Student Solutions Manual

~

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 1-1- 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.

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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; ) .

'·

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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.,.. a--a 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=

(x-1) 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

¢?

rsin8-3rcosB = 1

r s in B = 2

¢?

¢?

sin8

¢?

r(sin8-3cos8)=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(r-2ccos8) = 0

¢?

r = Oorr=2ccos8.

= 2c cos 8 when 8 = ~ + n1r, so the curve is represented by the single equation r

~

2c cors ().

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~

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

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SECTION 10.3

39. r

POLAR COORDINATES

= 1--2sin8 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

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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 r-values,) 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 c-valu~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.

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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)+_('-:-v-2 -:::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 intei-val 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 =>

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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

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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

{2-rr

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('

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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(1-cos89)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

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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/2-7r/6) =4\1'3-411'/3

27. 3cosB=1+cosB {::} cosB=~

=> B=t or-i.

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 0-values 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).

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=}

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 2a-by-2b 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 5-3=a(I-2) 2

'*

a =2,soanequationisy -3 =2(x-2? .

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 =64-25=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) + (y-2 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 =36-9x 2

1

~

36 for y to get an equation for the top half of the ell ipse:

y 2 = ~(4-x 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'4-x 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.

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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.

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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 right-hand focus moves to the right while its left-hand 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[d-rcos(7r-9)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 ®

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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 ·

TRUE-FALSEQUIZ 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.

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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),0-0 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 three-leaved 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 t-value 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- t-3)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 x-intercept of the curve is always -c.

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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(.2-t3 )=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 d0-2}0 2

-.! ("/ 2

2d0-2}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

=-

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=>

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/Z-dt ~ [ 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 x-values, 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 y-axis, 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 + 1-2 sin2 9

[but sin B # - 1 for 0 $ 9 $

H

=?

If sin 9 = ~, then B =

=>

j~ = -(2sin9-1)(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 =>

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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 y-va 1Ue IS - b-

=-

a

ad '-t-be

= - -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 --

be-ad) (ad+be + -ad-b 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 2n-1

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

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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

n-co

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

n-too

n-oo

""

(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 = e1imn-oo( l /nl = e0 = 1. Converges

,.,_ex:>

n-oo

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 n-oo 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" - e-n -+ oo. Converges

2x 11 2 H = lim - = lim - = 0, it follows from Theorem 3 that lim an= 0 . Converges X-tCO eX X--tOO e:t n-oo

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/ = t-o+ 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

!t-oo

( 2)"' = 1+ -

X

.

lim e 1n Y = e 2 , so by Theorem 3, lim

:r-oo

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,.

= 1.08Pn-t -

= 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

= :c-oo lim -=..._ :lb ~ ( r-:Z: X-+00

-

In\

T T-%

= lim ___!__I ."' :::: 0, so lim nr" %-+00 nT n-oo

~

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 rn-Fn 2 y 2, y. I 2 y rn-Fn 2 v 2, .. . } , a1· -_ 2 112 , a2 -_ 2 3/4 , aa -_ 2 7/B , ... , so an -_ 2 (2"-1)/2"

n-oo

=

= 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 - - > 3-an+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 an-1 pairs already present. Thus, an= qn- 1 + a n-2, so that {an}= {fn}, the Fibonacci sequence. ( b) an= f,.+ 1

f ,.

fn fn-1 + fn - 2 an-1 = - - = fn - 1 fn- 1

then L = . lim an- 1 and L n-+oo

= 1 + .f,.- 2 = 1 + f n-1

= nlim a,.- 2, soL must satisfy -+oo ·

1

L = 1 + -L

1

1

/ ·= 1 + --. If L = lim an, f n- 1 fn-2 an-2 n-oo

=> 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 n-oo

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 = a-Ha+ 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

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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. n-oo

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 = n-oo 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

=

= n--too 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}.

TI-86 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 n-oo

n-oo

(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)n-l 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· 1-0.9 0.1 23.

f: (~ 3~n-l 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

n-oo

lim -n - 1 = -1 r = n-oo 3n - 1 3

....£

0.

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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

= n-oo lim o/2 = lim 21 / n = 2° = 1 =/= 0. n-oo

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 n-3

. 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 -1-1- -. 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) = n-oo n -oo

(1 + 21 + 31 -

)

[using partial fractions). The latter sum is

t+3

\

= 6 n= l

n

+!2 - -n-1 -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 n-2 n

(-1 - l) + (-1 - _1 ) + (-1 - _1 ) + (.!. - --4) n-3

n

n -2

1 n+1 1 1 + 21 + 31

- +1 n

1

- + 2 1l

-

n + 1

1

n+ 2 -

n-1

1

n+3

[e t l . senes ' ] escopmg

1 ) = 1 + 21 + 31 = 11 . 6

..,----+ n 3

n+2

converges

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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) )

n-oo

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 = -X-2, 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,. n-oo

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 150-mg 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 + 1-1+· · · . 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

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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~r-l. 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

n-oo

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

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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 is-convergent, 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 p-series with p = -/2 > 1, so it converges by (1 ). . .

n=1

1 Thi . 1 1 'th p = 3 .. ++ ···= ~ s tsap-seneswL 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

ft-2 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 p-senes . w1'th p 3!2

n

E ~n is a constant multiple of a convergent p-series with

p

3

= 2

> 1..

= 2 > 1, so it converges. The sum of two

n= l

convergent series_is-convergent, 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 11n-diverges. 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

t-oo

1

1

·

X

t --t-oo

.

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 + 1-ln2] =0 + 0 + 1-ln 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)1-p] 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 p-series 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: n-1.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 x-l.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 p-series, 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 p-series 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.

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SECTION 11.4

n +1 5. -

> - n-

nVn

7.

= -

1

Jn

nVn

~ n for all n 2: 1, soLn=l

p-series 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 p-series 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 p-series 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 ap-series 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 n-oo Jn 2 + 1

n-oo

=

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 !._ · n-oo 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

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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 n-oo

n

n-oo

f: e-n = 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! - 2n-l

[lr I = ~

1], so

<

an

+ ~) n

2

= 1

> 0. Since

2

e-n 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 p-series 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 p-senes

.

(n 3 12 Jn+2)l(n 3 12 n--.oo (2n2 + n + 1)ln 2 .

= lim

is a convergent p-series 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 +-+- + .. · + -cos5-1010 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

x-oo

1

1 X

= lim x = oo, x-oo

. 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

x-oo

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

[p-serics 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 n-oo

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.) ©

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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 n-oo n-oo n + 4 n-too 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 n-oo

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

n-op

_

+ 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 !

n-ioo

= 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" =

"' n-7;: 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)n-l 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~n-l

= 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.

® 2012 Ccngnge Learning. All Rights Reserved. May not be scanned, copied, or dupJicatcd, or posted to a publicly actcss ible website, in whole or in part.

CHAPTER 11

INFINITE SEQUENCES AND SERIES

68

D

35.

2: b2 n = 2: 1/(2n)2 clearly converges (by comparison with ~e p-series for p = 2). So suppose that 2: (-1r-I bn .

.

converges. Then by Th~orem 11.2.8(ii), so does 2: [(-1

t-. 1 bn + bn] ::: 2( 1 + l + i + · · · ) = 22: -2n 1-1 . 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 n-oo

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 n-too

an

.

.

1 5 1 1 1 lan+l l = n-oo lim In+ · " 1 = lim 1.!. · n+ 1 =.!.lim + /n = .!. (1) = .!. < 1,sotheseries E ~is an 5n+l n n-too 5 n 5 n-too 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-+: ) k-oo 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 p-series [p = 3 3 n

00

e l /n n3

> 1], L: n= l

converges, and so

is absolutely convergent.

® 2012 Cengagc: Learning. All JUghts Reserved. May not be scan~. copied. or duplicated, or posled 10 a publicly accessible website, in whole or in p3l1.

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 n-oo

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 . I-an+ll . I(n + 1) C!1QO"+l · n! . I- lim . 100 1) hm - - Inn - (n -+10 a,. -n- co (n + 1)! n 0100" .-n-oo n+ 1 n -

n-oo

. 100 lim - ( 1 + -1 )100 n-co 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)! (2n-1)! I lim l an+ll = lim ~ ( - 1)"~1·3·5 .... ·(2n -1)[2(n +1)-l]_ a,. n-oo (2(n+1) - 1]! (- 1)"- 1 ·1·3·5·····(2n-1)

n - oo

=lim ~ (-1)(2n+l)(2n- 1)! 1 = lim _.!_=0 < 1, n -oo (2n + 1)(2n)(2n - 1)! n-oo 2n.. so the given series is ahsolutely convergent and therefore convergent.

® 2012 Ccngagc Lc.lm..ing. All Rights Resen·cd. May not be scanned. copiL-d. 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 n-oo

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

n--400

an

n-oo

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 x--1 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

n-too

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

(-1t-1

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

p-series 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 n-oo n + 2 n--<00 n -
5. lim Jan+II = lim n-oo 00

an

.

1 2 I(n -5)n+l + ?

n

n-
•

(-

(

.!)

5)" I = lim 2(n+ 1) 2 = ~ (1 + n - oo 5n2 5 n-oo n

llm

n 2 2 n-l

2

=

~(1) = ~ < 5 · 5

1, so the series

n2 2n-l

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 e-k

=

lim t .... oo

X

d~ =

lim [2J!nX] t = lim ·(2 villt- 2

y lnx

2

t_.oo

t-oo

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 p-series 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 n-oo

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 2k-13k+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.

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SECTION 11.7 STRATEGY FOR TESTING SERIES

17

lim

· n-oo

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 6-1 vn

Alternating Series Test. 21 . lim lan. l = lim IC- l )ncos(l/n2 )1= lim icos(1 /n n-oo

2

).1=cos O=

n -'oo

n-oo

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

n-oo 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"L-H>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

k-oo

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 n-too

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 n-tOO

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

n-H)O

n

n-00

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 n-oo 2

lim J (n + 1)2 x"+l . 2 2n+l n x"·

n-CXJ

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

I-a.,- 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 p-series

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 1-n1n ' 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)" l-n 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

n-too

= [-

¥, - ¥).

= 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 H-tOO

'U-tOO

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 ~~~(X2-x -) )"+ 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 p-series (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- 71--73--:::-5-· - -_-;(72n---,1")"(2::-n---cf--::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 n-oo 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. S2n-l = 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

= S2n-1 + 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 S2n-1 both 1

. . 1 + 2x oo. Thus, the mterval of convergence IS ( - 1, 1) and f (x) = - - 2 . 1- x

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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

n--too

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 1-xl < 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

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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) =

-~~ = 5-x

Putting x = 0, we get C 1 17. We know that 1 + 4X

= ln 5.

=

1- - 4X

f(x) = {1+x4x)2 = · -4x. for J-4xJ

< 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 l-x2/ 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

10-9 . 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 10-8 . 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(2n-1)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

Xn-1

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

n-oo

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 convergentp-series (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

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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: - -1-x" = 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 ) =

n-oo

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

n-oo

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(x )

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 n-oo

I a,. I= I (n + = (In = < a ..+!

2 lim ) !xi n-oo n -1-1

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 ?) (x- 2)" n.

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

c:E> :!012 Cengnge Lc:unin~. All Rights Rcscn1:d. ~ tay not be SCnMCd. copk-d. or duplicated, or postct.l loa publicly ncccs.~iblc wcbsilt:, in whole or in part.

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)

lx-21

":~ n + 1 - 2-

2.

n. ·

0

2e6

4e 6

+ l ! (x- a? +. 2! (x 8e6

4e0

l(- l)(x-2)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-

n-oo

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

{

(x-7r)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.

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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 1-xl < 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) = ·c-3)( - 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

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SECTION 11.10

TAYLOR AND MACLAURIN SERIES

D

87

1.5

- 1.5 t--r.'-----+---~ .. :--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 ·· .. · (2n-1)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.

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X

w-G

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 H-4) (o.4) 9 H-t)(-~) (o.4) 13 H-4)(-~)(-~) (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 x-0

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

x-o

sinx- x

+ l6 x 3

x5

since power series are continuous functions. .

_59. From Equation 1 I, we have e-x 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 th-degree 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.

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=>

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(k-1)(k-2) .. ·(k-n+1)(k-n) 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)···(k-n+ 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 (x)

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?

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D

91

I

92

0

INFINITE SEQUENCES AND SERIES

CHAPTER 11

9.

J(o)

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(x)

f {n)(4)

0

Vx

2

1

4

2

lx- 1/2 2 - l x - 3/2

3

1x-5/2

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) -·~(x-I?+ 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 (fa-t

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.)

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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=x-kx

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).

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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 first-degree 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

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= 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

p-se~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 n-oo

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.

TRUE-FALSE 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

n-oo

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)n-l = __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

.:c-oo

n---.oo

~ lim ~ ( -~) - 1/ (4x2)

. IliD

x--+oo

an

< 2. Note first that 1 < a2

fo r n = 2. Now assume that ak-1 < 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,

= n-oo lim an+l =>

=> L = 2. 00 n ·· 1 --g--- converges by the Comparison Test with the convergent p-series 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 n-oo 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=_:
= t-oo 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)n-1 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 p-series 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 n-Hxl

= 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)n-l (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

n-oo

= lim [(tan- 1 2- tan- 1 1) + (tan- 1 3- tan- 1 2) n-oo

= lim

n-oo

[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 n-oo

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,.

n-oo

soR = ~-

I:

n=O y

../n + 4

n-oo

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 n-oo

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) 4-x

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

xn-1

1 = - + X

00

xn-1

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

- ±x-3/2

3

l!x- 5/2·

4

-wx - 7/2

©

xn-1

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

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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 1-tan- 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

n-too

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"

n-l = 4 3

(4)n-l 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 one-n 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 pos1l-d 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 =

Sn-1

4n-1

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) tana-tanb tan(arctan x)-tan(arctan y) x-y 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))

x-y

= 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 by-yin 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 1-5. 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 -n-1 - - In - n - ) fork ~ 2. Then n n+ 1

3) + · · · + (In - -1 - In -k+1- = . (In-12 -In -+k-1 ) = 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

k-oo

13. (a)

k

1 - In - k - so In2 k+ 1 '

·I

ln 1 -In 2 - In 1 = - In 2.

The x-intercepts 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- (n-l)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- (n-1)"/

5 -

e-n"/ 5 ] =

;g;

2

[telescoping sum].

n=l

Another method: If the volume in part (a) has been written as Vn =

2

~g; e-n"/ 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+ 3-1

L = v'3r+r+(n-2)(2r)+r+v'3r=r(2n-2+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+../3-1) 2../3

n(n + 1) 1r A = (n+../3-1) 2 2.;3 1 + 1/n =

=>

An

1r

1r

-> - - asn ->oo

2

[1+(J3-1)/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 term-by-term:

= (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

t-o+ 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)"] t-o+ n

.

t x"(Inxf-

o- _ n _

2r

~-----L------~

1

n+ 1 } 0

+1

1

t

lim ( ___!!:__ x"(1nx f- 1 dx

t-o+

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

= tan-1

(

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

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CHAPTER 11

21. Let f (x ) denote the left-hand 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 ft-hand 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 Three-Dimensional Coordinate Systems

1. We start at the origin, which has coordinates (0, 0,. 0). First we move 4 units along the positive x-axis, affecting only the x-coordinate, bringing us to the point ( 4, 0, 0). We then move 3 units straight downward, in the negative z-direction. Thus only the z-coordinate is affected, and we arrive at {4, 0, - 3). 3. The distance from a point to the yz-plane is the absolute value of the x-coordinate of the point. G(2, 4, 6) has the x -coordinate with the smallest absolute value, soC is the point closest to the ·y z-plane. A( - 4, 0, -1) must lie in the xz-plane since the distance from A to the xz-plane, given by the y-coordinate of A, is 0. 5. The equation x

+ y = 2 represents the set of all points in y=2 - x

JR3 whose x- andy-coordinates 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 xy-plane 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 yz-plane 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

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SECTION 12.1 THREE-DIMENSIONAL 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 xy-plan.e , its radius is the distance from its center, (2, -3, 6), to the xy-plane, 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 yz-plane, 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 xz-plane, 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 yz-plane and 5 units in front of it.

25. The inequality y

< 8 represents a half-space 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 xy-plane) 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 z-axis 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 they-axis 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 y-axis. 35. Thi s describes all points whose x-coordinate 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 .

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114

0

CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE

39. (a) To find the x- andy-coordinates of the point P , we project it onto £2 I

and project the resulting point Q onto the x- and y-axes. To find the z-coordinate, we project P onto either the xz-plane or the yz-plane (using our knowledge of its x - or y-coordinate) and then project the resulting point onto the z-axis. (Or, we could draw a line parallel to

QO from P to the z-axis.) The coordinates of Pare (2, 1, 4). (b) A is the intersection of £ 1 and £ 2, B is directly below the y-intercept of L2, and 0 is directly above the x-intercept 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 u-w-v

--->

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=b-a

= b - ( ~ b - ~a)

=

d=!{b-a) = 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 l-3 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 x-component ofv is v1

= Ivi cos(1r/ 3) = 4 ·. ~ =

V2

= Ivi sin(1r / 3) = 4 · .,;; = 2 v'3. Thus

y

2 and they-component 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

· · x-ax1·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 y-direction, 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 y-axis and the line x = 3, and the current flows in the negative y-direction. The boatman wants to reach the point (3, 2). Let() be the angle, measured from the positive y-axis, 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'C--819)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.

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1-19

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 + 6-2 = 0. Thus QP and QRare orthogonal, so the angle of the triangle at vertex Q is a right angle.

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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 cos-y = ~· 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 cos-y = 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 · 2i-J2T

(

•

•

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 ft-lb.

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 x-liXis [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- C-H 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 - 9-6 = 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 right-hand 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 = ( -1-6) 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

I-13

11 · j + I 3

0

- 1

21 k= - i-j +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=l-16kl = 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

1-12 111 = 1(4-2)-2(- 4-4) + 3(-1-2) = 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 Rescrn-d. 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)= 3-1

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 z-axis, yet the x- andy-axes 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 yz-planes are not parallel, yet they are both perpendicular to the xz-plane. (e) False; the x- and y-axes 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 x-axis.

(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.

® 2012 CcngJOgc Lc:uning. All RightS Rcsc:"·cd. Moy not be SCllO!lcd. copiL-d, or duplicated, or 1""1cd ton publicly accessible wcbshc, in whole or in prut.

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 xy-plane when z = 0, so we need

Y

~ 5 = 2 =>

we need x

=0

y

=>

x- 1 0-6 x- 1 - = --=3 or -=1' = -=1' = -y+5 2

= -1. Thus the point of intersection with the xy-plane is (1

the x z-plane, we need y

= y ~ 5 = z ~36 =0

=>

=> y

=-

3, z

2

=>

x

=-

1,

1, - 1, 0). S imilarly for the y z-plane,

= 3. Thus the line intersects the y z-plane at {0, - 3, 3). For

=>

x

= - ~, z = - ~ . So the line intersects the xz-plane

([) 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(x-1) -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, 8-18,6 -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 x-intercept we set y and obtain 2x = 10 =>

---

=z =

0 in the equation 2x + 5y + z = 10

x = 5 so the x-intercept is (5, 0, 0). When

x = z = 0 we get 5y = 10 => y = 2, so they-intercept is (0, 2, 0). Setting x = y = 0 gives z = 10, so the z-intercept 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 xy-plane; 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

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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 · 0-2·0+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 z-axis.

(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 x-axis.

Thus, the rulings of the cylinder are parallel to the x-ax.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 they-axis.

cylinder with rulings parallel to the x-axis.

.1'

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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 z-axis.

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 xy-plane, 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 y-axis. 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 y-direction.

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

x-direction. We recognize the graph as an elliptic parabofoid with axis the x-axis and vertex the origin.

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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 x-axis 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 x-intercepts ±1, y-intercepts ±i 0

-

and z-intercepts ±~. So the major axis is the x-axis 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 y-axis, hence the correct graph is II.

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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 .

xz-plane. 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 they-axis. 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 xz-plane 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 y-axis.

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 z-range 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 y-axis and opens to the right. Thus the trace in the yz-plane 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 z-axis at z = ± 6356.523 and the 'e quator intersects the x- andy-axes 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 xy-plane 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 real-valued 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 non-parallel 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 yz-plane) by setting x = k and determining the curve represented by the resulting equation. Traces in the planes y = k (parallel to the xz-plane) and z = k {parallel to the xy-plane) are found similarly.

19. See Table I in Section 12.6.

TRUE-FALSE 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 = 1-v 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 three-dimensional 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 x-coord 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 yz-plane 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 right-hand 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.

l-2- (-24)1 - 22 . - ~· + I2 + (-4)2 v26

y'32

= z represents a plane perpendicular to

the xz-plane and intersecting the xz-plane 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 y-ax.is. For

4x 2

IYI > 2, traces parallel to the xz-plane 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 x-axis. 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 three-dimensional 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 three-dimensional 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 cross-section 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 x-axis. (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)

J-L•• 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/1-s

=?

'*

sin fJ = mg sin fJ - mgJJ-. cos(}

- JJ-.• cos'(} ) 1tmin = mg (sin8 cos (}+ 1-t. 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 - /1-s ( hmax sin 8 + mg cos B) = mg sin 8 hmax

'*

cos(} - hmn.x/1-s sin(} = mg sin(} + mgJJ-. cos

e '*

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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

(e-st 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 - e-2 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 - e-2 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 xy-plane is y = x

2

,

y

I

> 0, a half parabola.

I

I I

Onto the xz-plane z

= x 3 , z > 0, a half cubic, and the yz-plane, y 3 = z 2 •

,I

,,,-

,'

y= xl

,,' X

15. Tbe projection of the curve onto the xy-plane is given by r (t) = (t , sin t, 0} [we use 0 for the z-component] whose graph

= 0. Similarly, the projection onto the xz-plane is r (t) = (t, 0, 2 cost), whose graph is the cosine wave z = 2 cos x, y = 0, and the projection onto the yz-plane is r (t) = (0, sin t, 2 cost) whose graph is the ellipse is the curve y =sin x, z

2

y

- I

-2

xy-plane

xz-plane

yz-plane

From the projection onto the yz-plane 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 x-direction, 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 they-axis. 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 they-axis. 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 z-axis. 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 xy-plane 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 two-dimensional 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 xy-plane 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.

xy-plane 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 real-valued 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 t-a. \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 real-valued 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)] t-a

(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) t-a

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 t-a \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)) t-a

t-a

t- a

=

t--+a

t.-a

t-+a

t-+o.

t-ta.

([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), t-a.

b = lim r (t) t-+a

t-+a

~ ( t-a 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 t-a t -+ a t-a t -+a

and lim h(t) = b3 • But these are lim its of real-valued functions, so by the definition of limits, for every c t-Hl

> 0 there exists

0, 02 > 0, oa > 0 so that ifO < It - a l < 01 then lf(t) - b1 l < c/3, ifO < it- a l < 02 then lg(t) - b2l < c/ 3, and ifO < It- al

<·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 \ t-a t - •a t- a

t-+c&

®

~0 1 2

t- a

r (t) = (b1, ~. ba) = b.

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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

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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, e-t), so

r' (t)

= (e-t(- sin t) +(cost)( -e-t), e-t cost+ (sin t)( -e-t), (-e- t)) = ( - e-t(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 poste-d 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 xy-plane 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) = (cost-tsint, 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 z-cornponent,- 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 43-46, 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 real-valued 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 public-ly 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 - e-tk

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.

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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 xy-plane 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,-e-t)

=}

Jr'(t)J=v'2 +e2L + e 2t =..J(et+e-t)2= et+e-t.

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 ')

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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 + e-t -

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 + (-2x-3)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),

= (-.J2e-t, ../2(t + 2)e1 , 2t + 3),

= . l(t + v

,../2t)

1)2e2t + e zt + 2,

( ) and

"'

t

r"(t) = ((t+2)et,e-t, 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 + e-2t + 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).

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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 / dt-ds 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

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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 Serret-Frenet 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!

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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 1-0 •

-

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

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.

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 + e-t 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 z-axis, 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.

®

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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 z-component 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- fG-t 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--=:::==::::::::;::>o-f'...::::::::=:=::::~ 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 +e-t k

=?

r"(t) = et i + e-t 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].

TRUE-FALSE 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 x-component 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 x-axis. 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 xy-plane 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 2-rr, 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 y-component. 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 + e-t 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 + e-t 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 ,; e-t R (t). So, as in parts (a) and (b), v = r '(t) = e-t R '(t)- e-t R (t) = e-t[R '(t) - R (t)) => a = v' = e-t [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 + e-t R.

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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 <=>

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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 x-coprdinate 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=vo-gt 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

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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 (1-smo) 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 wind-chill 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 wind-chill 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 wind-chill 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 wind-chill 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 40-knot 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 30-knot 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 30-knot 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)l-1::;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 yz-plane 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 z-values 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

z-values 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 three-fourths 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 y-axis.

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 yz-plane (such as the left-front trace in the graph above) are parabolas; those parallel to the xz-plane (such as the right-front 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- andy-axes. 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 z-plane is the parabola z

= 1-

2

x and the trace in the yz-plane 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 x-axis and

radius ,fk. When k = 0 the level surface is the x-axis. (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 xy-plane. (d) T he graph of g(x, y) = - f(x , y) + 2 is the graph off reflected about the xy-plane and then shifted upward 2 units.

71. f(x,y)

= 3x -

x4

-

4y

2

-

20r-----------------.

lOxy

10

0

y

y

Three-dimensional 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.l-scn'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 they-axis 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 they-axis, 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 x-direction curl downward and approach the xy-plane (while the vertical

trace x = 0 remains fixed), giving a saddle-shaped 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 a-hyperbolic paraboloid for c

"'

ln : = lnb+ aln(~) (b) We list the values for ln(L/ K) and

ln(P/ K) for the years 1899-1922.

(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

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> 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 Cobb-Douglas production function is

= bL"' ](1-a = 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 x-axis. Then.f(x, 0) = x /x

0. Now approach (0, 0) along they-axis. 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 they-axis 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 x-ax1s,

{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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to a publicly accessible website. in whole or in part.

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(x-v) = 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 + ex-y 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) l-1 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 t-o 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

$

limx-a 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( = h-0

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 wind-chill 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 x-direction, the graph off increases. Thus f ,(1, 2) is positive. (b) If we start at (1, 2) and move in the positive y-direction, 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 y-direction. / 11 is negative at ( - 1, 2) and if we move in the positive y-d.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 y-direction, 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 x-direction.

b traces out a line with negative s lope, while c traces out a parabola opening downward. This tells us that b is tlle x-derivative 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 xz-plane. 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 yz-plane 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 h-0 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)(1-xy)-(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 '

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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)(1-xy)-(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

1-xy

(

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 · h-o 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 1-41 + C (Ko), where G_(Ko) can depend on K 0 • Then .

-

IPI = e<> lniLI-f;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 - (V-n'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

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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 h-o 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 x-axis, fxv(x, y)-> 1 while~

(x, y) -> (0, 0) along they-axis, 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(x-y) 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 y-5 (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

+ ~·

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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

+ e-2"' (-

.

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.

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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) 6-x + /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 =>

8z-dy= 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

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t

208

D

CHAPTER 14

PARTIAL DERIVATIVES

5. w=xevl =, x = t2, y = l-t, 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

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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 °

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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 x-iy 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=---x-y =

- 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.

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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 8-8 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 second-order 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 postl-d 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

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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 = h-0 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 _ 4-313 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.

® 2012 Ccngt~gc Lcnming. All Rights Rcscn·ccJ. M11y not be scanned, copk-c.J, or duplico.tcd. or postl"li lou publicly occcssiblc wcbsilc, in whole or in pan.

=>

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 .

@ 2012 Ccn&_:'lle Le:uning. All Rights Reserved. May not be scanned, enpicd. ordupl iealcd. or posted 10 o publicly occessible " '"""ile, in whole or in p:ul.

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

x-3 . y-2 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 z-intercepts 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 z-intercepts 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 i-16 j-12k.

-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

).

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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:

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-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 y-2 , 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 y-4 • 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 real-valued 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

Of-----1~ - 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 =f-ie, y =

1 12 c

= ~ex- x 2

-

2xy,

and z = ic- x- y = f-ie. 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 .

©

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), (±V-2, ±~, - );) . 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

only-critical 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 = - ye-xu, / 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 non-negative, 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.

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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.,(1-a)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, -2-Xz), 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

= 2-Xx, 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 .

'

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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 + J-L\lh

=>

+ i'(V, x + z, y), so yex-z = 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 real-yalued 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,

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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 · .. Xn-1) ~

= 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 y-plane. 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 first-order 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].

TRUE-FALSE 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)

h-o

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

74-80 =~ = -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 h-o , h-o 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

= kxk-lylzm., !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 yl-1zm-1 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

.

.

x-1 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, = ·2x-y+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 \"''<--1-l-1

-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 xy-z 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 , 3-1 / 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

W-y

+ (L- x) 2(W -

= [x2 + (L- x?JIY2 + (W -

v?

+ x 2(W- y) 2

L-x

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 cross-sectional 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)

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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 semi-circular gutter of radius r, we would have w

2

=

~7r.

../3-w-, 1t• would be better to bend the metal ·mto a gutter WI'th a sem1c1rcu · · Iar cross-section. . >-

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 y-value 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) + ~e-318(~) + ~e-318(~) + ~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

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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(b-a) (d -c)=k(b-a)(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] = sinO-sin~-,- (sin(-~) - sinO] 0

=0-1-(-1-0) = 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

= ~ e-6 + 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 xy-plane.

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, copk-d. 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 Integrals-over 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 ::~v-2 dy = I~l y2 [y - (- y- 2)] dy 1 = J-11 (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 ~ (y-2)(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 =

(! +4-4) -

(-~ +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'4-x2dx=

+ 2x v'4- x2 + H4 -

-H4-x2)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 J1-a: ( 1 _., 1-

10

1 1

=

[ ((1-

x

2

x+2y ) dydx=

) -

x(1- x

2

)

f1 [

10

+ (1 -

2

y -l -x y-xy + 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 _ ( 2-x-y) dydx 2 1 1

J

2

( = J•l_1 J1-x .,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

= J-1 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] 1-1

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 xy-plane. 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 d-X=

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

-Vl-x2

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 ocos-lv

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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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'B-1)=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•x-1 x 0

2

dydx

+

101-x-1 -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

1-x

~ {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 cross-section 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 cross-section 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)

-%(-1-1) · 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]"'/ [-le-r ) = -'11'/2 2 0

1r{-l)(e-4 2

-

e0 )

= 1!.(1e-4 ) 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) 1-rr/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

®

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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 Jl-T2 -

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)

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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 y-direction, so we are given that the depth is constant in the x -direction and the depth increases linearly in the y-direction from f(O, -20) = 2 to f(O , 20) The slope of this line is 20 ~(_?20 ) ~

= 7. The trace in the yz-plane 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(b-a) 2 o , rn=11'(b2 - a2) 2 7r -a = (b+a)( b - a) = a+b

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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 =·xe-x dx. Then du = dx and v = - ~e-x , 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

(_.!.te-t2) + .!. roo e-x 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 ·

t-oo

dx

.::?

= 2udu

I; .../Xe-"'dx = #

.::?

I

2

lim ,/f.ue-" t-oo 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·3-x( 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

= Jof2J3-v( ., 12 xy + y 2)dy dx

Hencem

7. m =

r2.[12xy2 + 3Y 1 3)11=3-x 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=l-x 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 = J-1 1 Jor1 -"'~ ky 2 dydx = kf-1

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 they-axis]

= ffo yp(x, y)dA = fo,.t
M:r

H ence (-x, -y ) -_

7-rrk/ :J

-

'

@ 2012 Ccnpgc Learning. All Rights Reserved. May not be seaMed, co pied, or duplicated, or posted to • publicly accessible website, in whole or In pori.

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 kf-1 (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

1-ka6 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. copie-d. 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 e-0.5:re-0.2y dy dx = O.l fooo e-o.sx dx f ooo e-0.2y dy = 0. 1 lim

rt

L--+oo Jo

e- 0·5 "' dx lim

rt

t-oo Jo

e- 0·2 Y dy = O.l. lim (- 2e- 0·5 "'] t lim t.--+eo

0 t --+oo

[- se-0·211 ] 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 e-0 · 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

t-oo

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(e-0 · 2 ' t-oo

0 t-oo

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 xe-O.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

e-0 · 5 "' dx {; 00 ye- 0 · 211 dy .0

= 001 t-oo. lim {;t e-o.sx dx lim rt ye- 0 · 2 Y dy o t-oo _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 t-oo 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(e-o. 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-(x-45) /0.s

1

0.5 ..,l2ir -

0.1 ..,f2ir

= lQe- 2(x-45) 2 -so(v-2~)2.

"

.

< y < 25) = rso r2s f(x y) dy dx = !.!!. fso f25 e - 2(x-45)2-so(y-2o)2 dy dx J 4o J2o . ' .,. . 40 J 2o ·

Then P(40 <X< 50 20 -

2 e-(y-20) /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

lQe-2(x - 45) -so(y- 20)

D

2

dA =

"

l 45+1/v'2 . lQ . "

45-1/.,/2

2D+tn/2- 4(x-45)2

120-ftp,/2-

Using a CAS or calculator to evaluate the integral, we get P( 4(X - 45?

2 2 e-2(:.;-45) -so(y-20)

dydi.

4(x-45)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 xy-plane 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

® 20 12 Cengage Looming. All Rights Reserved. May not be sc:uu>ed, copied, or duplicated, or posted too publicly occcssiblc website, in whole or in port.

267

-I

268

0

CHAPTER 15 MULTIPLE INTEGRALS 2

2

2

2

13. z = f(x, y) =e-x - v , f.,= -2xe-x - v , fv

II

A(S)=

2

= -2ye-x - v

2

. Then

If

v( - 2xe- xLv2 )2+(- 2ye- x2 -v2 )2+ 1dA =

x2+y2_54

.j4(x2-t-y2)e- 2(x 2 +v 2 l-t- ldA.

x2+ 11 25-1

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 -t--4y2 , 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

- vh-x.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 xz-plane, then the surface lies "above" the,' disk x 2 We have y = f(x, z)

=x

2

A(S) =

+ z2

~ 25 in the x z-plane.

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 - (y-z)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 (4-y)dv= ~ [4v - h 2 ]~ · = ~ (16-8 - 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 fot-x 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 yz-plane 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 xy-plane 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: fo4-2x-v 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 xy-plane 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 xz-planes, 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) 1-1 ::; 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

r4-x2 JJ4-x2-yf2

I

f4 - 4z

=

I - tJo

=

f _2

2

2

-~/ /

(

JV4-y-4z2

(

-.J4 _

11

_

2

z 4 2

r;--:;

)

,

f x, y, z dx dy dz

J~/2 r4-x2 -4:2' -y4 - :z:2/2

r4 f-14-ii JJ4-:z:2-y j 2

)

Jo

(

)

x,y,z dzdydx=:Jo --14-ii -~ 12 ! x,y,z dzdxdy 2

J V4-y-4z ( = Jof4 f-/4-!i/2 --14-ii/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 r4-o: ~Jo

-y4-4z2

-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 xz-planes, 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~ 2-ty, -vv~ x~

= {(x,y,z)

l-2

~ 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= J-2 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 xz-planes. 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 xz-planes 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 xz-plane, 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 xy-plane 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)

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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 xy-plane 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 xy-plane, 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 half-plane

half-line 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 =

xy-plane, so (} =

+ 12 =

0

i~cluding

2 ; ,

3).

the z -axis and intersecting the xy-plane 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 z-axis, 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 xy-plane, so we have a half-cylinder.

13. We can position the cylindrical shell vertically so that its axis coincides with the z-axis and its base lies in the xy-plane. 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 xy-plane 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 xy-plane (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'

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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 xy-plane 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 xy-plane 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 mountain-we 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

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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.

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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 xz-plane.

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.

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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 =

i sin 2¢]~/Z [sin(})~/2

32)

"i· 0 ~ 4> ~ ~}.Thus

Io"'12 sin2 1/>d¢ J;12 cos(} d(} j 01 p3eP

J: -I:

2

2

peP dp)

= p2 , dv = peP2 dp] 2

[

~p2 eP

2

-

~eP J: = (i- 0) (1- 0) (0 + ~) = i

27. The solid region is given byE= { (p, 0, cf>) I 0 ~ p ~ a, 0 ~ 0 ~ 21r, ~ ~ cf> ~ ~}and its volume is

V

= ffis dV = I:/63 J~"' I; p2 sin cf>dpd(}drjJ = I:/: sincf>dl/> I~11' dO I; p2 dp [-coscf>J;~: [0)~11' [~p3 )~

.=

= ( -~ +

4) (27r) (ta3) =

1

3

' \- 7ra

29. (a) Since p = 4 cos 4> implies p2 = 4p cos 1/>, the equation is that of a sphere of radius 2 with center at (0, 0, 2). Thus

(b) By the symmetry of the problem Mv=

= M x= = 0. Then

M, 11 = I:" I; 13 I 04 co• 4> p3 cos cf> sin cf> dp d¢ d(} = J~" I; 13 cos cf> sin r/> ( 64 cos4 rjJ) dcf> dO --

2

f "' 64 Jo

[-16 cos() -•] =71'13 dO -'~' ¢=0

2

f 71' .ll Jo 2

dO -- 21 7r

Hence (x,y,z) = (0,0, 2.1).

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dp

SECTION 15.9 TRIPLE INTEGRALS IN SPHERICAL COORDINATES

0

283

31. (a) By the symmetry of the region, My: = 0 and Mx: = 0. Assuming constant density K, m =

iJ< (from Example 4). Then ·

fffE K dV = K fffE dV =

M .,y = Jjj~ z K dV = K

J;,. }~,. 14 j~o• ¢>(pcos ¢) p2 sin r/J dp drpd8 =

K

J;" J0,.14 sin ¢cos¢ [tP4] ==~os¢ d¢ d8

=if< J:w f0"' 14sin¢cos¢ (cos4 ifJ) .d(j)d(} = iK J:" d(} J;14 cos5 {j}sin (j)d¢ = iK

(8]~" [-i cos6 t/>] ~14 =

[ c~r -

iK(27r)( -i)

.. (- __) (My: !vfu M.,u) ( 77rK/96 ) m'm'm = 0, 0, 1rf(jB

I: ~

+ y2

= p2 sin2 ¢ and

JJJE (x2 + y2) J( dV = =

K

J;" fo"/4 focos¢(p2 sin2 ifJ) p2 sin¢ dpd¢d8 =

iK J:" J; 14sin

3

= tK

¢cos"' ¢d¢ d8

K

=if< .r:w d(} j~w/

J:.,.. fo"'/4sin3 t/> [kPsJ::~os ¢ d¢d8 4

cos5 ¢

(1 -

cos2 ¢) sin¢d¢

[8]~"' [- t cos6 ¢ + ~-cos 8 ¢]~ 14

= 1J<(27r) [-16 5

(:ii) 2

6

+

1(:ii) 8

8

2

+ 16 -

1] 8

=

b!.]( 5

(..ll.) 384 -

u wK 960

33. (a) The density function is p(x, y, z) = K, a constant, and by the symmetry of the problem M .,z = My:

M.,11

;~ K

7) = (0,0, 12.

Thusthecentro1d1s x,y,z = (b)' As in Exercise 23, x 2

1] = -fiK (-~) =

= J:"' f0" 1 2 J; Kp3 sin ¢ cos ¢ dpdrjJd8 = !1rKa4 J;1 2 sin¢ cos ¢ d¢ =

the hemisphere)= ~1rKa3 , so the centroid is

= 0. Then

t1rKa4 • But the mass is K(volume of

(0, 0, ~a).

(b) Place the center of the base at (0, 0, 0); the density function is p(x, y , z) = K. By symmetry, the moments of inertia about any two such diameters will be equal, so we just need to find I.,:

I., =

J:"I; ' 2 I ;(Kp2 sin ¢) p2 (sin 2 ¢ sin2 (} + cos2 ¢) dpd¢d8

J0" 12 (sin3 rjJ sin2 8 + sin r,b cos2 ¢) (ka5 ) d¢ d8 2 2 3 3 iKa5 I 0 " [sin 8 (- cos¢+ t cos ¢) + (- t cos ¢)] ::~

= K I:.,.. =

5 [1(18- l sin28) = lKa 5 3 2 4

35. In spherical coordinates z V =

= ..jx2 +

J:" f0" 14 f01 p2 sin ¢ dpd¢d8 =

M.,11 = I:"'

2 + 18] " 3 0

.r:r. dO f0.,..14 sin t/>dr/J .{

(o,o,

( 8 2

1 0

27r[--! cos2¢] ~

+ 1(27r0)] = 3

2

p dp = 27r ( 14

{t)

d(} = i Ka5 j~" [~ sin2 8 +

i) dO

5 ..i.J(a 7r 15

'i· Then

y 2 becomes cos¢ = s in ¢ or¢ =

J;14I: p3sin¢cos ¢ dpd¢ d8 =

Hence (x,y,z) =

5 = 1Ka [1(7r - 0) 5 3

12

=

t

-4 + 1) (t) = %1r(2 -

v'2),

and by symmetry My: = M x: = 0.

~ v'2)).

37. In cylindrical coordinates the paraboloid is given by z = r 2 and the plane by z = 2r sili Band they intersect in the circle

r = 2 sin 8. Then JJJEz dV

=

g f 02 s in o.r:;•inO rzdz dr dO = 5ti'

[using a CAS].

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284

0

CHAPTER 15 MULTIPLE INTEGRALS

= .Jx 2 + y 2 and below the sphere x 2 + y 2 + z 2 = 2 in the first

39. The region E of integration is the region above the cone z octant. Because E is in the first octant we have 0 and 0 ~ p ~

~

B~

%. The cone has equation 1/> = "i (as in Example 4), so 0 ~ 1/> ~ .;f,

../2. So Ute integral becomes

f0rr14 .fa"''12 fov"i (p sin¢ cos B) (p sin c/> sin t1) p 2 sin¢ dp dB d¢ · =

J~-r/ 4 sin3 1/>d¢ J 0rr/ 2 sinB cost1dB .J0v"i p4 dp =

-- [13 cos3

A.. -

'~'

cos A..] rr/4.. 1. '2) '~' o 2 l5 ( V"'

41 . The.region of integration is the solid sphere x 2 + y 2 2 p2 sin2 c/> + (p cos lj> - 2)

= p2 -

4pcos + 4 ~

4 5 [~ 12 - ~ 2 - ( la - 1) ] . M5 -~ fl15

-

-

+ (z - 2? ~ 4 or equivalently p ~ 4cos¢, so 0 ~ 8 ~ 211', 0 ~ c/> ~

4 =>

0:::; p ~ 4cos ¢.Also (x2 + y 2 + z2) 312 = (p2) 312

J0" 1 2 J:" J04. cos