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.,Problem Set 35
19. Let '.,.'1 =;:.:: 2 ::. in':: 1/;:'::).
x-.03:-.02UJi..01.02.03:
X=0
20. eX = y
x = Iny
21. (sin -x{ cos (~ - x)][ see? (~ - x)]
= -sin x sin x csc2 x = -sin2 x csc2 x = -1
In x22. y = log x = --
2 In 2
23. y2 = 4x2 + 16 represents a hyperbola.
Let ',,·'1=..f'::4::<2+1E,) and '.,.'2= -V1.
\V/1\
Vertices: (0,4), (0, -4)
24. 7!4! 3!
= 7 . 5 = 35 ways
125. mLD = -x2
1mLB = -y2
1mLCED = mLB + mLD = -(x + y)2
72
-11
PROBLEM SET 351. L = w
SA = 2Lw + 2wH + 2LH
100 = 2L2 + 2LH + 2LH
100 = 2L2 + 4LH
4LH = 100 - 2L2
H = 25L-1 - .!.L2
v = LwH
= L(L) (25C1- ±L )
= (25L - ~L3 ) cnr'
2. S = S ektt 0
60 = 50ek
6 k- = e5
k = In 1.2
S = 50e(ln 1.2)(6)6
= 50(1.2)6 = 149.2992 fathoms/s
3. f 3 sin x dx = 3 f sin x dx = -3 eos x + C
4. f 2t-l12 dt = '2 f rll2 dt = 2(2tIl2) + C
= 4t1/2 + C
f 1 113 1 f V3 -_ -21(-43U4l3) + C5. "2u du ="2 u du --
= ~U4/3 + C8
6. f 3x dx = 3 f x dx = 3( ±x 2 ) + C = ix 2 + C
7.
(2x2) dy + y(4x) + 2y dy = -sin xdx dx
dy (2x2 + 2y) = -sin x - 4xydx
dydx = -sin x - 4xy
2x2 + 2y
Calculus, Second Edition
Ii~
8. u2 + v2 = 2uv
du dv dv du2u- + 2v- = 2u- + 2v-
dt dt dt dt
du dv dv- (2u - 2v) = 2u - - 2 v -dt dt dt
dv2(u - v)-du = dtdt 2(u - v)
du dv- --dt dt
9. y2 _ x2 = 1
dy2y - - 2x = 0dx
dy x- =-dx y
dyl _ Q _ 0dx (0,1) 1
y = 1
10. The graph of y = .J(x + 2)2 crosses the x-axis atx = 0 and touches the x-axis at x = -2. When x islarge and positive, y is large and positive. When x islarge and negative, y is large and negative.
The correct choice is A.
110/\ 0'(\", n 0'(\12. s(t) = tI/3 - 2tI12 + 3t-I
s'(t) = _!'r2l3 - r1l2 - 3t-23
s"(t) = _3..t-513 + _!'r3l2 + 6t-39 2
s"(2) = _3.. (2)-513 + !(2)-312 + 6(2)-39 2
s"(2) "" 0.8568
ex+Llx _ eX1im
Llx-tO Llx- !!.-ex = eX- dx13.
1. x2 + X - 61m =
x-t2 X - 2
= 1im (x + 3) = 5x-t2
1im --'-(x_+_3----')----'-.(x_-_2---=-)x-t2 X - 2
14.
Calculus, Second Edition
Problem Set 35
15. 1. x + 11m ---
x~oo X
1 +!= 1im x..
x-t~ 111 + 0
1
16. Y
-2I ! I I I •• X
(_00,00)
17. 82x-I = 4(23)2x- 1 = 22
26x-3 = 22
6x-3=2
5x = -
6
18. =lJ 2 0 1 k~ -2 2 -32 -2 3 [ill
k - 3 = 0
k = 3
19. Y
t Jf
x2 3
20. g(x) = -3 + 2f( x - ~)
= -3 + 2 sin (x - ~)Y
x---A-1
-2Jr -3 I {-7r .11"1 < I }
-5
73
Problem Set 36
21. 2 sin2 x - 3 sin x + 1 = 0
(2 sin x - l)(sin x-I) = 0PROBLEM SET 36
1. Circumference of circle:
C = 2,.r = 20,. em1sinx =
2sin x = 1
x =,. 5,.6'6
Circumference of cone:
2,. - x---(20,.) = 10(2,. - x) cm2,.
x =,.2
I" .::...f..:....(x_+_h~)_-~f--,,(x....:..)22. f'(x) = imh--+O h
= lim [2(X + h)3 + 3(x + h) - 4 - 2x3h--+O h
+ -3xh+ 4J
= lim [2x3 + 6x2h + 6xh2 + 2h3 + 3x + 3hh--+O h
+ ~4 - 2x3
h- 3x + 4]
= lim 6x2h + 6xh2 + 2h3 + 3hh--+O h
= lim (6x2 + 6xh + 2h2 + 3) = 6x2 + 3h--+O
Radius of cone:
2m = 10(2,. - x)
r,. = 5(2,. - x)
10,. - 5xr =
12. y = -x3 - x
3
y' = x2 - 1
o = x2- 1
o = (x + l)(x - 1)
23. x2 - 2x + y2 + 4y = 4
(x - 1)2 - 1 + (y + 2)2 - 4 = 4
(x - I? + (y + 2)2 = 9
Critical numbers: x = -1,1
This represents a circle centered at (1, -2).
Y1=-2+(9-(X-l)~)Y~=-2-(9-(X-l)~) Local maximum: (-1, ~)
Local minimum: (1, -~ )
"""'-f ••
\U3. f(x) = x3 - 2.x2 + 6x + 3
2
f'(x) = 3x2 - 9x + 6
o = 3(x2 - 3x + 2)o = (x - 2)(x - 1)
Critical numbers: x = 1, 224. A=
A =
52 Local maximum: (1,~1)
Local minimum: (2, 5)
74 Calculus, Second Edition