Problem Set 56

PROBLEM SET 561. (a) 2x + y = 16

y = 16 - 2x

V = x2y

V = x2(16 - 2x)V = 16x2 - 2x3

Window: ::-::p"lin = [1, ::-::r·ol.3):: = 10'.,.'p"lin = [1, '.,.'p"la::{ = 2~X1

v '" 151.7037 ft3max

Di . 16 ft 16 ft x 16 ftunensrons: 3 x 3 3

(b) V' = 32x - 6x2

o = 2x(16 - 3x)

x = 0 ~, 3

V" = 32 - 12x

v"C36) < 0

16Thus x = - is a maximum"

3V '" 151.7037 ft3max

Di . 16 ft x 16 ft x 16 ftunensions:3 3 3

dy = 4x3 dxdt dtdydt

dydt

= 4(216)(2)

it 4= 1728 um S

S

3. x(t) = t3 + 2t2 - 7t + 4

vet) = 3t2 + 4t - 7

o = 3t2 + 4t - 7

o = (3t + 7)(t - 1)

7t = -"3' 1

120

v

2

7At rest when 1 = - 3,1

7Moving left when - 3 < 1 < 1

Moving right when 1 > 1 and when 1 <73

4. h(t) = -16t2 + 40t + 100

vet) = -32t + 40

o = -32t + 40

t = 1.25 s

Maximum height at h(l.25) = 125 ft

5. ITr 2 sin x dx = -2 cos xeTr/2 = 2-Tr/2

6. r\1I2 dx = ~x3/2]4J1 3 1

16 2- --3 3

143

8. f ~ sin 6 t cos t dt = ~ sin 7 1 + C4 28

f x dx 1 f 4x dx9. 2x2 + 1 ="4 2x2 + 1

= !. In (2X2 + 1) + C4

11. f 4 cos (3t) sin2 (3t) dt = ~ f 3 cos (3t) sin2 (3t) dt

= ~ sirr' (3/) + C

12. f ~os (ax) dx,}1 + sin (ax)

~ f a cos (ax) [1 + sin (ax)]-1/2 dx

~[1 + sin (ax)]112 + Ca

Calculus, Second Edition

13. y =sin (2x + 1) 2

2 + tanxx + 2

(x2 + 2) cos (2x + 1)(2)(x2 + 2)2

sin (2x + 1)(2x) 2

- (x2 + 2)2 + 2 sec x

2 cos (2x + 1) _ 2x sin (2x + 1)x2 + 2 (x2 + 2)2

+ 2sec2x

Problem Set 56

1 2 1 217. -x --y =19 42 1 dy-x - -y - = 09 2 dx

2-xdy _ .2dx-1

-y2

dy = 4xdx 9y

4(~) 18 2~m---=--=-- 9-15 9-15 5

18. y1

f

y' =

y'=

14. y = In \sin x + x\ + csc (2x)

I cosx+ly =. - 2 csc (2x) cot (2x)

smx + x

15. n f(n)(x) f(n)(o)

0 _2x2 - 7x + 2 2

1 -4x - 7 -72 -4 -43 0 0

I \1 I I I II • X

4x2

p(x) = 2 - 7x - -2

p(x) = 2 - 7x - 2X2

19. y

16., "--. x.-/' ~ II,...

n f(n)(x) f(n)(o)

0 cosx 1

1 -sin x 0

2 -cosx -1

3 sinx 0

4 cosx 1

5 -sin x 0

6 -cosx -1

'( r:;-;:) [ . . ~ . . • " •20. f -y13 "" n ler' 1 I...•( •.::.:···S1n (,:':,)

+sin(cos(X»,X,[(13»"" -0.3208

1 r I 221. - "\14 - x dx2 -2

~ [1. 5f"t'"IIrlt.(.f(4-::·:::::::':a ::-:::a -2:a 2)

"" 3.1416

2 4 x6X ~ __ + ...

p(x) = 1 - 2! + 4! 6!

ee X2n

cos X = p(x) = L (_I)n (2n)!n=O

elf

22. f ~3 x + cos x dx.J3

""fnInt([(3AX+cos(X»,X,[(3),e·····:I1: )

"" 603,448.4645

23. Jim f(x) = lim x2

- 9x-+3 x-+3 X - 3

1. (x + 3)(x - 3)rm -'----'-'------'-

x-+3 x - 3

= lim (x + 3) = 6x-+3

Calculus, Second Edition 121

Problem Set 57

In x24. log x = --3 In 3

25. 90° The diameter does not matter, the measure of aninscribed angle is always half the measure of theintercepted arc.

PROBLEM SET 57

. 1. 32

3y = --x + 6

2m=

A = xy

A = X( -%X + 6)A = _2x2 + 6x

2

A' = -3x + 6

3x = 6

x = 2

(2,3)

2. n f(n)(x) f(n)(o)

0 2x3 + 4x2 - 2x + 6 6

1 6x2 + 8x - 2 -2

2 12x + 8 8

3 12 12

4 0 0

8x2 12x3p(x) = 6 - 2x + - + --

2! 3!

p(x) = 6 - 2x + 4x2 + 2x3

3. n f(n)(x) f(n)(o)

0 eX 1

1 eX 1

2 eX 1

3 eX 1

x2 x3p(x) = 1 + x + - + - + ...

2! 3!

122

4. x2 + y2 = 25

dx dy2x- + 2y- = 0dt dt

dx -y dy- =dt x dtdx -3- = -(-3)dt 4

dx 9 units- = ---dt 4 s

5. (a) V = x(0.3 - 2x)2

Window: ::<:P'Iin = [I, ::<:P·I.:t>:: = 0. 15,'/P'I in = 0, •.•.'nax = t1. ~305

Size of square = 0.05 m

v = 0.002 m3max

(b) V = x(0.09 - 1.2x + 4x2)V = 0.09x - 1.2x2 + 4x3

V' = 0.09 - 2.4x + 12~

o = 12x2 - 2.4x + 0.09

2.4 ± .)5.76 - 4.32

24x = 0.15,0.05

x =

x = 0.15 is a minimum.

x = 0.05 is a maximum.

Size of square = 0.05 m

V = 0.002 m3max

6. s(t) = -12t + t3

vet) = -12 + 3t2

3t2 = 12

t2 = 4

t = -2,2

Calculus, Second Edition