problem set - s3.amazonaws.com file13. y = sin (2x + 1) 2 2 + tanx x + 2 (x2 + 2) cos (2x + 1)(2)...
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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
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