cpm calculus chapter 08 solutions

Chapter 8 Solutions Calculus

Chapter 8 Solutions 1. a. A thin cylinder. b. Answers vary. c. It should be a semicircle. 2. a. The thickness times the area gives the volume of each slice if they are modeled as

cylinders, and the summation notations add up the volumes of n slices.

b. By integration: A(x)dxL

R

! , from the left side of the line to the right, where A(x) is the area of each cross-sections, which takes the limit of estimates with smaller and smaller slices.

3. a. ! 11.8

2( )

2

+ 14.32

( )2

+ 16.52

( )2

+ 182

( )2

+ 192

( )2

+ 18.12

( )2

+ 15.32

( )2

+ 112

( )2

+ 8.42

( )2

+ 72

( )2( ) " 5

# 8259.404cm3

b. It would be more accurate. c. Integrate the areas, which takes the limit of using thinner and thinner sections of the

vase. 4. a. dx represents an infinitesimal change in x. b. It represents the area of an infinitesimally thin rectangle. 5. See graph at right. a. height = x , width = !x = i

n(9 " 5)

in a Riemann sum or dx in an integral.

b. x dx = x3 232

4

9

!9

4=23(27 " 8) = 38

3


6. a. = 13(5x2 !17x)!2 3(10x !17)

b. = (12x2 + 2)(8 ! 7x3) + (4x3 + 2x2 ! 5)(!21x2 )= 96x2 ! 84x5 +16 !14x3 ! 84x5 ! 42x3 +105x2

= !168x5 ! 56x3 + 201x2 +16

c. = cos(sin t) !cos t d. ln(cos2 x) sin x ! 7 ln(49x2 ) 7. = ! (4x4 " 8x2 + 4 " (x4 " 2x2 +1))dx

"1

1

# = ! (3x4 " 6x2 + 3)dx"11

#

= ! ( 35x5 " 2x3 + 3x)

1

"1= ! 1 3

5" ("1 3

5)( ) = 16!5

8. a. = e3x sin(5x) + C b. = 1

4e4x+2

3

5=1

4e22!1

4e14

c. = sec x tan xdx!1 =" sec x + C d. U = sin2 x, dU = 2 sin x cos xdx = sin(2x)dx; sin(2x)2sin2x dx!

= 2U dU = 11n2

2U + C =! 11n2 "2sin2x

+ C

e. = (16x3 !16x2 + 4x)dx = 4x4 ! 163x3+ 2x2

2

0=

0

2

" 4 #16 ! 163 #8 + 8 = 72 !1283

= 29 13

9. a. 250 !1.1211 = $869.64 b. 1

11250 !1.12x dx

0

11

" = 111 !250 !1

1n1.12!1.12x

11

0=250(1.1211#1)11 ln 1.12

$ $497.06 10. a. At x = 2 there is a hole. b. At x = !4 and 1 there is no limit, and at x = 2 g is undefined. c. At x = !4,1 , and 2, because it is not continuous, and at x = !2 there is a cusp. 11. a. 3 b. 2 c. 2 d. Does not exist.


12. a. !x by f (x) = x b. The slice is a thin cylinder with depth !x and radius f (x) , and its volume is

V = !x "# f (x)2 = !x "# ( x )2 .

c. dx !" f (x)2 =0

8

# " ( x )2dx08

# = " ! xdx08

# = " x2

2

8

0= 32"

13. a. Again, each slice is a thin cylinder, with thickness !x (or dx). b. !x "# f (c)2

c. ! f (x)2dxa

b

" 14. See graph at right. a. A slice should be a disk with radius x2 and width !x .

b. ! (x2 )2dx = ! x4dx0

3

" =!03

" ! x5

5

3

0= ! 3

5

5=2435

! 15. ! ( 6 " x )2

2

5

# dx!+!! (x " 4)258

# dx =!! (6 " x)25

# dx!+!! (x2 " 8x +16)58

# dx =

! 6x " 12x2( ) 5

2+ ! 1

3x3 " 4x2 +16x( ) 8

5== 28.5! $ 89.54 !units3

16. See graph at right. a. See sketch at right. Radius = x; x2 = y or x = y

b. ! y( )0

4

"2

dy = ! x( )0

4

"2

dx = ! x0

4

" dx = 12 ! x24

0= ! (8 # 0) = 8! !un3

17. a. ! 1

2x3 + 3( )

2

0

2

" dx b. ! 2y " 63( )2

dy3

7

#

x

y


18. a. U = 6x4 ! 3x, dU = 24x3 ! 3 :8x3!1

6x4 !3xdx =

13

(24x3!3)dx

6x4 !3x=

13dU

U""" =13ln U + C = 1

3ln 6x4 ! 3x + C

b. = ! (x+1)33

" 9x#$%&'(5

2= ! 216

3" 45( ) " ! 273 "18( ) = 27! " ("9)! = 36!

c. = 5 ! x22

" 2x + 5 ln x + 3 + C

d. U = 11x2 ! 3,!dU = 22xdx : 2" x sin(11x2 ! 3)dx = 111 sin(11x2 ! 3) ! 22xdx"=111

sinU dU = ! 111cosU + C = ! 1

11cos(11x2 ! 3) + C"

e. U = x + 2, x =U ! 2, dU = dx :2xx+2

dx =!1

0

"2(U!2)U

du =!1

0

" 2 ! 4U( ) du = 2U ! 4 ln U!10

"0

!1

= 2(x + 2) ! 4 ln x + 20

!1= (4 ! 4 ln 2) ! (2 ! 4 ln1) = 2 ! 4 ln 2

f. U = 3x, dU = 3dx : 31!9x2

dx =du

1!U2= sin!1" U + C =" sin!1(3x) + C

19. It will look like a cone with a ridge. See diagram at right. 20. a. = lim

x!2( 12"x

+x"12"x) = lim

x!2

1+(x"1)

2"x= limx!2

x

2"x Does not exist.

b. = 018!18+1

=0

1= 0

c. = !"

d. limx!4

(2" x )

2" x= limx!4

2 + x = 2 + 2 = 4

e. limx!

1"5

x2+8

x3

24

x2+54

x3

= #

f. = limx!2

(x"2)(2x"1)

(x"2)(5x+3)= limx!2

2x"15x+3

=313

21. !R2 " ! r2 22. B: The middle integral computes a different area, but I and III compute the area of the same

region.


23. C. If you take the derivatives of I and II you get (ln x)3x

. 24. a. 10 inches b. 0.5 inches c. ! (radius)2(thickness) " ! (radius of hole)2(thickness) d. ! (10 " 0.5i)2 (2) " ! (0.5)2(2)

e. ! 10 " 0.5i( )2 " ! 0.5( )2( )i=0

9

# $2 = 2! 10 " 0.5( )2" 0.25( )

i=0

9

#

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!or 2! 5 + 0.5i( )2" 0.25( )

i=1

10

#

f. 1237.5! = 3887.721!in.3 25. a. A cylinder with a cone cut our of it: V = 4 !" ! 42 # 1

3! 4 !" ! 42 = 64" #

64

3" =

128"

3un3

b. ! x ! ("R2 # " r2 ) where R is the bigger radius and r is the smaller radius: ! x ! (" ! 42 # " ! x2 ) = ! x(16" # x2" )

c. (16! " x2! )dx = 16! x " x33!4

0= 64! " 64

3! = 128!

30

4

# un3 26. a. See graph above right. ! (" y + 3)2

0

4

# dy " ! (y " 3)234

# dy

b. See graph below right. ! (x + 3)2

0

1

" dx + ! (x # 3)413

" dx 27. a. They will be washers. The shape is egg-

like with a spherical hole. They are perpendicular to the x-axis.

b. ! x ! (" (#2x2 + 2)2 # " (#x2 +1)2 )

c. (! ("2x2 + 2)2 " ! ("x2 +1)2"1

1

# )dx

d. 2 (! ("2x2 + 2)2 " ! ("x2 +1)2 )0

1

# dx

x

y

x

y


28. a. Vertical rectangles: A = ( x ! (!x))dx = x

3 2

32

1

4

" + x2

2

4

1= 8

3 2+8( ) ! 23 + 12( ) = 163 + 243 ! 76 = 806 ! 76 = 736 un2

b. Horizontal rectangles. Intersections are (0, 2) and (3, 1):

A = ((!y2 + 4

!1

2

" ) ! (2 ! y))dy = (!y2 + y + 2)dy = !y3

3+y2

2+ 2y

2

!1!1

2

"

= !83+ 2 + 4 ! 1

3+ 12! 2( ) = 103 ! !

76( ) =

206+ 76! 276= 92= 4.5 un2

29. y = 3 1! x2

4 and x goes from 2 to 2: A = 2 ! 3 1" x2

4"2

2

# dx = 3"22

# 4 " x2dx $ 18.850 un2 or 6! using a calculator or noticing that it is a semicircle.

30. y = 3 1! x2

4: the horizontal cross-sections are disks with radius 3 1! x2

4 so

V = !"2

2

# 3 1" x2

4

$%&

'()2

dx = 9! x " 9!4

* x3

4x3

2

"2"2

2

#= (18! " 6! ) " ("18! + 6! ) = 12! " ("12! ) = 24! un3

31. a. lim

x!3

x2"x"6x"3

= limx!3

(x"3)(x+2)

x"3= limx!3

(x + 2) = 5 , so f (3) must be 5, h = 5 .

b. Except h = 1 does not work, notice we squared the original equation, which left room for !3+ h and 3+ h to be opposites, so h = 6 .

c. a: When graphed, the function looks like the line x + 2. h = 5, fills in the hole at

(3, 5). Therefore the function is differentiable for all x. b: !f ( x + 6) = 1

2 x+6=

1

2 3+6=16

at x = 3. !f ("x + 6) = "1 at x = 3. Therefore the function is not differentiable at x = 3.

!3+ h = 3+ h

(!3+ h)2 = 3+ h

9 ! 6h + h2 = 3+ h

h2! 7h + 6 = 0

(h ! 6)(h !1) = 0!!"!!h = 1 or 6


32. a. Use the reverse Chain Rule. U = ex , dU = exdx : ex(2!ex )20

1

" dx = dU(2!U )21

x=0x=0

1

"

=1

2!ex

1

0=

12!e

! 12!1

=12!e

!1 =1!(2!e)2!e

=e!12!e

b. Use the Chain Rule with g(x) = f (x)dx4

x

! and h(x) = x2 :

ddx

f (x)dx = h1(x) ! g1(h(x)) = 2x ! f (x2 )4

x2

"

c. U = sec x, dU = sec x tan xdx : sec x tan x1+sec2 x! dx =

dU

1+U2! = tan"1U + C = tan"1(sec x) + C

d. U = x2, dU = 2xdx : x1!x4

" = 12" #2xdx

1!x4=12

dU

1!U2" = 12 sin!1U + C =

12sin!1(x2 ) + C

e. = sin x 2cos x 2

dx,0

!2" U # cos x2 , dU = #

12sin( x

2)dx :

sin x 2

cos x 20

!2" dx = #2dUUx=0

!2"

= #2 ln U! /2

x=0= #2 ln(cos x

2)

! /2

x=0= #2 ln(cos !

4) + 2 ln(cos 0)

= #2 ln( 22) + 0 = #2 ln(2#1 2 ) = (# 1

2) $ (#2) ln 2 = ln 2 % 0.693

33. !f (x) = (x"3)("1)"(1)(1"x)

(x"3)2=3"x"1+x

(x"3)2=

2

(x"3)2, f 11(x) = "2 # 2

(x"3)3=

"4

(x"3)3, which is positive

when x < 3 : in (!", 3) . 34. Intersections are when

3 = x = 25 ! y2 , 9 = 25 ! y2, y2 = 16, y = 4; (3, 4)

V = (" ( 25 ! y2 )2 ! " (3)2 )dy!4

4

# = " (25 ! y2 ! 9)dy!44

# = " (16 ! y2 )dy!44

#

35. Yes (Rolles Theorem) or !f (x) = 2x "1 ; !f 1

2( ) = 0 36. a. Horizontally, washers. b. Horizontally, prisms. c. Vertically, disks.


37. a. A disk (thin cylinder), with thickness !y or dy . b. y c. ! ( y )2dy = ! ydy = ! " y2

2

10

00

10

#010

# = 50! un3

d. The inverse functions is y = x : ! ( x )2dx = ! " x22

10

00

10

# "50! un3 38. a. Since most of the region is closer to the y-axis than the x-axis, one should suspect the

volume of the rotation about the x-axis to have greater volume b. Each figure is the shape of a mostly hollow cone. For rotation about the x-axis, the inner

and outer radii are x2 and 2x and the width is !x :U = (" (2x)20

2

# $ " (x2 )2 )dx

= ! (4x2 " x4 )dx = ! # 43x3 " ! x

5

50

2

$2

0=32!3

" 32!5

=64!15

% 13.404 un3 .

For rotation about the y-axis, the inner and outer radii are y2

and y and the width is

!y = V = ((" ( y )2 # " (y

2)2

0

4

$ )dy = " (y #y2

4)dy = "

y2

2# "

y3

12

4

00

4

$!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!= 8" # 16

3" = 8

3" % 8.378 un3

c. Most of the region is closer to the y-axis and further from the x-axis. So even though the 2nd integral is twice as long (0 to 4), the radii are about 1

2 the size of the radii of

the 1st integral, and within the integral the radii are squared. 39. These are washers with inner and outer radii of 2 ! x and 4 ! x2

and width !x :V = (" (4 # x2 )2 # " (2 # x)2 )dx#1

2

$

= " (16 # 8x2 + x4 # 4 # 4x # x2 )dx#1

2

$

= " (x4 # 9x2 + 4x +12)dx#1

2

$

= " ( x5

5# 9x

3

2+4x2

2+12x)

2

#1

= " (14.4 # (#7.2)) = 21.6" % 67.858 un3


40. a. ! (x3 +1)2dx"1

1

# = ! (x6 + 2x3 +1"11

# )dx = ! ( x7

7+ 2 $ x

4

4+ x) 1

"1

= ! ( 17+12+1) " ! (" 1

7+12"1) = ! $ 23

14+ ! $ 9

14=3214

! = 167! % 7.181un3

b. A washer. c. The inner radius is 1 (the distance between y = !1 and the x-axis) and the outer

radius is x3 +1! (!1) = x3 + 2 (the distance between y = !1 and f (x) ).

d. (! (x3 + 2)2 " ! #12 )dx = ! x6"1

1

$"11

$ + 4x3 + 4 "1)dx = ! ( x7

7+ x

4+ 3x)

1

"1

= ! ( 17+1+ 3) " ! (" 1

7+1" 3) = 29

7! + 15

7! = 44

7! % 19.747 un3

e. Basically the integral is the same and gives the same answer: (! (x3 + 2)2 " ! #12 )dx =

"1

1

$ 44!7 % 19.747 un3 41. For a disk, it is as if it were a washer with an inner radius of 0. 42. a. U = 9 ! 2x, dU = !2dx : 1

9!2xdx

2

4

" = ! 121UdU = ! 1

2ln U

4

x=2x=2

4

"

= ! 12ln 9 ! 2x

4

x=2= ! 1

2# (0) + 1

2# ln 5 = 1

2ln 5

b. = 2 ! x66+ 3 ! x

3

3+ x

3

"3=

x6

3+ x

3+ x

3

"3= 243+ 27 + 3" (243" 27 " 3) = 60

c. = (x!1)(x+1)x!1

dx = ! (x +1)dx2

3

"32

" = ! x2

2+ x( ) 32 = ! 92 + 3( ) + (2 + 2) = ! 72

d. = 4x3 2dx1

4

! = 4 " x5 2

5/2

4

1=85" x5 2

4

1=85" (32 #1) = 8"31

5=2485

e. U = sin x, dU = cos xdx : sin3 x cos xdx! 4

! 2

" = U 3dUx=! 4! 2

"

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!=U4

4

! /2

x=! /4=sin4 x4

! /2

! /4=14#( 2 2)4

4=14# 116

=316


43. a. = (3x+1)!6"3!(6x"2)(3x+1)2

=18x+6"18x+6

(3x+1)2=

12

(3x+1)2

b. = sec(5 ln x) tan(5 ln x) ! 5x

c. x32 cos(2x)!3x2 sin(2x)x6

=2x cos(2x)!3 sin(2x)

x4

d. = 2k !93k"1 + k2 ! (ln 9) !93k"1 ! 3 = 2k !93k"1 + (3 ln 9)k293k"1 e. = (2 sin x cos x) cos x + sin2 x(! sin x) = 2 sin x cos2 x ! sin3 x

or 2 sin x(1! sin2 x) ! sin3 x = 2 sin x ! 3sin3 x 44. a. s(0) = 136 ft b. !s (t) = "32t +120 , so initial velocity is !s (0) = 120 ft/s, which is just the t term. c. !s (t) = "32t +120 = 0,120 = 32t, t = 120

32=308=154= 3 3

4sec

d. s(154) = !16(15

4)2 +120(15

4) +136 = !225 + 450 +136 = 361 ft

e. 0 = s(t) = !16t2 +120t +136, 0 = !2t2 +15t +17 = (!2t +17)(t +1),t =

172

: "s (172

) = !32(172

) +120 = !272 +120 = !152 m/s: it will explode.

45. Let x be the total price deduction, so the charge is 159 ! x , the occupancy is 265 + 2x , and

the revenue is (159 ! x)(265 + 2x) .

0 = d

dx((159 ! x)265 + 2x)) = d

dx(159 "265 ! 265x + 318x ! 2x2 )

=d

dx(159 "265 + 53x ! 2x2 ) = 53! 4x, x = 53

4= $13 1

4

So they should charge 159 !13 14= $145.75 .

46. a. k = 1

2 (by symmetry) b. k = 1

c. k = !1 d. k = 1


47. a. y = x ! x3 , "y = 12(x ! x3)!1 2 # (1! 3x2 ) = 1!3x

2

2 x!x3 or 1!3x2

2y

b. When the denominator from part (a) is 0. and y = 0 at each of these points.

c. x = 0,1, !1 48. The vertex is (0, 0): the distance is d(t) = (y ! 0)2 + (x ! 0)2 = (x2 )2 + x2 = x4 + x2 , so

d

dtd(t) = d

dtx4+ x

2 , !d (t) = 12(x4 + x2 )"1 2 # (4x3 + 2x) # x1 (by the extended Chain Rule.

!d ("3) = 12

(81+ 9)"1 2 # ("4 #27 " 6) # (0.5) = 12 90

# ("114) # 12=

"57

2 90=

"57

6 10$ "3.004 units/s

or 3.004 units is also appropriate. 49. a. The shape is a cylinder with a whirlwind cut out of it. A slice is a washer with inner

and outer radii of x2 and 4 and width !x .

V = (! " 42 # ! (x2 )2 )dx

0

2

$ = (16 # x4 )dx = ! (16x # x5

5)2

00

2

$

= ! (32 # 325) = 128!

5% 80.425 un3

b. The shape is like a bowl. A slice is a disk with radius y and width !y . V = ! ( y )2dy = ! ydy

0

4

"04

" = ! #y2

2

4

0= 8! $ 25.133un3

c. The shape is almost like a quarter of a donut and will be hard to draw. A slice is a washer with inner and outer radii of 1+ x2 and 5 and width !x .

V = (! "52 # ! (1+ x2 )2 )dx = ! (25 #1# 2x2 # x4 )dx

0

2

$02

$ = ! (24x # 2 " x3

3# x

5

5)2

0

= ! (48 # 163# 325) = ! (544

15) = 544!

15% 113.935 un3

d. Again this is like a quarter donut. A slice is a washer with inner radius r ! y and outer radius 3 and width ! y .

V = (! " 32 # ! (3# y )2 )dy

0

4

$ = ! (9 # 9 + 6 y # y)dy04

$ = ! (6 y # y)dy04

$

= ! (6 "y3 2

3 2#y2

2)4

0= ! (4 " y3 2 #

y2

2)4

0= ! (32 # 8) # 0 = 24! % 75.398 un3

0 = 2 x ! x3

0 = x ! x3

0 = x(1! x2 )

0 = x(1! x)(1+ x)

x = 0,1, !1


50. a. The radius of the outside cylinder is 5. The radii of the inside disks are y = 2x +1 . ! 52

0

2

" dx # ! (2x +1)202

" dx = ! (25 # 22x # 2 $2x #1)02

" dx = ! (24 # 22x # 2x+1)dx02

"

= ! 24x # 22x

2 ln 2# 2

x+1

ln 2( ) 20 = ! (48 # 162 ln 2 # 8ln 2 ) # (0 # 12 ln 2 # 2ln 2( )= ! 48 # 27

2 ln 2( ) % 28.524 un3

b. The radius is x, but y = 2x +1 or x = ln(y!1)ln 2

.

! x22

5

" dy = !ln(y#1)ln 2( )2

5

"2

dy $ 17 un3

c. ! (52 " (5 " (2x +1))20

2

# )dx = ! (25 " (16 " 8 $2x + 22x )02

# )dx = ! (9 + 8 $2x " 22x02

# )dx

= ! 9x + 8$2x

ln 2" 2

2x

2 ln 2( ) 20 % 41.804! % 131.333un3

d. The radius is x + 1, but y = 2x +1 or x = ln(y!1)ln 2

.

! (x +1)2 "12( )2

5

# dy = ! ln(y"1)ln 2 +1( )2

"12$%&

'()2

5

# dy * 15.75! * 40 un3

e. The radius of the outside cylinder is 7. The radii of the inside disks are y + 2 = 2x + 3 . ! 72

0

2

" dx # ! (2x + 3)202

" dx $ 135.75 un3 51. a. This is a sideways bowl. A slice is a disk with radius x and width ! x . V = ! ( x )2dx = ! xdx = ! " x2

20

4

#04

#4

0= 8! $ 25.133un3 .

b. This is a cylinder with a whirlwind cut out. A slice is a washer with inner and outer radii of y2 and 4 and width dy.

V = (! " r2 # ! (y2 )2 )dy

0

2

$ = ! (16 # y4 )02

$ dy = ! (16y #y5

5)2

0= ! (32 # 32

5) = 128!

5

% 80.425 un3

c. It is like a quarter donut. A slice is a washer with inner and outer radii of 2 + y2 and 6 and width ! y .

V = (! "62 # ! (2 + y2 )2 )dy

0

2

$ = ! (36 # 4 # 4y2 # y4 )dy02

$ = ! (32y # 4 #y3

3#y5

5)2

0

= ! (64 # 323# 325) = 704!

15% 147.445 un3

d. It is like a quarter donut. A slice is a washer with inner and outer radii of 5 ! x and 5 and width ! x .

V = (! "52 # ! (5 # x )2 )dx

0

4

$ = ! (25 # 25 +10 x # x)dx04

$= ! (10 x # x)dx

0

4

$ = ! " (10 " x3/2

3 2# x

2

2)4

0! (20

3"8 # 8) = 136!

3% 142.419 un3


52. a. ! (3x +1)2dx0

2

" # 166.421un3

b. The outer radius is 2, the inner radius is x. Since y = 3x +1 , x = ln(y!1)ln 3

. The outside cylinder is solid from y = 0 to y = 2, so we only need to subtract out the inside from y = 2 to y = 10.

! 22dx0

10

" # !ln(y#1)ln 3( )

2

dy2

10

" $ 40! #16.5! = 23.5! $ 73.827 un3

c. The outer radius is (3x +1) +1 = 3x + 2 , the inner radius is 1. ! (3x + 2)2dx

0

2

" # ! (1)2dx02

" $ 73.54! # 2! $ 224.74 un3

d. This would give an unwanted middle term as well as the limits of integration would be incorrect.

53. This is the inside half of a donut (half torus). It will have a cylindrical exterior and a

donuts interior. A slice is a washer with inner and outer radii of 2 ! 1! x2 and 2 and width ! x . V = (! "22 # ! (2 # 1# x2 )2 )

#1

1

$ dx = ! (4 # 4 + 4 1# x2 # (1# x2 ))dx#11

$

= ! (4 1# x2 #1+ x2 )dx#1

1

$ % ! " 4.950 % 15.550!un3

54. a. Vertical slices: disks with radius f (x) and width dx. V = ! ( f (x))2dx

"1

2

# .

b. Vertical slices: washers with inner and outer radii of 1 and 1+ f (x) and width dx. V = (! (1+ f (x))2 " ! (1)2

"1

2

# )dx or ! (2 f (x) + f (x)2 )dx"12

# .

c. Horizontal slices: disks with radius g(y) and width dy. V = ! (g(y))2dy1

3

" .

d. Horizontal slices: washers with inner and outer radii of 5 ! g(y) and 5 and width dy. V = (! "52 # ! (5 # g(y))2 )dy

1

3

$ or ! (10g(y) " g(y)2 )dy13

# . 55. a. A typical slice is a horizontal disk with radius 6

y!

5

y2 and width dy.

V = !1

13

" (6y #5

y2)2dy = ! (36y#2 # 60y#3 + 25y#4 )dy

1

13

"

= ! (36 $y#1

#1# 60 $

y#2

#2+ 25 $

y#3

#3

13

1= ! (#36

y+

30

y2# 25

3y3)

13

1% 36.875 in.3

b. Height is 12: V = ! r2h = ! r2 "12 = 36.875, r2 = 36.87512!

, r = 36.87512!

# 0.989 in.


56. a. y = h ! hrx (slope is 0!h

r!h=

!h

r) or hx + ry = rh or x = r ! r

hy .

b. Using x = r ! rhy as the radius of a disk, V = ! (r " r

hy)2dy

0

h

# .

c. Yes: V = ! (r " rhy)2dy

0

h

# = ! (r2 " 2 r2

hy + r

2

h2y2

0

h

# )dy = ! r2 (1" 2y

h+y2

h2)dy

0

h

#

= ! r2(y "y2

h+

y3

3h2)h

0= ! r2(h " h

2

h+

h3

3h2) = ! r2(h

3) = ! r

2h3

57. Now we have washers with inner and outer radii of 1 and 1+ 6

y!

5

y2

V = (! (1+ 6y" 5y2

)21

13

# " ! $12dy = ! (1+ 36y2113

# + 25y4 +12y" 10y2

" 60y3

"1)dy

= ! (12y"1 + 26y"2 " 60y"3 + 25y"4 )dy1

13

# = ! (12 ln y + 26"1 y"1 " 60 $y"2

"2+ 25 $

y"3

"3)

13

1

= ! (12 ln y " 26y"1 + 30y"2 " 253

)13

1% 104.572 in.3

58. a. The graph of x ! 4 is symmetric about x = 4 , so

x ! 4 dx0

8

" = 2 (x ! 4)dx = 2( x2

2! 4x)

8

44

8

" = 2 # (32 ! 32 ! (8 !16)) = 2 #8 = 16 (This is easily found as the area of two triangles.)

b. = (w3 ! 2)dw = w44

! w2

2+ C"

c. = ! !7dx!5

!1

" = 7dx = 4 # 7 = 28!5!1

" , area of a rectangle.

d. This is a quarter circle: ! "524

=25!

4.

59. Each doll is like a slice of the largest doll, so that when you add up all their volumes,

ideally you will get the full volume of the largest doll. 60. sum = 2! xdx = 2! " x2

2

r

0= ! x2

0

r

#r

0= ! r2 = area of a circle.

61. Rotating y = r2 ! x2 about the x-axis yields disk slices with radius r2 ! x2 and width dx:

V = ! ( r2 " x2 )2dx ="r

r

# (r2 " x2 )dx = ! (r2 $ x " x3

3)

r

"r"r

r

# = ! ((r3 " r3

3) " ("r3 + r

3

3))

= ! (2r3 " 23r3) = ! $ 4

3r3 =

43! r3


62. a. Use washers, but with 2 sections (2 integrals). b. y = 0.5x2, 2y = x V = (! (3)2 " ! (1)2 )dy

0

0.5

# + (! (3)20.54.5

# " ! ( 2y )2 )dy

= 8! y0.5

0+ ! (9 " 2y)

0.5

4.5

# dy

= 4! + (! (9y " y2 )4.5

0.5)

= 4! + ! (812" 814) " ! (9

2" 14)

= 4! + 814! " 17

4! = 4! + 64!

4= 20! un3

63. a. Positive slope, concave up: b b. Negative slope, concave down: d c. Positive slope, concave down: c d. Negative slope, concave up: a, e 64. a. Doll shells b. Circumference strips c. He used the circumference of a strip and dx as the rectangles length and width. 65. Note: These are the actual volumes using the shell method. 2! x " f (x)dx

a

b

#

1A. 2! x(x " 2)2dx0

2

# = 2! (x3 " 4x2 + 4x)dx02

# = 2! 14 x4 "43x3 + 2x2( ) 2

0= 83! !un3

1B. In this case the radius is 2 ! x . 2! (2 " x)(x " 2)2dx

0

2

# = 2! ("x3 + 6x2 "12x + 8)dx02

# =

2! " 14x4 + 2x3 " 6x2 + 8x( ) 2

0= 8! !un3

1C. In this case the radius is x +1 .

2! (x +1)(x " 2)2dx

0

2

# = 2! (x3 " 3x2 + 4)dx02

# = 2! 14 x4 " x3 + 4x( )2

0=

2! (4 " 8 + 8) = 8! !un3

2A. 2! x(4 " x)0

4

# dx = 2! (4x " x2 )04

# dx = 2! 2x2 " 13 x3( )4

0= 64

3! !un3

Solution continues on next page.


65. Solution continued from previous page. 2B. In this case the radius is 4 ! x .

2! (4 " x)(4 " x)dx0

4

# = 2! (16 " 8x + x2 )04

# dx = 2! 16x " 4x2 + 13 x3( )4

0=

2! 64 " 64 + 643( ) =

1283

! !un3

2C. In this case the radius is 2 + x . 2! (2 + x)(4 " x)

0

4

# dx = 2! (8 + 2x " x2 )04

# dx = 2! 8x + x2 " 13 x3( )4

0= 160

3! !un3

3A. 2!0

4

" x2dx = 2! 13 x3( )4

0= 128

3! !un3

3B. In this case the radius is 1+ x . 2! (1+ x)(x)

0

4

" dx = 2! (x + x2 )04

" dx = 2! 12 x2 +13x3( ) 4

0= 176

3! !un3

3C. In this case the radius is 4 ! x . 2!

0

4

" (4 # x)(x)dx = 2! (4x # x2 )04

" dx = 2! 2x2 # 13 x3( )4

0= 64

3! !un3

4A. 2!0

4

" x xdx = 2! x3/204

" dx = 2! 25 x5/2( )4

0= 128

5! !un3

4B. In this case the radius is 4 ! x .

2! (4 " x) x

0

4

# dx = 2! (4x1/2 " x3/2 )dx04

# = 2! 83 x3/2 "25x5/2( ) 4

0=

2! 643" 645( ) =

25615

! !un3

4C. In this case the radius is 5 ! x . 2! (5 " x) x

0

4

# dx = 2! (5x1/2 " x3/2 )dx04

# = 2! 103 x3/2 "25x5/2( ) 4

0=

2! 803" 645( ) =

23615

! !un3

66. a. A thin box with length 2! x , width f (x) , and thickness dx. b. 2! x " f (x)dx

c. 2! x " f (x)dxa

b

#

d. 2! x(0.5x2 )dx1

3

" = ! x3)dx13

" = ! x4

4

3

1= ! 81

4# 14( ) = 20! $ 62.832 cm3


67. a. 2x= 1+

1

y

2

x!1 =

1

y

2!xx

=1

y

x2!x

= y

!y = 1"(2#x)#x(#1)(2#x)2

=2

(2#x)2 At (1, 1) it is 2.

b. x2y ! y3 = 82xy + x2 "y ! 3y2 "y = 0

2xy = 3y2 "y ! x2 "y

2xy = "y (3y2 ! x2 )

2xy

3y2 !x2= "y

At (3, 1) it is 1.

68. a. = 2 ! t3/2

3

2

+ t3

3

4

1= 43!8 + 64

3"

4

3+ 13( ) =

96

3"5

3= 91

3

b. U = x2 +1,!dU = 2xdx : x

x2 +1dx! = 12 "

2xdx

x2 +1! =12

dU

U! =12ln U + C = 1

2ln(x2 +1) + C (using substitution)

c. U = cos u,!dU = ! sin du : tan udu!" /4

" /4

# = !! sin u ducos u!" /4

" /4

# = ! dUU!" /4" /4

#= ! ln U

" /4

!" /4=! ln cos u

" /4

!" /4= ! ln 2

2( ) + ln 22( ) = 0

The best way to realize it is 0 is by symmetry (tangent is odd). d. ln(e2x )! dx = 2x dx! = x2 + C 69. E: 2x2dx

1

2

! = 2 " x#1

#1

2

1=

#2x

2

1= #1+ 2 = 1

70. B: This is ! (2x)2 " ! (2x2 )2( )

0

1

# dx = (4x2 " 4x4 )dx01

# 71. D: At x = 0, lim

x!0f (x) = 1 (by graphing of lHopitals rule), so I is false.

As x!",! f (x)! 0 , so II and III are true. 72. D: I is false because !f (b) does not exist. II and III are true.


73. E: The average is 1e!1

(ln x2 )dx1

e

" # 1.164 Without a calculator if one notes that f is concave down in this interval, with endpoints at

(1, 0) and (e, 2), then it is apparent that its average value must be above the average of 0 and 2, i.e., greater than 1.

74. D: (x2 +1)dx0

c

! = (x2 +1)c2

! dxx3

3+ x

c

0=

x3

3+ x

2

c

c3

3+ c =

83+ 2 "

c3

3" c

c3+ 3c = 8 + 6 " c3 " 3c

2c3 + 6c = 14

c3+ 3c " 7 = 0!!!!!#!!!!!c $ 1.406!by!using!a!calculator

75. a. !(x ! 3)2 + 4( ) dx1

4

" = (!x2 + 6x ! 5)dx14

" = ! x3

3+ 3x2 ! 5x

4

1

= ! 643+ 48 ! 20( ) ! ! 13 + 3! 5( ) = 9 un2

b. 2! x("(x " 3)2 + 4)dx1

4

# = 2! ("x3 + 6x2 " 5x)dx14

# = 2! " x4

4+ 6x

3

3" 5x

2

2( ) 41= 2! ("64 +128 " 40) " " 1

4+ 2 " 5

2( )( ) = ! 48 " " 32( )( ) = 99!2 $ 155.509 un3

76. a. It would have two different parts of f (x) as its bounds (radii). b. Shells would only requite one integral, instead of 2 with washers/discs. c. x ranges from 0 to 3. d. radius = 1 + x, height = f (x) = !x2 + 2x + 3

e. 2! (1+ x)("x2 + 2x + 3)dx0

3

# = 2! ("x3 + x2 + 5x + 3)dx03

#= 2! " x

4

4+ x

3

3+ 5x

2

2+ 3x( ) 30 = 2! " 814 + 273 + 452 + 9( ) = 81!2 $ 127.235 un3

77. 2! (x +1)("x2 + 2x + 3" (x +1))dx

"1

2

# = 2! (x +1)("x2 + x + 2)dx"12

# Note: The height is the difference of the two functions.


78. a. See graph at right. Graph is scaled by 10s. 5x0.5 ! 0.1x1.5 = 0

5x0.5 = 0.1x1.5

5 "x0.5

x0.5

= 0.1 "x1.5

x0.5

5 = 0.1x

x = 50 ft

b. V = ! (5x0.5 " 0.1x1.5 )2dx = ! (25x " x2 + .01x3)dx0

50

#050

# = ! (25 $ x2

2" x

3

3+ .01 $ x

4

4)

50

0

= ! (31250 " 1250003

+15625) = ! $5208 13% 16362.462 ft3

c. The 120 aliens take up 10% !16362.462 ft3 " 1636.246 ft3 and the average weight is 13.635 ft3 !

3.5lbs

ft3" 47.724 lbs.

79. a. Vavg = total displacementtotal time = 1200m96sec = 12.5 m/sec

b. d(t) ! 0.000145x4 " 0.0246x3 +1.315x2 "15.808x + 5.582,#d (t) ! 0.000579x3 " 0.0734x2 + 2.629x "15.808

#d (96) ! 69.952 m/s ! 0.069952 km/s $602s/hr ! 251.829 km/hr

Which might be too fast. c. See graph at right. It is a cubic. Between 50 and 70

seconds the competitor does not go far, but then he speeds up again.

80. a. f (x)dx =

c

a

! " f (x)dx = "( f (x)dx +ab

!ac

! f (x)dx) = "(2.5 " 5) = 2.5bc

!

b. 2 f (x)dx = 2( f (x)dx + f (x)dx)b

c

!ab

!ac

! = 2(2.5 " 5) = "5

c. = ! f (x)dx = !( f (x)dx + f (x)dx + f (x)dxc

d

"bc

"ab

"ad

" ) = !(2.5 ! 5 +1.5) = !(!1) = 1

d. 0 81. B: !f (x) = d

dx(x3 "10x2 + 21x) = 3x2 " 20x + 21 = 0, x =

20 400"4(3)#21

6=20 148

6,

max is at x = 20! 1486

=10! 37

3 and is about 12.597 .

t

v(t)

20

20

40 60 80

40


82. C: x2 = 8 + 2xyx2 !82x

= y

y = x2!4

x

!y = 12" 4 # ("1)x"2 = 1

2+

4

x2

!y = 12+

44=

32

at x = 2

83. A: !!!!!!!!! x 1! x2dx

0

1

" !!#!!U = 1! x2, dU = !2xdx

= ! 12

(!2x) 1! x2dx0

1

" = ! 12 UdUx=01

"

= ! 12U3/2

32

= ! 13U 3/2 = ! 1

3(1! x2 )3 2

1

0= ! 1

3(0) + 1

3$1 = 1

3

84. D: !f (x) = "e"x " 3x2 + 6x "10

!!f (x) = e"x " 6x + 6

, which is zero and changes sign at about 1.06 .

85. B (At horizontal tangents the derivative is 0.) 86. a. Shells: V = 2! x(x +1" (x2 " 6x + 7))dx

1

6

#

or 2! x("x2 + 7x " 6)dx1

6

#

b. Washers: V = (! (3+ x +1)2 " ! (3+ x2 " 6x + 7

1

6

# )2 )dx

or ! ((x + 4)2 " (x2 " 6x +10)2 )dx1

6

#

Graph is scaled by 2s. c. Washers: V = (! ("y2 + 3)2 " ! (y2 +1)2 )dy

"1

1

#

or ! ("8y2 + 8)dy"1

1

#

Solution continued on next page.


86. Solution continued from previous page.

d. Shells: V = 2! (2 " y)("y2 + 3" (y2 +1))dy"1

1

#

or 2! (2 " y)("2y2 + 2)dy"1

1

#

e. Disks: V = ! (cos x + 2)2dx

0

2!

" f. Shells: V = 2! x(cos x + 2)dx

0

2!

" g. Washers: V = (! (6 " x)2 " ! (2x )2 )dx

0

2

#

= ! ((6 " x)2 " 4x )dx0

2

#

h. Shells: V = 2! x("x + 6 " 2x )dx

0

2

#


87. a. g(2) = f (t)dt0

2

! = 2 (triangle), g(4) = 2 ! "#22

2= 2 ! 2" (semicircle)

b. No, it is the opposite.

c. f (x)dx =!2

2

" f (t)dt + f (t)dt02

"!20

" = !g(!2) + g(2)

d. Yes, in fact, its derivative is ddx

f (t)dt = f (x)0

x

! in this interval.

e. This is where its derivative changes from positive to negative: x = 2 . f. g(4) = 2 ! 2" and g1(4) = f (4) = !2 , so the tangent line is

y = !2(x ! 4) + 2 ! 2" = !2x +10 = 2" . g. This is where the derivative of f changes sign: x = 0 and 4. 88. a. = yey !ey

y2, using the Quotient Rule.

b. = 12(sin x)!1 2 " cos x , using the Chain Rule.

c. = dd!(5(cos2 ! + sin2 !)) = d

d!(5) = 0 , by simplifying first.

89. D: This is d

dx( x ) = 1

2x!1 2 .

90. A: There are twenty terms, all multiplied by 1

20, which is the width of a rectangle.

91. C: The intersection is at ex = !x + 3 or ex + x ! 3 = 0, x " 0.792 : A ! ("x + 3" ex )dx

0

.792

# = " x2

2+ 3x " ex

0.792

0! "0.145 " ("1) ! 0.855

92. C: It is flat at x = 0 and 2.

93. E: =x12x!1 2( )e x ! 12 x!1 2e x

x=ex (1!x!1 2 )

2x=ex (x! x )

2x2.

94. C: = f (x)dx

a

b

! + ("3)dx = 2b " a + ("3)(b " a) = 2b " a " 3b + 3a = "b + 2aab

!


95. D: I uses cylindrical shells and II uses disks. 96. a. Slice horizontally b. Slice horizontally, but you would get three different similar shapes. c. Slice vertically. d. No convenient way to slice.

97. b. A(i)!xc=1

n

"

c. With an integral: A(x)dxL

R

! The limit of the sums of volumes as the number of slices !" .

100. It does not matter when some cards in a deck protrude out of it; the deck maintains its

volume, as each card has the same height. 101. a. ! (1

2x cos x + 4)2dx

0

5

" b. 2! x(12 x cos x + 4)dx05

"

c. 2! (6 " x)(12x cos x + 4)dx

0

2

# d. (! (12 x cos x +11)2 " ! # 4905

$ )dx 102. D: = !1

1!(3x)2" 3 = !3

1!9x2 by the Chain Rule

103. B: Vavg = 1! (t + sin t)dt = 1!0

!

" ( t2

2# cos t)

!

0=1!(!2

2# (#1) # (#1)) = !

2+2!

. 104. C: !F (5) = 9 " 5 = 2 , but F(!7) is negative on the graph, and !!F (x) = "1

2 9"x is negative.


105. D: dPdt

= 0.02P + 357

dP

0.02P+357= dt

dP0.02P+357! = dt!10.02

ln(0.02P + 357) = t + C

ln(0.02P + 357) = 0.02t + C

0.02P + 357 = e0.02t+C

0.02P = Ce0.02t

0.02P = "357 + Ce0.02t

P = "17850 + Ce0.02t

!17850 + C = 18000C = 35850

P(6) = !17850 + 35850e0.02"6

# 22571

106. D: xy = x(3x !12) : d

dx(x(3x !12)) = d

dx(3x2 !12x) = 6x !12 = 0, x = 2,

minimum product = 2(6 !12) = !12 . 107. The derivative of g is f: D starts out in a positive direction and is flat at a and b. 109. ln xdx

a

b

! ; No, Cavalieris theorem. 108. 1(a): 2 ( 9 ! x2 )2

0

3

" dx = 36!un3

1(b): 2 12( 9 ! x2 )2

0

3

" dx = 18!un3

1(c): 2 2( 9 ! x2 )20

3

" dx = 72!un3

1(d): 2! 12( ) (

129 " x2 )2

0

3

# dx = 92 ! !un3

1(e): 2 12( 9 ! x2 )2

0

3

" dx = 18!un3

1(f): 2 32( 9 ! x2 )2

0

3

" dx = 18 3 !un3

1(g): 2 12

9 ! x2 + 129 ! x2( ) 12 9 ! x2( )0

3

" dx = 34 (9 ! x2 )dx03

" = 272 !un3

1(h): 2 !4( 9 " x2 )2

0

3

# dx = 9! !un3

1(i): 2! (129 " x2 )2

0

3

# dx = 9! !un3



108. Solution continued from previous page. Note: For number 2, x2 = 9 ! y2 .

2(a): (2x)(2x)dy0

3

! = 4 (9 " y2 )03

! dy = 72!un3

2(b): (2x)(x)dy0

3

! = 2 (9 " y2 )03

! dy = 36!un3

2(c): (2x)(4x)dy0

3

! = 8 (9 " y2 )03

! dy = 144 !un3

2(d): !2

(x)(x)dy0

3

" = !2 (9 # y2 )03

" dy = 9! !un3

2(e): 12

(2x)(2x)dy0

3

! = 2 (9 " y2 )03

! dy = 36!un3

2(f): 12(2x)(x 3)dy

0

3

! = 3 (9 " y2 )03

! dy = 18 3 !un3

2(g): 12(2x + x)(x)dy

0

3

! = 32 (9 " y2 )03

! dy = 27!un3

2(h): !4

(2x)2dy0

3

" = ! (9 # y2 )03

" dy = 18! !un3

2(i): ! (x)(x)dy0

3

" = ! (9 # y2 )03

" dy = 18! !un3

3(a): 2 (3! y)20

3

" dx = 2 (6 ! 2x)203

" dx = 72!un3

3(b): 2 12(3! y)2

0

3

" dx = (6 ! 2x)203

" dx = 36!un3

3(c): 2 2(3! y)20

3

" dx = 4 (6 ! 2x)203

" dx = 144 !un3

3(d): !2

12(3" y)( )

2

0

3

# dx = !8 (6 " 2x)203

# dx = 92 ! !un3

3(e): 12

(3! y)20

3

" dx = 12 (6 ! 2x)203

" dx = 18!un3

3(f): 12

3

2(3! y)2

0

3

" dx =3

4(6 ! 2x)2

0

3

" dx = 9 3 !un3

3(g): 12(3! y) + 1

2(3! y)( )

0

3

" 12 (3! y)( ) dx =38

(6 ! 2x)20

3

" dx = 272 !un3

3(h): !4

(3" y)20

3

# dx = !4 (6 " 2x)203

# dx = 9! !un3

3(i): ! 12(3" y)( )

2

0

3

# dx = !4 (6 " 2x)203

# dx = 9! !un3 Solution continues on next page.



4(a): (x ! (!3))2dx!3

3

" = (x + 3)2dx!33

" = 72!un3

4(b): 12

(x + 3)2dx!3

3

" = 36!un3

4(c): 2 (x + 3)2dx!3

3

" = 144 !un3

4(d): !2

12(x + 3)( )

2dx

"3

3

# = !8 (x + 3)2dx"33

# = 9! !un3

4(e): 12

(x + 3)2dx!3

3

" = 36!un3

4(f): 12

3

2(x + 3)2dx

!3

3

" =3

4(x + 3)2dx

!3

3

" = 18 3 !un3

4(g): 12(x + 3) + 1

2(x + 3)( )

!3

3

" 12 (x + 3)( ) dx =38

(x + 3)2!3

3

" dx = 27!un3

4(h): !4

(x + 3)2"3

3

# dx = 18! !un3

4(i): ! 12(x + 3)( )

2dx

"3

3

# = !4 (x + 3)2dx"33

# = 18! !un3

5(a): ( x +1)2dx!1

3

" = (x +1)dx!13

" = 8!un3

5(b): 12( x +1)( x +1)dx

!1

3

" = 12 (x +1)dx!13

" = 4 !un3

5(c): 2( x +1)( x +1)dx!1

3

" = 2 (x +1)dx!13

" = 16!un3

5(d): !2

12

x +1( )2dx

"1

3

# = !8 (x +1)dx"13

# = ! !un3

5(e): 12( x +1)( x +1)dx

!1

3

" = 12 (x +1)dx!13

" = 4 !un3

5(f): 12( x +1)(

3

2x +1)dx

!1

3

" =3

4(x +1)dx

!1

3

" = 2 3 !un3

5(g): 12!1

3

" x +1 + 12 x +1( ) 12 x +1( ) dx = 38 (x +1)!13

" dx = 8!un3

5(h): !4( x +1)2dx

"1

3

# = !4 (x +1)dx"13

# = 2! !un3

5(i): ! 12

x +1( )2dx

"1

3

# = !4 (x +1)dx"13

# = 2! !un3




6(a): 2 (3sin x)2dx0

3

! = 18 (sin x)203

! dx " 28.26!un3

6(b): 2 (3sin x) 32sin x( ) dx

0

3

! = 9 (sin x)203

! dx " 14.13!un3

6(c): 2 (3sin x)(6 sin x)dx0

3

! = 36 (sin x)203

! dx " 56.52!un3

6(d): 2 !2(3 sin x)2dx

0

3

" = 9! (sin x)203

" dx # 14.13! !un3

6(e): 2 12(3 sin x)(3 sin x)dx

0

3

! = 9 (sin x)203

! dx " 14.13!un3

6(f): 2 12(3 sin x)(

3 3

2sin x)dx

0

3

! =9 3

2(sin x)2

0

3

! dx " 12.24 !un3

6(g): 2 12(3 sin x + 3

2sin x)( 3

2sin x)dx

0

3

! = 274 (sin x)203

! dx " 10.60!un3

6(h): 2 !4(3 sin x)2dx

0

3

" = 92 ! (sin x)203

" dx # 7.065! !un3

6(i): 2 ! (3 sin x)2dx0

3

" = 18! (sin x)203

" dx # 28.26! !un3

Note: For number 7 the radius/length is (3! x) , where x = y2 !1 . Thus r = (4 ! y2 ) .

7(a): (3! x)20

2

" dy = (4 ! y2 )2dy02

" = 17 115 =25615

!un3

7(b): (4 ! y2 ) 12(4 ! y2 )( ) dy

0

2

" = 12 (4 ! y2 )2dy02

" = 25630 !un3

7(c): (4 ! y2 ) 2(4 ! y2 )( ) dy0

2

" = 2 (4 ! y2 )2dy02

" = 51215 !un3

7(d): !2

12(4 " y2 )( )

2dy

0

2

# = !8 (4 " y2 )2dy02

# = 3215 ! !un3

7(e): 12(4 ! y2 )(4 ! y2 )dy

0

2

" = 12 (4 ! y2 )2dy02

" = 25630 !un3

7(f): 12(4 ! y2 ) 3

2(4 ! y2 )( ) dy

0

2

" =3

4(4 ! y2 )2dy

0

2

" =64 3

15!un3

7(g): 12(4 ! y2 ) + 1

2(4 ! y2 )( ) 12 (4 ! y2 )( ) dy0

2

" = 38 (4 ! y2 )2dy02

" = 325 !un3

7(h): !4(4 " y2 )2dy

0

2

# = !4 (4 " y2 )2dy02

# = 6415 ! !un3

7(i): ! 12(4 " y2 )( )

2dy

0

2

# = !4 (4 " y2 )2dy02

# = 6415 ! !un3 Solution continues on next page.



8(a): (3! x)2dy!3

3

" = (3! y)2dy!33

" = 72!un3

8(b): 12

(3! y)(3! y)dy!3

3

" = 36!un3

8(c): 2 (3! y)2dy!3

3

" = 144 !un3

8(d): !2

12(3" y)( )

2dy

"3

3

# = !8 (3" y)2dy"33

# = 9! !un3

8(e): 12

(3! y)2dy!3

3

" = 36!un3

8(f): 12

3

2(3! y)2dy

!3

3

" =3

4(3! y)2dy

!3

3

" = 18 3 !un3

8(g): 12(3! y) + 1

2(3! y)( )

!3

3

" 12 (3! y)( ) dy =38

(3! y)2!3

3

" dy = 27!un3

8(h): !4

(3" y)2"3

3

# dy = 18! !un3

8(i): ! 12(3" y)( )

2dy

"3

3

# = !4 (3" y)2dy"33

# = 18! !un3

Note: For number 9 the radius/length is 2x = y + 3.

9(a): (y + 3)2!3

3

" dy = 72!un3

9(b): 12(y + 3)(y + 3)

!3

3

" dy = 36!un3

9(c): 2(y + 3)(y + 3)!3

3

" dy = 144 !un3

9(d): !2x2

"3

3

# dy = !212(y + 3)( )

"3

3

#2

dy = !8

(y + 3)2"3

3

# dy = 9! !un3

9(e): 12(y + 3)(y + 3)

!3

3

" dy = 36!un3

9(f): 12(y + 3) 3

2(y + 3)( )!3

3

" dy =3

4(y + 3)2

!3

3

" = 18 3 !un3

9(g): 12(y + 3) + 1

2(y + 3)( )

!3

3

" 12 (y + 3)( ) dy =38

(y + 3)2!3

3

" dy = 27!un3

9(h): !4(y + 3)2dy

"3

3

# = 18! !un3

9(i): ! x2"3

3

# dy = ! 12 (y + 3)( )"33

#2

dy = !4

(y + 3)2"3

3

# dy = 18! !un3


110. The radius is x. x = y ( y )20

9

! dy 111. ( x )2

0

4

! dx = x04

! dx = 12 x24

0=162

" 0 = 8 un2 112. 2 (2(4 ! x2 ))2

0

2

" dx 113. dl = 6,!dw = 2,!A = lw,!dA = (dl)w + l(dw) = 6(15) + 30(2) = 90 + 60 = 150 cm2 /sec 114. a. Use integration by parts. Let u = x!!!!!1dx and dv = (2x !1)!1/2dx!!"!!(2x !1)1/2 .

x(2x !1)1/2 ! (2x !1)1/2dx1

5

"

x(2x !1)1/2 ! 13(2x !1)3/2( ) 5

1= (15 ! 9) ! 1! 1

3( ) = 513= 16

3

b. Let u = 1! x2 !!" du = !2xdx . If x = 1, u = 0. If x = 0, u = 1. u ! 1

2du( )

1

0

" = ! u ! 12 du( )01

" = 122

3u3/2( ) 1

0= 13

c. Let u = x2 +1!!!!!du = 2xdx . If x = 2, u = 5. If x = 0, u = 1. 12u

!du1

5

! = 12 ln u5

1=12(ln 5 " ln1) = 1

2ln 5

d. Let u = !x2 !!"!!du = !2xdx . If x = 2, u = 4. If x = 0, u = 0. ! 32eu

0

!4

" du = 32 eu!40

" du = 32 eu0

!4=3

2! 32e!4

115. a. x

15=42

60!!!!!x = 10.5 cm3

b. V = 13! r3 !!"!!dV = ! r2dr 15 = ! (1.7)2dr !!"!!dr = 15

! (1.7)2= 1.652 cm/sec

c. Only if it is chocolate. 116. !g (x) = 1

!f (g(x))!!"!! !g (3) = 1

!f (g(3))=

1!f (1)

=12


117. The average rate is the average slope. y(8) = 12(8)2 + 2(8) +1 = 49 ,

y(0) = 12(0)2 + 2(0) +1 = 1. m = 49!1

8!0= 6

The slope is !y (t) = t + 2 . t + 2 = 6!!!!!t = 4 118. a. This is !y at x = 6 if y = ln(x ! 4) . !y = 1

x"4=

1

6"4=1

2

b. This is !y at x = 25 if y = 2x !1 . !y = 12 2x"1

=1

2 49=1

14

c.

limx!"

sin2 xx

= limx!"

sin xx

!i! limx!"

(sin x) = 0 because limx!"

sin x

x= 0

d. 3x > ex since 3 > e . Therefore limx!"

3x +7x

ex+10x

!" . 119. a. y = 4 ! x

3 (the slope is ! 4

12= !

1

3)

b. y22

!dx = (4 " x3)2dx

c. 12

(4 ! x3)2dx = 1

!2(!3)

(4!x 3)3

30

12

"12

0= ! 1

2(4 ! x

3)3

12

0= !0 + 1

243 = 32 cm3

120. a. !

2

x

2( )04

"2

dx b. 9 ! x2( ) 9!x22"#$

%&'0

3

( !dx

c. y2( )2

dy !y

2! 4y !16( )

4

8

"04

"2

dy 121. a. For a given y, y = 0.05(x + 50)2, 20y = (x + 50)2, 20y = x + 50, x = 20y ! 50 , so

the length is 2 ! (50 " 20y ) = 100 " 4 5y , so a slice has volume (100 ! 4 5y )2dy . b. (100 ! 4 5y )2dy = (10000 ! 800 5y + 80y)dy

0

12

"012

"

= 1000y ! 80 5y3 2

3 2+ 40y2

12

0# 76186 yards3

c. Full volume is (100 ! 4 5y )2dy = 10000y ! 1600 53

y3 2 + 40y220

00

20

"

= 109333 13

yards3, so there are about 109333 13! 76186 " 33147.5 yards3 left,

33147.5 yrds3

60 yrds3 /day= 552.5 days left, 1.514 years left, so no delay.


122. b. y = h ! hb 2

x = h ! 2hbx or 2h

bx = h ! y, x = b

2!

b2hy

c. [2(b2! b2hy)]2

0

h

" dy = (b ! bh y)2dy0h

" = ! hb #13(b ! b

hy)3

h

0= !0 + h

b# 13#b3 = 1

3b2h

123. b. The length and width of the squares is x. x = !2y + 20

(20 ! 2y)2dy = (400 ! 80y + 4y2 )dy0

10

" = 400y ! 40y2 + 43 y310

0=40003un3

0

10

" 124. !y = "1

(1+x2 )2#2x = "2x

(1+x2 )2 by the Chain Rule.

At x = !1 we have y = 12

and !y = 12

so the tangent line is y = 12(x +1) + 1

2=12x +1 .

At x = 2 , we have y = 15

and !y = "425

, so the tangent line is y = ! 425(x ! 2) + 1

3.

Therefore 12x +1 = ! 4

25(x ! 2) + 1

5, 25x + 50 = !8x +16 +10, 33x = !24, x = !24

33,

y = 12(!2433) +1 = 21

33: (!2433, 2133) " (!0.727, 0.636)

125. a. x ! 4dx = (x!4)3 2

3 2

8

44

8

" = 43 2

3 2=163un2

b. x ! 4dx = x ! 4

c

8

"4c

" dx,(x!4)3 2

3 2

c

4=(x!4)3 2

3 2

8

c,(c!4)3 2

3 2! 0 = 16

3!(c!4)3 2

3 2,

43(c ! 4)3 2 = 16

3, (c ! 4)3 2 = 4, c ! 4 = 43 2, c = 4 + 42 3 # 6.520

c. ! ( x " 4 )2dx4

8

# = ! (x " 4)dx = ! ( x2

2" 4x)

8

44

8

# = ! (32 $ 32 " (8 "16)) = 8! un3 using disks.

d. ! ( x " 4 )2dx = !d

8

#4d

# ( x " 4 )2dx

! ( x2

2" 4x)

d

4= ! ( x

2

2" 4x)

8

d

! (d2

2" 4d " ("8)) = ! (0 " (d

2

2" 4d))

d2

2" 4d + 8 = " d

2

2+ 4d

d2 " 8d + 8 = 0

d =8+ 64"4(8

2= 4 +

32

2= 4 +

4 2

2= 4 + 2 2 $ 6.828

126. A: V = ( 4 ! x )2dx

0

4

" = (4 ! x)dx04

" = (4x ! x2

2)4

0= 16 ! 8 = 8


127. C: ddx(x2 + y2 ) = d

dx(25)

2x + 2y !y = 0

2y !y = "2x

y ' = "xy

ddx( !y ) = d

dx(" x

y)

!!y =y("1)" !y ("x)

y2=

"y+x !y

y2=

"y+x(" x y)

y2

="y2 "x2

y3=

"(x2 +y2 )

y3=

"25

y3

128. For x in (2, 2), !f (x) = 1

4"x2# ("2x) = "2x

4"x2.

For x > 2 or x < !2 , ddxf (x) = d

dxln(x2 ! 4) = 1

x2 !4"2x = !2x

4!x2.

We get the same answer: D. 129. E: !f (x) = " 1

x2" 8x + 7, !!f (x) = "("2) # 1

x3" 8 = 2

x3" 8 , which is zero when

2

x3= 8, 2

8= x

3=14, x = (2!2 )1 3 = 2!2 3 , and undefined at x = 0 it is positive for x in the

interval (0, 2!2 3) . 130. C: 4ydy = (2x ! 3)dx

4ydy = (2x ! 3)dx""2y2 = x2 ! 3x + C

y2 = x2

2! 3xx + C

y = x2

2! 32x + C

2 = 12!32+ C

4 = !1+ C

C = 5

y = x2

2!32x + 5, f (2) = 2 ! 3+ 5 = 2

131. E: !y = 4x3 + 9x2, !y (0) = 0 , so the normal line is vertical; x = 0 . 132. The real length is 1+ (2x)2dx ! 33.637

"4

4

# , so hopefully students will get answers between 30 and 35, for example, by using secants.

133. The more secants that are used, the better they approximate the curve, so that the estimated

arc length becomes more accurate.

134. a. (! x)2 + (! yi )2 b. (! x)2 + (! yi )2i=1

n

!


135. a. As ! x! 0 , the secant line approaches a line tangent to f (x) and xi , so the slope of the secant line approaches the slope of f (x) .

b. dydx

at xi; !f (xi )

c. ! yi! x

! "f (xi ), ! yi ! ! x # "f (xi ) ! ! x #2xi (since ddx (x2 ) = 2x )

d. arc length ! ! x2 + (2 ! x " xi )2i=1

50

# = ( 850 )2+ [2 " 8

50" ($4 + 8i

50)]2

i=0

49

#

= ( 425

)2 + ( 825

)2($4 + 4i25

)2

i=0

49

# ! 33.646 units

136. a. limn!"

! x2+ (2! x # xi )

2

i=1

n

$ = limn!"

(8n)2 + (16

n# (%4 + 8i

n))2

i=0

n%1

$

b. limn!"

(1+ 4xi2)! x2

i=1

n

# = limn!"

1+ 4xi2

i=1

n

# ! x = 1+ 4x2$4

4

% dx

c. ! 33.637 , so the approximation was off by a little less than 0.01(! 0.009) .

137. a. limn!"

! x2 + ! yi2

i=1

n

# = limn!"

1+ (! yi! x)2

i=1

n

# ! x2 = limn!"

1+ (! yi! x)2

i=1

n

# ! x

b. 1+ ( !f (x))2a

b

" dx 138. a. 1+ cos2 x

0

!

" dx # 3.820 units (by calculator)

b. 1+ (x !1)1/2( )21

9

" dx = 1+ (x !1)19

" dx = x19

" dx = 23 x3/29

1= 54

3! 23= 52

3!units

139. a. !f (x) = ex : 1+ e2x

"2

2

# dx $ 9.010!units (by calculator).

b. !f (x) = 1x

x2 +1

x21

e

" # 2.0035!units (by calculator)


140. ln y = ln ex2 = x2 !!!!!x = ln y,

V = (2 ln y1

e

" )2dy = 4(ln y)dy = 4(y ln y # y)1e

"e

1

= 4(e # e) # 4(0 #1) = 4 !un3

141. xn+1 = xn ! f (xn )"f (xn ) (xn )

"f (x) = ! 1(1+x2 )

#2x = !2x(1+x2 )2

x2 = 1!1 2

1 2= 2

x3 = 3.25, x4 = 5.023,!...

The values keep getting larger, but the function has no root for Newtons Method to approach; only an asymptote of y = 0 .

142. a. Use washers: (! (2x +1)2 " ! (x2 +1)2 )dx

0

2

#

b. Use shells: 2! x(2x +1" (x2 +10

2

# ))dx

c. Use washers: (! (4 " x2 )2 " ! (4 " 2x)2 )dx0

2

#

d. Use shells: 2! (4 " x)(2x +1" (x2 +1))dx0

2

# 143. a. f = x, dg = 5x dx, df = dx, g = 1

ln 5!5x :

x5x dx = xln 55x " 5

x

ln 5dx = x

ln 5!5x " 5

x

(ln 5)2## + C =5x

ln 5(x " 1

ln 5) + C

b. u = ln x +1, du = dxx: ln x+1

xdx

1

e

! = udu =x=1e

! u2

2

e

x=1=(ln x+1)2

2

e

1= 2 " 1

2=32

c. u = x2 ! 4, du = 2xdx : duu=" 12 ln u + C =

12ln x2 ! 4 + C

d. u = x2 + 4, du = 2xdx : xx2 +4

dx =! 12 ln u + C =12ln(x2 + 4) + C


144. a. Vavg = 120 ( (10000 +100t)dt + 10t dt)1020

!010

! = 120 (10000t + 50t210

0) + ( 1

20 ln 1010t

20

10)

= 100250 + 120 ln 10

(1020 "1010 ) # 2.1710 $1018km/hr

b. The Mean Value Theorem does not apply because v is not continuous, but v does equal Vavg at 10t = 2.171 !1018, t = log(2.171 !1018 ) " 18.337sec .

c. aavg = Vfinal!V020seconds = 1020

!1000020

" 5 #1018km/hr/sec d. The Mean Value Theorem does not apply, but a = aavg at

(ln10) !10t = 5 "1018, t = log( 1ln 10

!5 !1018 ) # 18.337sec .

cpm calculus chapter 08 solutions

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