Thomas Calculus 13th [Solutions]

314 Chapter 4 Applications of Derivatives
100. dr cos
d
r 1 sin ( ) C ; at r 1 and 0 we have 1 1 sin( 0) C C 1 r 1 sin ( ) 1
101. dv 1 sec t tan t 1
dt 2 v 2 sec t C;
at v 1 and t 0 we have 1 1 sec(0) C C 1 v 1 sect
1
2
2 2 2
102.
dv
dt
2
8t
csc t
2 2
v 4t cot t 7
2
v 4t cot t C;
at v 7 and
t we have
2
2
7 4 cot
2 2
C
C
7
2
103.
dv
dt
v
t t
3
2
,
1
t > 1
1
3sec t
1
v 3sec t C ; at t = 2 and v = 0 we have
1
0 3sec 2 C C =
104. dv 8 2 1
1
sec t v 8tan t tan t C ; at t = 0 and v = 1 we have 1 8 tan (0) tan(0) C C = 1
dt 2
1 t
1
v 8 tan t tan t 1
105.
2
d y
dy
2 dy
2
2 6x 2x 3 x C
2
1;
at 4 and x 0 we have 4 2(0) 3(0) C
dx
dx
dx
1 C1
4
dy
2 2 3
2 3
2x 3x 4 y x x 4 x C 2;
at y 1 and x 0 we have 1 0 0 4(0) C
dx
2 C2
1
2 3
y x x 4x
1
2
d y dy dy
dy
106. 0 C
2 1;
at 2 and x 0 we have C
dx dx dx
1 2 2 y 2 x C
dx
2;
at y 0 and x 0 we have
0 2(0) C2
C2 0 y 2x
2
107. d r 2 3
2t
dr 2
t C
2 3
1 ; at dr
2
2
1 and t 1 we have 1 (1) C
dt t
dt
dt
1 C1
2 dr t 2
dt
1
1
1
r t 2 t C 2;
at r 1 and t 1 we have 1 1 2(1) C2 C2
2 r t 2t 2 or r 1 2t
2
t
108.
2
3
2
3
2
d s t ds t
dt 8 dt 16
s 4 and t 4 we have
C 1;
at ds 3
dt
3
and t 4 we have
4 4 C
16 2 C2
0 s t
16
2 2 3
3 3(4)
C 3
16 1 C1 0 ds t s t C
dt 16 16 2 ; at
3
109.
3 2
d y
3 2 1
2
d y
d y
6 6 x C ; at 8 and x 0 we have 8 6(0) C
2 1 C1
8 d y 6x
8
2
dx dx
dx
dx
dy 2
dy
2 dy 2
3x 8 x C 2;
at 0 and x 0 we have 0 3(0) 8(0) C
dx
dx
2 C2
0 3x 8x
dx
3 2
3 2 3 2
y x 4 x C 3;
at y 5 and x 0 we have 5 0 4(0) C3 C3
5 y x 4x
5
2
3 2
2
110. d 0 d C
3 2 1;
at d 2 and t 0 we have d 2 d 2 t C
2 2 2;
at d 1 and t 0 we have
dt dt
dt
dt
dt
dt 2
1 2(0) C 1
2
2 C d 1
2
2 2t
t 1 t C
2 dt 2
2 3;
at 2 and t 0 we have
2 1
2
2 0 (0) C
2 3 C3 2 t 1 t 2
2
2
111.
(4)
y sin t cost y cost sin t C 1;
at y 7 and t 0 we have 7 cos (0) sin (0) C1 C1
6
y cost sin t 6 y sin t cost 6 t C 2;
at y 1 and t 0 we have
2
1 sin (0) cos (0) 6(0) C2
C2 0 y sin t cost 6t
y cos t sin t 3 t C 3;
at
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