Thomas Calculus 13th [Solutions]

Section 11.2 Calculus with Parametric Curves 811
11.
3
t x sin , y 1 cos 1 1 1 ; dx 1 cos t,
3 3 3 3 2
3 2 2 dt
dy dy dy/
dt sin t
sin t
dt dx dx/ dt 1 cos t
dy
dx t
3
sin
1 cos
3
3
3
2
1
2
3; tangent line is
1
3 3
y 3 x y 3x
2;
2 3 2 3
1
2 2
1 1 cos t 1
2 2 2 2
dy (1 cos t)(cos t) (sin t)(sin t) d y dy / dt d y
dt (1 cos t) 1 cos t dx dx/ dt 1 cos t (1 cos t)
dx
t
3
4
12.
t
x cos 0,
2 2
2
y 1 sin 2; dx sin t,
2 dt
dy
cot 0; tangent line is y 2;
dx t
2
dy
dy cos t
cos t
dt dx sin t
cot t
2 2
2
csc t t
2 2
dy 2 d y 3 d y
csc t
csc 1
dt
dx sin t
dx
t
2
13. t 2 x 1 1 , y 2 2; dx 1 ,
2 1 3 2 1 dt 2
( t 1)
y 9x
1;
2 3 2 3
dy 4( t 1) d y 4( t 1) d y 4(2 1)
dt 3 2 3 2 3
( t 1) dx ( t 1) dx
t 2
(2 1)
2 2
dy 1 dy ( t 1) dy (2 1)
dt ( t 1) dx ( t 1) dx
t 2 (2 1)
2 2 2
108
9;
tangent line is
14.
0
t 0 x 0 e 1,
0
y 1 e 0; dx t
1 e ,
dt
y 1 1
2 x 2 ;
2 2
dy t
t
0
e d y e d y e
dt t
t
0
1 e dx 1 e dx
t 0 1 e
2 2 3 2 3
dy t dy t dy
0
e e e 1 ; tangent line is
dt dx t
0
1 e dx
t 0 1 e 2
1
8
3 2 2 2
15. x 2t 9 3x dx 4t 0 3x dx 4 t dx 4t
;
dt dt dt 2
3x
3 2 2
2y 3t 4 6y dy 6t
0
dt
dy 6t t
dx ; thus
2 2
6 y y
x
1;
t
y
2
2
t 3x
2
4t
2 2
3x
2
dy dy/ dt 3x
;
dx dx/ dt y ( 4 t) 4 y
3 2 3 3
t 2 2y 3(2) 4 2y 16 y 8 y 2; therefore
3 2 3 3
t 2 x 2(2) 9 x 8 9 x 1
dy
dx t
2
2
3(1) 3
2
4(2) 16
16.
1/2
1 1 1/2
x 5 t dx
1
2 5 t t
dt
2
;
4 t 5 t
( t 1) dy 1 dy
y
2 1 2 y t
dt y 2 dt ( t 1)
; thus
t 2t t 2 t
2 1 2y t 5
1 t
dy
dx t
4
t
;
1
t
y ( t 1) t y ( t 1) dy 1
dt 2
t
dy
1 2 y t
dt 2t t 2 t
dx 1
dt 4 t 5 t
dy 1 2 y t 4 t 5 t
dx
2 t ( t 1) 1
t 4 x 5 4 3; t 4 y 3 4 y 2 therefore,
3
2 1 2
2
3
4 5 4
10 3
1 4 9
1/2
3/2 2 1/2 1/2
x 2x t t dx 3x dx 2t 1 1 3x dx 2t 1 dx 2t
1 ; y t 1 2t y 4
dt dt dt dt 1 3x
17.
1/2
dy 1 1/2 1 1/2 dy dy y dy
t 1 y ( t 1) 2 y 2t y 0 t 1 2 y t 0
dt 2 2 dt dt 2 t 1
y dt
Copyright
2014 Pearson Education, Inc.