Thomas Calculus 13th [Solutions]

Section 4.8 Antiderivatives 315
2 2
y 1 and t 0 we have 1 cos (0) sin (0) 3(0) C3 C3
0 y cos t sin t 3t
3
3
y sin t cos t t C 4;
at y 0 and t 0 we have 0 sin (0) cos (0) 0 C 4
3
C4 1 y sin t cos t t 1
(4)
112. y cos x 8sin(2 x) y sin x 4cos (2 x) C 1;
at y 0 and x 0 we have
0 sin(0) 4cos(2(0)) C1 C1 4 y sin x 4cos(2 x) 4 y cos x 2sin(2 x) 4 x C 2;
at
y 1 and x 0 we have 1 cos(0) 2sin(2(0)) 4(0) C2 C2
0 y cos x 2sin(2 x) 4x
2
2
y sin x cos(2 x) 2 x C 3;
at y 1 and x 0 we have 1 sin(0) cos(2(0)) 2(0) C3 C3
0
2 1 2 3
y sin x cos(2 x) 2x y cos x sin(2 x) x C
2 3 4;
at y 3 and x 0 we have
1 2 3 1 2 3
3 cos(0) sin(2(0)) (0) C
2 3 4 C4
4 y cos x sin(2 x) x 4
2 3
113. m y 3 x
1/2 3/2
3x y 2 x C;
at (9, 4) we have
3/2
4 2(9) C C
3/2
50 y 2x
50
114. (a)
2
d y
dx
2
dy 2
2
6x 3 x C 1;
at y 0 and x = 0 we have 0 3(0) C
dx
1 C1
0
y x 3
C 2 ; at y = 1 and x = 0 we have 3
C2 1 y x 1
(b) One, because any other possible function would differ from
of the initial conditions
dy
dx
2
3x
3
x 1 by a constant that must be zero because
115.
dy
dx
4 1/3 4 1/3 4/3
1 x y 1 x dx x x C ; at (1, 0.5) on the curve we have
3 3
4/3 4/3
0.5 1 1 C C 0.5 x x 1
2
116.
dy
dx
2
x 1 y ( x 1) dx x x C ; at ( 1, 1) on the curve we have
2
2 2
( 1)
1 ( 1) C C 1 y x x 1
2 2 2 2
dy
117. sin x cos x y (sin x cos x) dx cos x sin x C ; at ( , 1) on the curve we have
dx
1 = cos( ) sin( ) + C C = 2 y = cos x sin x 2
118.
dy
dx
we have
1 1 1/2 1 1/2 1/2
sin x x sin x y x sin x dx x cos x C ; at (1, 2) on the curve
2 x
2 2
1/2
2 1 cos (1) C C 0 y x cos x
119. (a)
ds
2
2
9.8t 3 s 4.9t 3 t C ; (i) at s = 5 and t = 0 we have C = 5 s 4.9t 3t
5;
dt
displacement = s(3) s(1) = ((4.9)(9) 9 + 5) (4.9 3 + 5) = 33.2 units; (ii) at s = 2 and t = 0 we have
2
C = 2 s 4.9t 3t 2; displacement = s(3) s(1) = ((4.9)(9) 9 2) (4.9 3 2) = 33.2 units;
2
(iii) at s s 0 and t = 0 we have C s0 s 4.9t 3 t s0;
displacement s(3) s(1) ((4.9)(9) 9 s0 ) (4.9 3 s0
) 33.2 units
(b) True. Given an antiderivative f(t) of the velocity function, we know that the bodys position function is
s = f(t) + C for some constant C. Therefore, the displacement from t = a to t = b is
(f(b) + C) (f(a) + C) = f(b) f(a). Thus we can find the displacement from any antiderivative f as the
numerical difference f(b) f(a) without knowing the exact values of C and s.
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