR CALCULUS
3
=  3
:U (y3 ) ]
[ :., ( x3 ) 
= ffv
9. fc y dx x dy
J:" dO. .r: r 3 dr =
dA = ff0 ( 3x2

J:"" J:(3r2 ) rdrdO
3y2 ) dA =
3(27!)(4) =  2471"
11. F (x, y) = (y cosx  xysinx, xy + xcos x) and the region D enclosed by Cis given by
{(x, y) I 0
~
x
~
2, 0 ~ y ~ 4 2x}. Cis traversed clockwise, soC gives the positive orientation.
f c F · dr =  J_ 0 (ycosx xys.i nx )dx + (x~ + xcosx) dy =  ffv [ :., (xy + xcosx)  :Y (ycosx xysinx)] dA =
ffv(Y  xsin x
+ cosx : cosx + xsin x)dA =

J02 .{04  2
:r
4 2 2 2 2 2 2 = Jo f [ ly ] v=  "' dx = f 1(4  2x) dx   f (B  Bx 2 v=O Jo 2 Jo
=  (16  16 + 1: 13. F (x, y)

o)
= 
ydydx
2 3 2 + 2x2) dx  (Bx 4x + ~x a )o
16 3
= (y cosy, x sin y) and the region D enclosed by Cis the disk with radius 2 centered at (3, 4).
C is traversed clockwise, so C gives the positive orientation .
.f0 F · dr =
 J_ 0 (y  cosy)dx + (xsiny) dy = 
JJ0
(xsiny)
[ :.,
:u(y cosy)] dA
=  ffv(siny  1  sin y) dA = ffv dA = area of D .= 7r(2? = 47r y
15. Here C = C1 + C2 where
01 can be parametrized as X = t, y = 1,  1
~
t
~
1, and
C2 isgivenby x=t, y=2  t 2, 1 ~t~ l. Then the line integral is
f
. 2
y e"' dx
~~
1
+ x 2 ev dy = J_ 1 [1 · et + t 2 e · 0] dt
+ J~ 1 [(2 t 2? et( 
1) + ( t) 2e2 
.
2t)] dt
t· ( 
I
0
X
= J~ 1 [et  (2  t 2 ) 2 e t  2t3 e 2  t ] dt =Be+ 4Be 1 2
according to a CAS. The double integral is
.! r (aQ
. JD
aP)dA = ;·1111r2_,2 (2xeY .2ye"') dy dx =  Be + 4Be 1'
ax  8y
17. By Green's Theorem, W = f c F · dr = f c x(x + y) dx + xy 2 dy = J(0 (y 2
verifying Green's Theorem in this case.

x) dA where C is the path described in the
question and D is the triangle hounded by C . So
W=
J01 J;"'(y2 
• = [
f.I (1  x) 4
x) dy dx = 
f01 [ty 3 
~x 2 + ~x 3 ] ~
J; (i(1 
xy]~ = ~"' dx =
= ( ~ + ft )  ( f2) = 
19. Let C1 be the arch of the cycloid from (0, 0) to (27r, 0), which corresponds to 0 ~
(271", 0) to (0, 0), so 02 is given by X
=
271"..:. t, y = 0, 0 ~ t :::; 271". Then c =
x) 3

x(1  x)) dx
1 12
t
~
271", and let C2 be the segment from
cl u 02 is traversed clockwise, so  C
oriented positively. Thus C encloses the area under one arch of the cycloid and from (5) we have
A
=
f_0 y dx = .{0 1 y dx + f 0 2. y dx = f~" (1 cos t )(l
= J~" (1 
2 cost+ cos 2 t) dt + 0
 cost) dt + f~" 0 ( dt)
= (t 2 sin t + ~t + ~sin 2t] ~" =
371"
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is
SECTION 16.4
GREEN'S THEOREM
0
315
21. (a) Using Equation 16.2.8, we write parametric equations of the lin_e segment as x = (1  t)x1 + tx2, y = (1 t)y1 + ty2 , 0 ~ t ~ 1. Then dx = (x2 xl) dt and dy = (y2 yl) dt, so
f c x dy y dx =
f01 [(1 
t)x1 + tx2](y2 Yl) dt + [(1  t)y1 + ty2](x2 x1) dt
=
J; (x1(Y2 
=
f01 (x1y2 x2yl) dt =
Yl)  y1(x2 xl) + t((y2 Y1)(x2 x1): (x2  x1)(y2 Y1)]) dt
x~y2 X2Y1
(b) We apply Green 's Theorem to the path C = C1 U C2 U · · · U Cn, where Ci is the line segment that joins (xi , Yi) to (xi+l,Yi+ l) fori= 1, 2, ... , n 1, and Gn is the linesegmentthatjoins (x,.,yn) to (xl,Yl). From (5),
~ J~ x dy  y dx =
JJ0
dA , where D is tite polygon bounded by C. Therefore
area of polygon= A(D)= ffv dA = ~ J~ xdy ydx =
~ (J~ 1 x dy ydx + f c 2 xdy  ydx +
.... +
J~nl xdy ydx +fen xdy ydx)
To evaluate these integrals we use the formula from (a) to get
A(D) = ~[(x1y2 X2Yl) + (X2Y3  XaY2) + ' · · + (Xn;lYn XnYn 1) + (XnYl  XlYn)]. (c) A = ~ [(0 · 1  2 · 0) + (2 · 3  1 · 1) + (1 · 2  0 · 3) + (0 · 1  ( 1) · 2) + ( ~ 1 · 0 0 · 1)] = Ho+5+2+2) = ~ 23. We orient the quartercircular region as shown in the figure.
y
fc
a
2
A = {7ra sox = Here
c=
cl
1l'a! /2
2
x dy and y = 
+ c2 + Cs where cl: X =
t,
1ra!12
y=
fc
0, 0
2 y dx.
~
·t ~ a;
C2: x = a cos t , y =a sin t, 0 ~ t. ~ ~;and c3:
X
= 0, y = a  t, 0
2
+f c
t
~
c.
a. Then
a
X
J0" 12 (acost) 2(acost) dt + J; Odt 12 a 3 cos3 tdt = a 3 J0"/\1 sin 2 t) costdt = a 3 [sin t .i s in3 t]; = ~a~
f cx 2 dy = fc, x 2 dy = J~"/
~
2
2 2 x dy +.fc3 x dy = foa Odt +
J
_ 1 4a 2 so x = 7ra2 /2 Jc x dy = 311'. 12 fcy 2 dx = fc, y 2 dx + )~2 Y2 dx + fc 3 Y2 dx = foa 0 dt + f 0" (a sin t) 2(  a sin t) dt
J 12 (  a 3 sin 3 t) dt =
= 0" so 'ii = 
1l'a!/2
£y
2
dx
+ J0a 0 dt
12  a 3 J0" 1\ 1  cos2 t) sin tdt = a3 [~ cos3 t  cos t]; =  ~a3 ,
= ~;.Thus (x,y) =
( ::. :: )·
25. By Green's Theorem,  ~pfcy 3 dx =  kp J.f'v(  3y 2 ) dA = ffv y 2pdA = I ., and
tPfcx 3 dy
= ~p ff0 (3x 2 ) dA = ffv x 2 pdA = Iu .
27. As in Example 5, let C' be a counterclockwiseoriented circle with center the origin and radius a, where a is chosen to be small enough so that C' lies inside C, and D the region bounded by C and C'. Here 2 22 2 2 2 3 EJP = 2x(x + y ) .  2xy · 2(x + y ) · 2y = 2x  6x~ and => oy (x2 + y2)4 (x2 + y2)3 ® 2012 Cc:nsugc: Lcnrning. All Rights Rcscned. May nol be sciUUlCd., copied, or dupHc::ued, or posted to 3 pubJicJy ncccs.~ible '"'cbsite, in whole or in pnl1~
316
0
CHAPTER 16
VECTOR CALCULUS 2
'* .. {)Q {)X = 2x(x
L
=
2:~;3
6xy 2 . a . Thus, as m the example,

(x2 + y2)
IL(~~ ~:) = IL
0 dA = 0
dA
.fc, F · dr. We parametrize C' as r (t) =a cost i +a :sin t j , 0:::; t:::; 21r. Then
1
F · dr =
c
(X 2 + y 2 ) 4
+ j_ c, P dx + Q dy =
P dx + Q dy
and J c F · dr =
+ y 2 ?  (y 2  .x2 ) • 2(x2 + y 2 ) · 2x
1
2
F · dr
=
C'
1
"
o
2
2
2
a o
2
·
(

. . ·) a sm t 1 + a cost J dt
l 2" (cost sin t cost (1  sin t))
=. 1 l2~ (cost sin2 t  cos3 t) dt = 1 a o a o
11 = 
2
2(acost)(asint)i+ (a sin. t  a cos t) j 2 (a2 cos2 t + a 2 sin2 t)
~ costdt =  1 sint] a
2
2
dt
2 "
o
=0
.29. Since C is a simple closed path which doesn' t pass through or enclose the origin, there exists an open region that doesn't contain the origin but does contain D. Thus P
= y/(x2 + y 2 ) and Q =
xj(x 2
+ y 2 ) have continuous partial derivatives on
this open region containing D and we can apply Green's Theorem . But by Exercise 16.3.35(a), aPjay = aQ j8x, so
fc F · dr = ffv OdA ~ 0.
.
.
31. Usmg the first part of(S), we have that I In dx dy = A(R) = Ion X dy. But X = g(u, v), and dy =
&
au du + M 8v dv,
and we orient as by taking the positive direction to be that which corresponds, under the mapping, to the positive direction alongaR, so
{
i on
X dy =
{
l as
8 g(u, ·u) ( ah du + {)ah dv)
u
= ± .ffs =
[ :v.
v
(g(u, v) g~)

± ffS (!L9.. Uu !ll! Dv + g(u > v)
=
r
.los
g(u, v) {)ah du + g(u, v) Bah dv
u
:v (g(u, v) ~~) ) dA
02 " l:lul:lv
= ± .ffs (~~ ~ Z~ ~ ) dA
81 !l.!l l:l·v Vv.'

g(u > v)
v
[using Green's Theorem in the uvplane] 82 " ) 8vUu
dA
[using the Chain Rule]
_[by the equality of mixed partials] =
± Jj~ ~~::~\ dudv
The sign is chosen to be positive if the orientation that we gave to 8S corresponds tothe usual positive orientation, and it is negative
othe~ise. In either case, since A(R) is positive, the s ign chosen must be the same as the sign of ~i:: ~?.
ThereforeA(R) = l l dxdy=
16.5
lfsl~~::~~ ~dudv.
Curl and Divergence
1. (a) curlF
= 'il x
F
=
8j 8x x
=
+ yz
k
8 / 8y
8 /8z
y + xz z
+ xy
[:Y(z + xy) ...:. :z(y + xz)] i  [:x(z + xy) :z(x + yz)] j + [:x (y
= (x x) i  (y ®
j
y) j
~ xz) 
:y
(x + y z)] k
+ (z z) k = 0
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SECTION 16.5
8 0X
CURL AND DIVERGENCE
D
317
8 8 (y+xz)+  (z + xy)=1 + 1+1 =3 0Z 0y
(b)cliv F = 'V · iF= (x+yz) + 
k
j
8/8x 8/8y 8/8z = (ze" 0) i  (yze"'  xye• )j xyez 0 yze"'
3. (a) curlF = \1 x F =
= ze"' i
+ (xye" 
yze"' ) j  xez k
8
(b) clivF = 'V · F =ox (xye")
5. (a) curlF = \1 x F =
= (
X2
+ (0  xe") k
8
8
+ oy (0) + oz (yze'" ) = ye• + 0 + ye"' = y(e• + e"' ) j
k
8/8x
8/8y
8/ 8z
X
y
z
.jx2 + y2 + z2
.jx2 + y2 + z2
.
?l [( yz+ yz) i (xz +xz)j + ( xy+xy) k ]=O ) +y + Z 2 312
(b) div F = \1. F
= ..E._ ( ox
.jx2
x
+ y2 + z 2
+~(
+ z2)3/ 2 +
.jx2 + y2 + z2
) +..E._ (
az
Z
.jx2 + y2
+ z2
)
k
j
7. (a) curlF = \1 x F =
Y
8y
x2 + y2 + z2 _ y2 x2 + y2 + z2 _ z 2 2x2 + 2y2 + 2z2 2 (x2 + y2 + z2)3/ 2 + (x2 + y2 + z2)3/ 2 = (x2 + y2 + z 2)3/ 2 = ..jrx=;2;=+=y:;:= 2 =+=z=;;: 2
x2 + y2 + z2 _ x2 = (x2 + y2
)
= {0 
8 / ax 8/8y 8/ 8z e"' sin y eY sin z ez sin x
eV cos z) i  (e"' cosx  O)j + (0 e'" cosy) k
= (e11 cosz,  e" cosx, e"' cosy) (b) cliv F = \1 · F =
!(
e"' sin y)
9. If the vector field is F = .Pi+ Q j
8P P = 0, hence 0X .
In
8P
+ ~ (eVsin z) +
!(
e" sin x) = e'" sin y
+ Rk, then we know R =
aP
8R
= a = aZ = 0 X y
8R 0y
= 
+ eYsin z + e" sin x
0. ln addition, the x component of each vector ofF is 0, so
aR
.
.
8Q
·
= aZ = 0. Q decreases as y mcreases, so 0 y < 0, but Q doesn't change
h d' . . oQ oQ 0 t eX Or Z · 1rectJons, SO ax = az = .
.
8P 8Q 8R 8Q +  +a =0 +  + O < O Z 0X 8y 0y
(a)cliv F = 
(b) curlF =(. oR  aQ) i + (oP oR) j 8y 8z az OX 11. If the vector field is F = Pi + Q j
8Q Q = 0, hence ox . th
e
X
.
=
8Q 8y
=
8Q az
+ (oQ  ap) k ax
ay
= (0 O) i+ (0  O)j + (0  O)k = 0
+ R k,' then we know R = 0. ln addition, the ycomponent of each vector ofF is 0, so =
8R aR 8R ox = oy = az
.
.
= 0. P mcreases as y mcreases, so
8P . oy > 0, but P doesn't change m
d' . ap ap O or Z  lreCtiOns, SO ax = az = .
ap aQ + 8y ax
(a) d1v F = 
aR
+ 8z
= 0+0+0
=0
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318
0
CHAPTER 16
({)R  {)Q) i + ({)P  {)R) j + ({)Q  {)p) k = (0 0) i t (0  O) j + (o {)P) k = {)P k 8y .{)z {)z 8x ax 8y 8y 8y
(b) curlF =
fJP : S mce ;:;uy
13. curiF
VECTOR CALCULUS
> 0, 
= "il x F
8P k .IS a vector porntmg . . m . the negative . zd"1rect1on. . .q,, vy
• ·
= 8f8x
j
k
8f8y
8f8z
= (6xyz
2

6xyz 2 ) i  (~y 2 z2

3y2 z 2 )j + (2yz3

2yz 3 ) k
=0
and F is defined on all oflR3 with component functions which have continuous partial derivatives, so by Theorem 4, F is conservativ~. Thus, there exists a· function f such that F
= "il f. Then f., (x, y , z) = y 2 z 3 implies
f(x,y,z) = xy2 z 3
+ g(y,z) and f 11 (x,y,z) =
2xyz3 + g11 (y,z). But f 11 (x,y,z) = 2xyz3 ,sog(y,z) = h(z) and
f(x, y,z) = xy 2 z 3
+ h(z) . Thus fz(x, y, z) =
3xy2 z 2 + h'(z) but f:(x, y, z) = 3xy2 z 2 so· h(z) = K, a constant.
Hence a potential function for F is f (x, y, z) = xy 2 z 3
15. curl F
= "il X F =
8f8x
j
k
8f8y
8f8z
+ K.
3xy2 z 2 2x2 yz3 3x2 y2 z 2
= (6x2 yz 2  6x 2 yz 2 ) i  (6xy2 z 2  6xy 2 z) j .+ (4xyz3 = 6xy2 z(I z) j + 2xyz2 (2z 3) k =1 0
6xyz 2 ) k
so F is not conservative.
k
j
17. cur!F
= "il x F =
8f8x 8f8y
8f8z
F is defined on all oflll3 , and the partial derivatives of the component functi~ns are continuous,'so F is conservative. Thus there exists a function f such that "il f =F. Then f,(x, y, z) = e11 = implies f(x, y, z) = xe11=+ g(y, z)
fu(x, y, z) = xzev= + g11 (y, z). But / 11{x, y, z) = xze11=, so g(y, z ) = h(z) and f(x, y, z) = xe11z
f(x , y, z) =
+ h(z).
= xye11z + h'(z) but f:(x, y, z) = xye11= so h(z) =Kanda potential function for F xe11=+ K .
Thus f:(x, y, z)
.19. No. Assume there is such a G. Then div(curl G )
='!
uX
(x sin y) +
~
uy
(cosy) +
~
uz
=>
is
(z  xy) =sin y siny + 1
which contradicts Theorem 11 . j
k
21 . curlF = 8f8x 8f8y 8f8z
f(x)
g(y)
= {0 0) i + {0 O)j + {0 
0) k = 0. Hence F
= f(x) i + g(y) j + h(z) k
h(z)
is irrotational. ® 2012 Ccnl!llgC Lcnming. All Rights Reserved. Mny not be sclllUlcd. copied, or duplicated. or posted ro a publicly accessible websire, in whole or in par1.
f 0,
SECTION 16.5
CURL AND DIVERGENCE
0
319
ForExercises2329, 1etF(x,y,z) = P1 i+ Qd + R1 ~and G(x,y,z) = P2 i + Q2j + R2 k.
23. div(F +G)= div(Pt + P2 , Q1 + Qz,R1 + R 2 ) = o(P1 + P2) + o(Q 1+ Qz )" + o(R1/ R2 ) • 0X 0y Z = oP1 + oPz + oQ1 + oQ2 + oR1 + oR2 = (oP1 + oQ1 + oR1) + ( oP2 + oQ2 + oR2 ) ox ox oy oy oz oz ox oy oz ox oy oz
= div(P1 ,,Q1, R1 ) + div(Pz , Qz , Rz ) = div F + div G
27. div(F X G) ='iJ · (F x G)=
=
[Q
1
oR2 ox
+ R2 oQ1 _ Q2 oR1 _ R1 oQ2] _ ox ox ox
[p oRz + Rz oP1 _ Pz oR1 _ .oy oy ay 1
R 1 oPz]
ay
+ [Pt oQ2 + Q2 oP1 _ p2 oQ1. _ Q1 oP2 ] oz oz . oz oz = [p2 (oR1 _
oy
oQ1) +Q 2 (oP1 _ oR1) +Rz(oQ1 _ 8P1 )] oz oz ox OX ay
_[p (oRz _ oQz) f_ Q (oP2 _ oRz) + R1(oQz ~ aP2 )] oy az az ox ox 1
= G · curl F
1
8y
 F · curl G
29. curl{curl F) = 'iJ x ('iJ x F) =
ajax
j
k
a;ay
a;az
Now let's consider grad(div F )  'iJ2 F and compare with the above. (Note that 'iJ 2 F is defined 'on page J J 19 [ET I 095).) [continued]
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0
CHAPTER 16
. F) d(d IV gra
VECTOR CALCULUS
2 2 n 2p  [ ( EPH+fiQ1+fiR1)·1 +( 8 P1 +8 Ch

V
OX2

OXOY
OXOZ
oyox
2 8 R1)·
oy 2 + [)y[)z
(8OZOX
J+ ,
2 P1.
2 8 Q1
2 8 R1)k]
+ Ozoy + OZ 2
2 2 2 2 2 2 _ [( 8 P1 + 8 P1 + B P1) i + (8 Q1 + 8 Q1 + 8 Q1) j 8x2 8y2 8z 2 8x 2 8y2 8z 2
Then applying Clairaut's Theorem to reverse the order of differentiation in the second partial derivatives as needed and comparing, we have curl curl F = grad div F  \l 2F as desired. 31. (a)\lr =V.Jx2+y2 + z2= x i+ y j + z k = xi + yj+z k . .Jx2+y2+z2 .Jx2+y2+z2 .Jx2+y2+z2 .Jx2+y2 + z2 j
(b)\lxr= :x
:V
(c) \l
(.!) = r
\l (
.Jx2
r
k
:z =l:V(z)  !(v)]i + [:z(x)  :x(z)]j + [:x(y):V(x)]k = O
y
X
=~
Z
1
+ y2 + z2
) .
·
1
1
1
(2x) · (2y) (2z) . 2 .Jx2 + y2 + z2 . 2 Vx2 + y2 + z2 2 .Jx2 + y2 + z2 ~~~~~1 Jk x2 + y2 + z2 x2 + y2 + z2 x2 + y2 + z~ 
x i+ y j +z k
r
(x2 + y2 + z2)3/2 = r3 (d) \lin r
= \lln(x2 + y2 + x
.
:::::: 1
x2 + y2 + z2
33. By (13), fc f(\lg) · n ds =
z2)1/2 = ~ \lln(x2 + y2 + z2)
y • z k x i + y j +z k r + x2 + y2 + z2 J + x2 + y2 + z2 = x2 + y2 + z2 = 7.2
ffv div(f\lg) dA = JJ'o!f div(\lg) + \lg · \l f] dA by Exercise 25. But div(\lg) = \l2 g.
Hence ffv f\l 2 gdA = fc f(Vg) · nds  ffv \lg · \lf dA. 35. Let f(x, y)
= 1. Then V f
ffv \l2 gdA V 2g = 0
= 0 and Green's first identity (see Exercise 33) says
= f c (Vg) · n ds 
ffv 0 · \lgdA =>
ffv \l 2 gdA = J~ \lg · n ds. But g is harmonic on D, so
=> f c\lg· nds = OandfcDngds = fc(\lg· n)ds =O.
37. (a) We know thatw = vjd, and from the diagram sin (;I= d/r
=> v = d!JJ = (sinO)rw = lw x rl. But vis perpendicular
to both wand r, so that v = w x r.
®
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SECTION 16.6 PARAMETRIC SURFACES AND THE;IR AREAS
0
321
k
j
(b) From(a), v=w x r= 0 0 w =(O ·z wy)i + (wx~ O ·z) j+ (O · y x· O) k= wyi+ wx j X
y
Z
k
j
(c) cur] v
= "il X v =
ajax a jay ajaz wy
wx
0
= [~ (0)  ~ (wx) ] i + [~ ( wy )  ~ (o)] j + [~ (wx ) ~ (wy )] k ay az az ax ax ay = [w 
( w)] k
= 2wk =
2w
39. For any continuous function f on iR3 , define a vector field G (x , y, z) = (g(x , y , z) , 0, 0) where g( x, y, z ) = fox
a
Then div G = a
X
f (t, y, z) dt.
a a a x (g(x ,.y, z)) +a (0) +a (0) = ~ f0 f (t , y , z) dt = f (x, y, z) by the Fundamental Theorem of y
Z
vX
.
3
Calcu lus. Thus every continuous function f on R is the divergence of some vector field.
16.6
Parametric Surfaces and Their Areas
1. P (7, 10, 4) lies on the parametric surface r (u , v) = (2u + 3v , 1 + 5u  v , 2 + u + v) if and only if there are values for u and v where 2u + 3v = 7, 1 + 5u  v = 10, and 2 + u + v = 4. But solving the first two equations simultaneously gives
u = 2, v = 1 and t~ese values do not satisfy the third equation, so P does not He on the surface. Q(5, 22, 5) lies on the surface if2u + 3v = 5, 1 + 5u v
=
22, and 2 + u + v
= 5 for some values ofu and v. Solving the
first two equations simultaneously gives u = 4, v =  1 and these values satisfy the third equation, so Q lies on the surface.
3. r(u , v) = (u + v ) i + (3  v) j + (1 + 4u + 5v) k = (0,3, 1} + u (1,0, 4) + v(1, 1, 5). From Example 3, we recognize this as a vector equation of a pl.ane through the point (0, 3, 1) and containing vectors a = {1, 0, 4) and b = (1,  1, 5). If we j wish to find a more conventional equation for the plane, a normal vector to the plane is ~
X
b =
1
k
0 4 =4 i  j  k
1 1 5
and an equation of the plane is 4(x 0) (y 3) :._ (z 1) = 0 or 4x y z =  4.
5. r(s, t ) = (s, t ,
es
2
),
so the corresponding parametric equations for the surface are X
point (x , y , z) on the surface, we have z = y 2

=
s, y = t, z = t 2

s 2 . For any
x 2 . With no restrictions on the parameters, the surface is z = y 2
we recognize as a hyperbolic paraboloid.
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x
2,
which
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0
CHAPTER 16
VECTOR CALCULUS
The surf~e has parametric equations x
= u 2 , y = v 2 , z = u + v, ; 1 ~ u
~ 1, 1 ~ v ~ 1.
In Maple, the surface can be graphed by entering
pl ot3d([u"2,v"2,u+v ) ,u=l .. l,v=l .. l);. In Mathematica we use the Paramet r icPlot3D command. If we keep u constant at u 0 , x = uij, a constant, so the corresponding grid curves must be the curves parallel to the yzplane. If vis constant, we have y
= vij, a constant, so these
grid curves are the curves parallel to the xzplane.
9. r (u,v)
= (ucosv,usinv,u5 ).
The surface has pararnetrid equations x = u cos v, y = u sin v, z = u 5 , 1 ~ u ~ 1, 0 ~ v ~ 21r. Note that ifu
= u 0 is constant
then z· = u8 is constant and x = uo cos v, y = Uo sin v describe a
vconstant
luo 1. so the corresponding grid curves are circles parallel to the xyplane. If v = vo, a constant, the parametric
circle in x, y of radius
equations become x
= u cos vo, y = u sin vo, z =
u 5 . Then
= (tan v0 )x, so these are the grid curves we see that lie in vertical planes y = kx through the zaxis.
y
11. x = sin v, y Note that if v
= cosusin4v,
z = sin2usin4v, 0 ~ u ::; 21!',  ~ ::; v::; ~ 
= vo is constant, then x = sin vo is constant, so the
corresponding grid curves must be parallel to the yzplane. These are the vertically oriented grid curves we see, each shaped like a " figureeight." When u = uo is held constant, the parametric equations become x
= sin v, y =
cos u 0 sin 4v,
z = sin 2u0 sin 4v. Since z is'a constant multiple of y, the corresponding grid curves are the curves contained in planes
z = ky that pass through the xaxis. 13. r( '1.1;• v) = u cos v i + u sin v j
+ v k. The parametric equations for the surface are x = u cos v, y = u sin v, z =
v. We look at
the grid curves first; if we fix v, then x andy parametrize a straight line in the plane z = v which intersects the zaxis. If u is held constant, the projection onto the xyplane is circular; with z = v, each grid curve is a helix. The surface is a spiraling ramp, graph IV.
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SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS
15. r (u., v)
= sin v i+ cos U. sin 2v j
0
323
+sin u. sin 2v k. Parametric equations for the surface are x = sin v, y = cos u. sin 2v,
z = s in u. s in 2v. lfv = vo is fixed, then x = sin vo is constant, andy= {sin 2vo) cosu and z = (s in 2v0 ) sin u. describe a c ircle of radius !s in 2vo l, so eachcorresponding grid curve is a circle contained in the vertical plane x =sin v 0 parallel to the yzplane. The only possible surface is graph II. The grid curves we see running lengthwise along the surface correspond to holding u constant, in which casey = (cosu.o) sin 2v, z =(sin uo) sin 2v
~
z =(tan uo)y, so each grid curve lies in a
plane z = ky that includes the xaxis.
17. x =
3
= vo is held constant then z
cos3 u cos3 v, y·= sin3 u cos3 v, z = sin v. lf v
= sin3 v 0 is constant, so the
corresponding grid curve lies in a horizontal plane. Several of the graphs exhibit horizontal grid curves, but the curves for this surface are neither circles nor straight lines, so graph Ill is the only possibility. (ln fact, the horizontal grid curves here are
= acos 3 u , y = asin3 u. and are called astroids.) The vertical grid curves we see on the surface uo held constant, as then we have x = cos3 uo cos 3 v, y = sin3 u.o cos3 v so the corresponding grid curve
members of the family x correspond to u =
lies in the vertical plane y = (tan3 uo)x through the zaxis. 19. From Example 3, parametric equations for the plane through the point (0, 0, 0) that contains the vectors a = (1 , 1, O) and
b
= (0, 1, 1 ) are x = 0 +
u(1) + v(O)
= u., y = 0 + u.(  1) + v(1) =
2
~
2
21. Solving the equation for x gives x = 1 + y + !z2
x
= J1 +
v  u., z
= 0 + u.(O) + v( 
1)
=  v.
y 2 + tz 2 • (We choose the positive root since we want
the part of the hyperboloid that corresponds to x ~ 0.) lfwc let y and z be the parameters, parametric equations are y
= y,
z=z, x= J1+y2 +i·z 2 • 23. Since the cone intersects the sphere in the circle x 2 can parametrize the surface as x
+ y 2 = 2, z = .../2 and we want the portion of the sphere above this, we
= x, y = y, z = v'4 
Alternate solution: Using spherical coordinates, X
x2

y 2 where x 2 + y 2 ~ 2.
= 2 sin ¢ cos e, y =
2 s in ¢ sine, z
= 2 cos¢ where 0 s;
and
o s; e s; 21r. 25. Parametric equations are x
= x, y = 4cos8, z = 4sin0, 0 s; x s; 5, 0 :5 () :5 21r.
27. The surface appears to be a portion of a c ircular cylinder of radius 3 with axis the x axis. An equation of the cylinder is
. y2 + z 2
= 9, and we can impose the restrictions 0 :5 x :5 5, y :5 0 to obtain the portion shown. To graph the surface on a
CAS, we can usc parametric equations x
=
u, y
= 3 cos v , z = 3 s in v
w ith the parameter domajn 0 ~ u ~ 5, ~ ~ v :5
Alternatively, we can regard x and z as parameters. Then parametric equations are x = x, z = z , y = 0
s; x s; 5 and 3 s; z s; 3.
29. Using Equation.s 3, we have the parametrization X
z
.J9=?, where
= e"' sin8,
0
s; x s; 3,
0
=
X,
y = e:r. cos e,
s; 8 s; 27r.
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3; .
324
0
CHAPTER 16
VECTOR CALCULUS
31. (a) Replacing cos u. by sin u. and sin u. by cos u. gives parametric equations
x = (2 + sin v ) sin u., y = (2 +sin v) cos u., z
= u. +cos v. From the graph, it
appears that the direction of the spiral is reversed. We can verify this observation by noting that the projection of the spiral grid curves onto the xyplane, given by
x = (2 + sin v) sin u., y = (2 + sin v) cos u., z = 0, draws a circle in the clockwise direction for each value of v. The original equations, on the other hand, give circular proj ections drawn in the counterclockwise direction. The equation for z is identical in both surfaces, so as z increases, these grid curves spiral up in opposite directions for the two surfaces. (b) Replacing cos u. by cos 2u. and sin u. by sin 2u. gives parametric equations
x = (2 + sin v) cos 2u., y = (2 +sin v) sin 2u., z = u. + cos v. From the graph, it appears that the number of coils in the surface.doubles within the same parametric domain. We can verify this observation by noting that the projection of the spiral grid curves onto the xyplane, given by x = (2 +sin v) cos 2u., y = (2 + sin v) sin 2u, z = 0 (where v is constant), complete circular revolutions for 0 ~
original surface requires 0
~
u.
~ 1r
while the
u. ~ 21r for a complete revolution. Thus, the new
surface winds around twice as fast as the original surface, and since the equation for z is identical in both surfaces, we observe twice as many circular coils in the same zinterval. 2
33. r (u., v) = (u. + v) i + 3u. j + (u.  v) k.
r u = i + 6u.j + k and r , = i  k , so r ,. x r ,
=  6u. i +
2 j  6u k. Since the point (2, 3, 0) corresponds to u. = 1, v = 1, a
normal vector to the surface at (2, 3, 0) is 6 i + 2 j  6 k , and an equation of the tangent plane is 6x + 2y 6z 3x y
=  6 or
+ 3z = 3.
35. r (u.,v)=u.cosv i +u.sinv j +v k
=> r(1, ~)
= (~.4. i)·
r u = cos v i + sin v j and r , = u sin v i + u. cos v j + k , so a normal vector to the surface at the point ( ~ , r u ( 1, f) x r , ( 1, i') = ( ~ i +
4 j) x ( 4 i + ~ j + k) = 4 i 
(t .4,f)is4(x· t)  ~(y4) + 1 (z ~ )=0 37. r (u,v)=u 2 i +2u.sinv j +ucos v k
or
~j +
4 ,~) is
k. Thus an equation of the ta~gent plane at
~x  h +z =~ .
=> r (1,0)=(1,0, 1).
r u = 2u. i + 2sinv j + cosv k and r , = 2u.cosv j  usinv k ,
2
so a normal vector to the surface at the.point (1 , 0, 1) is r u(1, 0) X r ,(1, 0) = (2 i + k ) X (2j) =  2 i + 4 k. Thus an equation of the tangent plane at (1, 0 , 1) is  2(x 1) + O(y 0) + 4(z  1) = 0 or x + 2z
= 1.
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SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS
=
39. The surfaceS is given by z
f(x , y)
2y which intersects the xyplane in the line 3x + 2y
= 6 3x 
I
triangular region given by { (x, y ) 0:::; x:::; 2, 0 :::;
v$
D
= 6, soD is the
3  ~x}. By Formula 9, the surface area of S is I
!fv
A (S)=
= f(x, y) =
A(S) = =
43. z
(~Y + (~Y dA + (2)~ dA = Ji4J.(v dA = v'I4A(D ) = v'I4 (~ · 2 · 3) = 3 y'l4.
J1 + ( 3) 2
= ffv 41 . Here we can write z
1+
1~v
i : i x jy and Dis the disk x
1+
2
= ViA1r
= f (x, y) = ~(x3 1 2 + y 312 ) and D = {(x , y) I 0 $ A(S) = ffv J1 + (..fi) 1
=j
2
A(S ) = ffv Jl
J:::
dx
1
= j f0
~(x + 1)5/ 2]: =
[Hx+ 2)5/2 _ + y2 $
x $ 1, 0 $ y $ 1 }. Then f,
= x 112 , f v = y 112 and
+ (.jfj) 2 dA = f01 J; y'1 + x + ydydx
[tCx + y + 1)312
45. z = J(x, y) = xy with x 2
$ 3, so by Fonnula 9 the area of the surface is
(:~r + (:~r dA = 1~v J1 + (t/ + (~) 2 dA = 41 1~v dA
:lP A(D) = fl ·1r(y'3)
= fo
+ y2
2
1, so f , = y, fu = x
[(x+ 2)
312
+ 1) 312] dx
 (x
Ji(35/2 _ 2s; 2 _
2 s~2 + 1) = A( 31>/2 _
2112 + 1)
=>
+ Y2 + X 2 dA = J;"' J; .fr2TI rdrd(} = J:"' [~ (r2 + I?' 2
J:::
d(}
= J;"' i(2 V2  1) d8 = 2; (2 .j2  1) 47. A parametric representation of the surface is x
A(S) = =
J; J; ) 17 + 4z
2
dxdz
= J01 )
:::; 1: 0 :::; z :::; 1.
A(S) = ffv lr u x r vl dA =
=
J01 J;
.../v 4
( :~y + ( ~~y dA. Then
4I + ,t(ln(2 + v'2I) 
+ 4u2 v 2 + 4u4 dv du =
1
0 $ U u) $ 1
1+
1
ln v'I7]
(v 2 ,  2uv, 2u2 ). Then
= J0 J; (v2 + 2u2 ) dv du = J; (tv 3 + 2u2 vJ ::~ 51. From Equation 9 we have A(S)
JL
17 + 4z 2 dz
%(z ) 17 + 4z2 + ¥ lni2z + v'4z2 + 17 I)] ~ =
(2u, v , 0), r , = (0, u, v), and r u x r ,.
2
= z with 0 :::; x
~ r : = :~ i  j + ¥z k and A (S) =
Note: In general, ify = f(x , z ) then r ,
=
4x + z 2 , z
= (i + 4 j ) X (2z j + k ) = 4 i  j + 2z k.
Hence r :x: x r :
49. r .,
= x, y =
du
J; J; J(v
2
+ 2u2)2 dv du
= J01 (J + 4u2 ) du = [~u + ~u3 ]~ = 4
= ffv J l + (1, ) 2 + Uu) 2 dA. But if Jf, J:::; 1 and If vi :::; 1 then 0 :::;
(f.,? :::; 1,
=> 1 $ 1 + U:~Y + Cfu)2 $ 3 => 1 $ J 1 + (!:,) 2 + Uu) 2 $ v'3. By Property 15.3 .1 1,
ffv1dA $ ffvJl+(f,.)2+(f11 )2 dA $ ffv ../3dA =>
A(D) $ A(S)$../3A( D ) =>
.
1rR2 $ A(S) $ ../31rR2 •
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20 12 Ceognge L~iog. All Ri(!lus Reserved. Mny 001 be scanned, copied, or duplic:noed, or posted lo a publicly oc<essiblc wcbsilc, in whole or in pnrt.
325
326
D
CHAPTER 16
VECTOR CALCULUS
53. z = f( x , y) = e "'
2
2 "
+ y2
with x 2
IIoJ 1 + ( 2xe>= = I:1< I: v'1+ 4r e  r 2
A(S) =
2
2
2
IIo y'1 + 4(x2 + y2)e2(>= +u > dA r dr dB = I:"dB I: r y'1 + 4r2e 2r dr = 27r I: r V1+ 4r2e 2r
2 Y )
2 2
:=:; 4. 2
+ (  2ye"'  v
2
)
2
dA =
2
2
4x
2
2
dr ~ 13.9783
+ 4y 2
1 + (1 +x 2 + y2)4 dydx.
=
Using the Midpoint Rule with f( x, y) 3
A(S) ~
2
2: 2:
1+ (
.
+ 4y2 ) , m 2 1 +x +Y2 4 4x
2
f(xi ,'il; ) ~A = 4 (!(1, 1) + /(1, 3) +
= 3, n = 2 we have
/(3, 1) + / (3, 3) + /(5, 1) + / (5, 3)] ~ 24.2055
i = lj = l
11 6
(b) Using a CAS we have A(S) =
4
4 2 1 + (l x
4
Y:) 4 +x: +y
dy dx
~ .24.2476. This agrees with the estimate in part (a) 
to the first decimal place. 57. z = 1 + 2x
+ 3y + 4y2 , so
Us ing a CAS, we have
o r !!! 8
v'I4 + ll Jn n3 ~+ V'J! 16 0 + 70 .
59. (a) x = a sin u cos v, y
= bsin u sin v, z = ccos u =>
~~ IZfi
(b)
71.
2
z 0
and s ince the ranges of u and v ~ suffic ient to generate the entire graph,
2
'lii! ~
the parametric equations represent an ellipsoid. 2
(c) From the parametric equatio ns (with a = 1, b = 2, and c we calculate r u
= cosucosv i + 2cos usinv j 
r, = s in usin v i + 2sin ucos v j. So r ,. x r,
61. To find the region D : z = x 2 + y 2 implies z
= 6 sin 2 t! cos v i +
+ z 2 = 4z or z 2 
J4 

3sin u k and 3 sin2 'u sin vj + 2 sin u cos u k , and the surface
3z = 0. Thus z = 0 or z = 3 are the planes where the
surfaces intersect. But x 2 + y 2 + z 2 = 4z implies x 2 + y 2 + (z  2) 2 Thus (z  2) 2 = 4  x 2  y 2 or z = 2 +
y
= 3),
0
= 4. so 'z = 3 intersects the upper hemisphere.
x
x 2  y 2 • Therefore D is the region inside the c~cle 2 + y 2 + (3  2) 2 = 4,
© 2012 Ccngogc Lc:uning. All Rights Rcscn·ed. Moy not be scanned, copied, or duplicotcd, or posted too publicly accessible website, in whole or in p:ut.
SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS
that is, D
0
= {(x, y) I x 2 + y 2 :<:; 3}. A(S) =
Jl
.)1 + (( x)(4  :z;2 y2)l/2j2 + [(y)(4 _ x2 _ y2)1/ 2]2 dA
1 ""1,(3 J1+ 4 ~2 r o
o
1 2
2
=
2
rdrd8=
" 1
o
..f:i
o
1
2r dr ~dB= 4 r2 o
2
" [
2 1 2 ) r = ..f:i
2(4  r ) I
d8
r=O
2 2 = Jo f " (2+4)d8 = 28] o" =411" .
63. Let A(Si) be the surface area of that portion of the surface which lies a9ove the plane z = 0. Then A(S) = 2A(Si). Following Example 10, a parametric representation of sl is
X =
asin <,bcosO, y = asin ¢ sin8,
z = a cos ¢ and Jrq. x ro.l = a 2 sin¢. ForD, 0 :<:; <,b :<:; ~and for each fixed ¢, (x ~a) 2 + y 2 5 (~a) 2 or 2 [asin ¢ cos0  ~a] + a2 sin2 <,bsin2 0 :<:; (a/ 2) 2 implies a 2 sin2 ¢  a 2 sin ¢ cos8 :<:; Oor
sin <,b (sin <,b  cos 8)' 5 0. But 0 5 <,b 5 ~ . so cos 0 ~ sin¢ or sin(~ + 0) ~ sin <,b or <,b ~ 5 0 5 ~  <,b. Hence D = { (¢, 0) I 0 5 <,b 5 ~. <,b  ~ 58 5 ~  <,b }. Then
A(Si) =
f 0.,.12J~"'}~l;;j a 2 sin <,b dO d<,b = a 2 f 0" 12 (1r 2¢ ) sin <,b d¢
= a 2 [( 1r cos¢) 2(  ¢ cos¢+ sin¢)]~ 1 2 = a 2 (1r  2) Thus A(S) = 2a2(7r 2). Alternate solution: Working on 81 we could parametrize the portion of the sphere by x
Then
Jrz X r11 l =
Ja2
!!
A(S1) =
0 ~ (x (a/2))2
=
1f/ 2
= 4a
2
(~  1)
a dA = Ja2:z;2y2
2
r= O
a (1  lsin OI) dO = 2a2
 71'/2
J0..12(1 
y, z =
J a2 
:z;2  y2.
a and x2  y2
]r=acoso dO= J" 12
frr/ 2 a(a2  r 2 ) 1 12
= f~~~2 Thus A(S)
+ y2 ~ {a/ 2)2
= x, y =
j "/21acoa8 _ .,./2 o
a ~
r dr dO
a 2 [1 (1 cos2 8?12] dO
sinO) d8 = 2~2 ( ~ 1)
= 2a2 (1r 2).
Notes: (1) Perhaps working in spherical coordinates is the most obvious approach here. However, you must be careful
in setting up D. (2) In the alternate solution, you can avoid having to use Isin 81 by working in the first octant and then
multiplying by 4. However, if you set up S1 as above and arrived at A(S1 ) = a 21r, you now see your error.
©
2012 Ccngngc looming. All Righls R=rv
327
328
CHAPTER 16
0
VECTOR CALCULUS
16.7 Surface Integrals
= 0 and x = 2 have surface area 24 and centers (0, 2, 3), (2, 2, 3)0 The faces in y = 0 and y = 4 have surface area 12 and centers (1, 0, 3), (1, 4, 3), and the faces in z = 0 and z = 6 have area 8 and centers (1, 2, 0),
1. The faces of the box in the planes x
(1, 2, 6)0For each face we take the point P;j to beth~ center of the face and f(x , y, z)
= e ool (x+u+z), so by Definition I,
ffs f(x, y, z) dS ~ [f(O, 2, 3)](24) + [!(2, 2, 3)](24) + [f(l, b, 3)](12) + [f(l, 4, 3)](12) + [f(l , 2, 0)](8) + [f(l, 2, 6)](8)
0 30 We can use the xz and yzplanes to divide H into four patches of equal size, each with surface area equal to ~ the surface
area of a sphere with radius VSO, so D.S
= H4)7r(v'50) 2 = 257l'. Then (±3, ±4, 5) are sample points in the four patches,
and using a Riemann sum as in Definition l , we have
ffH f(x , y , z) dS ~ j(3, 4, 5) D.S + f(3, 4, 5) !:!.S + !( 3, 4, 5) !:!.S + f( 3, 4, 5) D.S
= (7 + 8 + So r(u, v)
= (u + v) i +
9 + 12}(257!')
(u  v) j + (1 + 2u + v) k, 0
r ,. x r 11 = (i + j + 2k)
(i  j + k ) = 3 i+j  2 k
X
~
u
= 9007l' ~ 2827
~ 2, 0 ~
iru
==?
X
v
~ 1 and
0
rvl = y'3 2 + 12 + ( 2)2 =
ff8 (x + y + z) dS = ffv(u + v + u v +1 + 2u + v) lru x r vl dA = .;I4
J; [2u
2
+ uv + u] ::~ dv
=f
1 0
1140 Then by Formula2,
f~(4u + v + 1) /I4dudv 0
= .;I4 J; (2v + 10} dv = .;I4 [v2 +
lOv] ~
= 11 v'l4
7o r (u,v) = (ucosv,usinov,v), 0 ~ u ~ 1, 0 ~ v ~ 7l' and
ru
X
rv = (cosv,sin v, O)
X
('usinv, ucosv, 1)
= (sin v, 
cosv, u ) · =?
!ru X rv! = Vsin2 v + cos 2 v + u 2 = ..)u2 + 1. Then ffsoY dS = ffv (usin v) lru x r vl dA = J; j~" (usinv) = [t(u2 + 1?/ 2 ] : 9. z = 1 + 2x + 3y so
8z
ox =
2 and
8z
oy =
[
cosv]~ = t(2 3 / 2 
0
vu
2
+ 1 dv du =
J; u..Ju
2
+ 1 du J0"
1) 02 = H2v'2  1)
0
30 Then by Fonnula 4,
= /14 J:j J;(x 2y + 2X3?i + 3x2y2) dy dx = /14 J; [~ x2y2 + x3y2 + x2y3]~=~ =
sin v dv
v'l4 J;(10x2 + 4x 3 ) dx = v'l4 [4fx 3 + x 4 ] ~
= 171
dx
v'f.1
110 An equation of the plane through the points (1, 0, 0), (0, 2, 0), and (0, 0, 4) is 4x 2y + z
= 4, so Sis the region in the
I 0 ~ x ~ 1,2x 2 ~ y ~ 0}. Thus by Formula4, 1 x y'(  4) 2 + (2) 2 + 1 dA = ...J2f J; J2°x_2 x dy dx = v'2f f 0 [xy]~~~x 2
plane z = 4  4x + 2y over D = {(x, y)
ffs
X
dS = ffv =
v'2f J01 ( 2x2 + 2x) dx = v'2f [ix3 + x 2 ]~ = v'2f ( i + 1) =
dx
4J.
© 2012 Crngnge Learning. All Rights Reserved. Mo.y not be scnnned. copjcd, or dupJicarcd. or posted to a publicly accessibh; website, in whole or in p.1rt.
SECTION 16.7
13. S is the portion of the cone z 2
= x2 +
SURFACE INTEGRALS
y 2 for 1 $ z $ 3, or equivalently, Sis the part of the surface z
0
329
= ...jx2 + y2 over the
region D = { (x, y) Jl $ x 2 + y 2 $ 9 }. Thus r~~
2 · Jfs x2z2dS= Jfv x (x +y ) 2
2
(~Y +(~Y + l dA
15. Using x and z as parameters, we have r (x, z) = x i+ (x 2 + z 2 )j + z k 1 x 2 + z 2 :::; 4. Then r x x r : = (i + 2xj )
ffs ydS =
X
JJ
(2zj + k ) = 2x i  j +2z kand Jr,
x2 +z2~ ~
(x 2 + z 2))1 + 4(x 2 + z 2) dA = ·
= 1+
[let u = 27T J:7 t(u l )JU. id·u =
X
r , J = J4x2 + 1 + 4z2 = ) 1 +4(x2 +z2). Thus
J;,. J; r 2 J 1 + 4r 2rdrdO = J;,. dB J; r 2 v'1 + 4r2 r dr .
47'
2 =>
7'
2
=
t(u 1) and idu =
1' dr]
ft1T f t17 (u3/2 ...:. ul/2) du
= ...!..7T (1u5/2  1u3/2)17 = ...!..7T(1(17)5 /2 1(17? / 2  1 + 1) = 2:.. (391 Jl7 + 1) 16.
5
3
1
16
:;
3
5
3
60
17. Using spherical coordinates and Example 16.6.1 0 we have r (¢,B) = 2 sin 4> cos Bi + 2 sin 4> sin Bj + 2 cos 4> k and
Jrq, x
r~l =
4 sin¢. Then
ffs(x 2 z +
2
2
y z) dS = .{0 "
J; 12 (4sin2 1p)(2 cos¢)(4 sin 4>) d¢ dB =
19. S is given by r (u,v) = ui + co~vj + sinv k , 0 $ u $ 3, 0 $ v $
ru
X
rv = i
X
( sin v j +cosv k ) =  cosv j sin v kand Jr,. .[[5 (z
+ x 2 y) dS
X
1Tj2.
167T sin4 ¢] ~ 12 = 167T.
Then
rvl = ) cos2 v +sin2 v = 1, so
= J; 12 J;(sinv + u 2 co~v)(l) dudv = f~.,. 12 (3sin v + 9 cos v) dv = [ 3cosv + 9sin v]~/ = 0 + 9 + 30 = 12 2
21. From Exercise 5, r (u, v) = (u + v) i + (u  ·u) j + (1 + 2u + v) k , 0 $ u $ 2, 0 $ v $ 1, and r u x r ., = 3 i + j  2 k.
Then F (r (u, v)) = (1 + 2u + v)e
= (1 +
2
2u + v)e"
2 "
2 112 i  3(1 + 2u + v)e"  j
+ (u 2 
+ (u + v)(u v) k v2 ) k
Because the zcomponent ofru x r v is negative we use  (r ,. x r v) in Formula 9 for the upward orientation:
ffs F · dS = J/~ F · ( (r .. =
./~1 f 02 2(u2 
v
2
X
)
J; J; [3(1 + 2u + v)e" ~ + 3(1 + 2u + v) e"' dudv = 2 .{ [tu uv J:::~ dv = 2 J; (~ 2v 2) dv 2
r u)) dA =
= 2 [~v  ~v3 ] ~ = 2
1 0
(! 
j) =
3

2
2
2  "
+ 2(u2  v 2))
d·1~ dv
2
4
= 4  x 2  y 2, and Dis the square [0, 1) x 1 yz(  2y) + zx] dA = .{0 J; [2x 2 y + 2y2 (4 x2 
23. F (x, y, z) = xy i + yz j + zx k, z = g(x, y)
Jfs F · dS = ff0 [xy( 2.7J) = J; (t x2 + 1} x  x3 + ~~) dx = i~~
[0, 1], so by Equation 10
y 2 ~ + ~(4  x2

y 2 )] dy dx
@) 2012 Ccngnge Lc::r.ming.. AH Rights Reserved. May nol be scanned, copied. or duplicated. or poslc:d to :l publicly access ible wdJsitc. in whole or in part.
330
0
CHAPTER 16 VECTOR CALCULUS
25. F(x, y, z) =xi  z j + y k , z = g(x, y) = .j4 x 2
{ (x , y) I0 ~ x ~ 2, 0 ~ y ~ v'4  x 2
=
8 has downward orientation, so by Fonnula 10,
}.
!1( v'
x2 4  x2  y2
D
y 2 and Dis the quarter disk

 v'4 
x2

=  IIv x 2 (4 (x 2 +y 2 )) 112 dA = 
.
y
)
v'4  x2  y2 + y dA I 0..,.12 I 02 (r cos B?(4 r 2 ) 112 rdr dB y2
·
2
·=  .[0..,.1 cos2 Bd8 I~ r 3 (4 r· 2 ) 1 12 dr
[let u = 4 r 2
= 
I0..,. 12 (~ + t cos 28) d8 I~ ~(4  u)(u)  112 du
= 
[~8 + t sin28]~12 (~) [8v'U 
=?
r 2 = 4  u and~ du
= rdr]
iu312]: = t(~)(16 + ¥) = ~1!'
27. Let 8 1 be the paraboloid y = x 2 + z 2 , 0 ~ y ~ 1 and 82 the disk x 2 + z 2 ~ 1, y
= 1. Since 8 is a closed
surface, we use the outward orientation. On S 1 : F (r (x, z)) = (x2 + z 2 ) j  z k and r , x r = = 2x i  j + 2z k (si~ce the jcomponent must be negative on S 1 ). Then
fis 1 F · dS =
o;2
II [(x 2 .+ z2 )  2z2 ] dA = I~..,. I;(r2 + 2r2 sin2 8) r dr d8 +=2 $1
= I~..,. I; r 3 (1 + 2sin2 8) dr d() =  ]~..,. (1 + 1 cos 28) dBI 0 r 3 dr 1
=  [28
t sin 28]~..,.
[~r·4 ]~
=  41!' . ~ =
OnS2 : F(r(x, z))= j z kandr.., x r , =j. Thenifs F · dS= 2
.,2
JI
+ ;:2 $
1r (1)dA=1r.
1
Hence'Jis F · dS =  1r + 1r = 0. 29. Here 8 consists of the six faces of the cube as labeled in the figure. On 81: F = i + 2yj + 3z k, rv x rz = i and IIs 1 F · dS = I~ 1 }~ 1 dydz = 4; 82: F =xi + 2j + 3z k, r= x r, = j and IIs 2 F · dS = I~ 1 I~ 1 2dxdz = 8;
8s: F = x i +2yj + 3k, r , x r v ·= kand fis3 F · dS = I~ 1 I~ 1 3dxdy = 12;
8 4 : F = i +2y j +3zk,r: x r v = i and IIs 4 F · dS = 4; 85: F = x i 2j+3zk, r., x r: ~  j andiJ~ 5 F · dS = 8; 86: F =x i+ 2yj 3k, r 11 x r ., = k and J/~0 F · dS = J~ 1 j~ 1 3dxdy = 12. 6
Hence ffs F · dS =
I: Jfs. F i=l
· dS = 48.
1.
31. HereS consists of four surfaces: S 1 , the top surface (a portion ofth~ circular cylinder y 2 + z 2 = 1); S2, the bottom surface (a portion of the xyplane); Ss, the front halfdisk in the plane X= 2, and
s 4.
the back halfdisk in the plane X = 0.
May not be sconned. copied, or uupliCAU:d, or posted.to a publicly accessible website, in whole or in p:>ll .
SECTION 16.7
On 8 1 : The surface is z = J1  y 2 for 0 S x
!"ls1f
1~o !
F . dS =
1
1
0
331
S 2,  1 S y S 1 with upward orientation, so
[x (0)  y h) + z dy dx = 1 2
2
2
2
SURFACE INTEGRALS
( 
1 y
2
]
o
·
.
1
/
(  1
h
1y2
+ 1 .
2
y ) dy dx
On 8 2 : The surface is z = 0 with downward orientation, so
On
8 3 : The surface is x
= 2 for  1 S y S
1, O.S z S yf1 y 2 , oriented in the positive xdirection. Regarding y and z as
parameters, we have r 11 x r.., = i and
On 8 4 : The surface is x = 0 for 1 :::; y :::; 1, 0 $ z $ yf1  y 2 , oriented in the negative xdirection. Regarding y and z as parameters, we use 
(r 11 x r z) = i and
ff54 F · dS = f~ 1 fo~ x2 dzdy = f~ 1 fo~ (0) dz dy = 0 Thus 33. z
ffs F
· dS =
= xeY =>
1+ 0 + 211" + 0 =
211" + ~·
8zj8x = e11 , 8zj8y = xe11 , so by Formula 4, a CAS gives
JJ5 (x 2 +y2 +z2 )d8 = J01 f 01 (x2 +y2 +x2 e211 )Je2 Y +x2 e2v + 1dxdy ~ 4.5822. 35. We use Formula 4 with z = 3 2x 2
3 2x2

y2
~ 0 are

y2
/i $ s ji X
=> 8zj8x
and
J3 
= · 4x, 8z j8y = 2y. The boundaries of the region
2x2 $ y $
J3 
2.7:2 • so we use a CAS (with precision reduced to
seven or fewer digits; otherwise the calculation may take a long time) to calculate
37. If 8 is given byy n
= h(x,z), then 8
is also the level surface f(x,y,z)
Vf(x,y,z))I =  h., i+jh: k , and = I~/( v x,y,z y'h~+l+h~
n
=y
h(x,z)
= 0.
. h · 1h · h 1ft N · IS t e umt norma t at pomts tot e e . ow we proceed as m the
derivation of (1 0), using Formula 4 to evaluate
where D is the projection of 8 onto the xzplane. Therefore
©
IIs
F · dS
=
Ii ( ~~
2012 Ccngnge Lenming. All Righ1S R="..1. Moy 001 be scanned, copied, or duplicolcd. or posted 10
P
•
 Q +R
~~) dA.
publicly oeccssible websilc, in whole or in port.
332
39.
0
CHAPTER 16
VECTOR CALCULUS
m =· ffs }( dS = }( · 47r(~a 2 ) = 21ra 2 K; by symmetry M xz = lvf11z = 0, and M:cv
= Jj~ z }( dS = }( f~" f 0 12 (a cos ¢)(a2 sin ¢) d¢ d() = 2711
Hence (x,y,z)
= (0,0, ~ a). I
(b) 1::
= ffs(x
2
+y
2
)(10 Jx
2
+y2 )
dS
=
.
1
2
ff
s :r2 + v2 s 16
43. The rate of flow through the cylinder is the flux JJ~ pv · n dS
r (u,v)
= 2cosu i +
(x + y
2
)(10
jx +y2 ) ../2dA 2
•
= ffs pv · dS . We use the parametric representation
2 sinuj + v k for S, where 0 $ u $ 27r, 0 $ v $ 1, so r ,.
=
2sinu i + 2cosuj, r v = k , and the .
outward orientation is given by ru x r v = 2 cos u i + 2 sin u j . Then
ffs pv · dS
f01 (vi+ 4sin2 u j + 4 cos2 u k) · (2cosu i+ 2sinu j ) dvdu = p f02" . J: (2v casu+ 8sin3 u) dv du = p J~" (casu+ 8 sin3 u) du = p[sin u + 8(i)(2 + sin2 u) casu] ~" = 0 kg/s
= p ]~"
45. S consists of the hemisphere 8 1 given by z = ja2

x2
y 2 and the disk 8 2 given by 0 $ x 2 + y 2 $ a 2 , z = 0.

On 81: E = asin¢ cosB i +a sin ¢ sinBj + 2acos¢ k ,
T q, x To = a 2 sin 2 ¢ cosB i +a2 sin2 sinl'.lj +a2 sin ¢ costpk. Thus
ffs 1 E · dS =
.r;" f0"1\a3sin
= J:" J~rf On S2: E =X i + y j , and r y
3
2
q) + 2a3sin <J; cos2 ¢) dq) dB
(a3 sin ¢+ a 3 sin ¢ cos2
.
= (27r)a3 (1 + .i)
3
= ~1ra
X r x = k so Jfs2 E . dS = 0. Hence the total charge is q = co ffs E . dS = ~7ra 3 co.
47. IC9u = 6.5(4yj + 4z k) . S is given by r (x, 1'.1)
= x i + y'6 cos(} j + v'6 sin 8 k and since we want the inward heat flow, we
=  v'6 cos 0 j  v'6 sin(} k. Then the rate of heat flow inward is given by ffs (K \i'u) · dS = f 02"' f04  (6.5)(  24) dx d() = (27r)(l56)(4) = 12487r.
use r x x ro
49. LetS be a sphere of radius a centered at the origin. Then ir l =~and F (r ) = cr / lrl 3 parametric representation for S is r( ¢, 0) = a sin¢ cos 0 i +a sin¢ sin 0 j r .p = a CDS by r 4> x ro =
= (c/a3 )
+ a cos
(x i + y j + z k). A
$ 1r, 0 $ 0 $ 27r. Then
+ a cos
CDS 0
i
ffs F · dS =
g
Jt'~ c3 (asin
+ acos¢k)
·. (a2 sin2
sin8 j +a2 sin k) dOd¢ = :3
f0"' .{02 " a 3 ( sin3 ¢J + sin cos2 ¢) dOd¢ =
c .{0"
f021r sin