Thomas Calculus 13th [Solutions]

Section 11.2 Calculus with Parametric Curves 813
24. (a)
(b)
2 6
x t , y t ,
3 9
x t , y t ,
0 t 1 A
1 1 6 1 7 8
1
y dx t 2t dt 2t dt 1 t 1 0 1
0 0 0 4 0 4 4
0 t 1 A
1 1 9 2 1 11 12
1
y dx t 3t dt 3t dt 1 t 1 0 1
0 0 0 4 0 4 4
2 2
2 2
25. dx
dy
dy
sin t and 1 cos t dx
( sin t) (1 cos t) 2 2 cos t
dt dt dt dt
Length
1 cos 2
sin
sin
0 2 2 cos 2 t
0 1 cos (1 cos ) 2 0 1 cos
2 t
t dt t dt t dt dt
t t 0 1 cos t
(since sin t 0 on [0, ]); u 1 cos t du sin t dt; t 0 u 0, t u 2
2 1/2 1/2
2
2 u du 2 2u
4
0 0
26.
2
2
2
2
2
dx
dy
dy
2 4 2 2
3t and 3t dx
3 t (3 t) 9t 9t 3t t 1 since t 0 on 0, 3
dt dt dt dt
3 2
2
Length 3t t 1 dt;
u t 1 3 du 3 t dt; t 0 u 1, t 3 u 4
0
2
4 1/2 3/2
4
3 u du u (8 1) 7
1 2
1
27.
2 2
dx dy 1/2 dy 2 2
t and (2t 1) dx
t (2t 1) ( t 1) | t 1| t 1 since 0 t 4
dt dt dt dt
Length
4
0 ( t 1) dt t t
2
(8 4) 12
0
2 4
28.
dx
dt
1/2
(2t 3) and
since 0 t 3 Length
dy
2 dy
2
2 2
1 t dx
(2t 3) (1 t) t 4t 4 | t 2| t 2
dt dt dt
3
( t 2) dt t 2t
21
0 2
0
2
2 3
29.
2 2
dx
dy
dy
2 2 2 2 2 2
8t cos t and 8t sin t dx
(8t cos t) (8t sin t) 64t cos t 64t sin t
dt dt dt dt
|8 t | 8t since
0 t Length
2
/2 2
/2
2
8t dt 4t
0 0
30.
dx
2
2
1
dy
dy
sec t tan t sec t cos t sec t cos t and sin dx t
dt sec t tan t dt dt dt
2 2 2 2
(sec t cos t) ( sin t) sec t 1 tan t tan t tan t since 0 t
3
Length /3 /3 sin /3 1
0 tan t dt t dt
0 cos ln |cos t
t
| 0
ln 2
ln 1 ln 2
2
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