Thomas Calculus 13th [Solutions]

804 Chapter 11 Parametric Equations and Polar Coordinates
25. For simplicity, we assume that x and y are linear functions of t and that the point ( x, y ) starts at (2, 3)
for t 0 and passes through ( 1, 1) at t 1. Then x f ( t), where f (0) 2 and f (1) 1.
Since slope x 1 2
t 1 0
y
Since slope 1 3
t 1 0
y 3 4 t, t 0.
3,
4.
x f ( t) 3t 2 2 3 t . Also, y g( t), where g(0) 3 and g(1) 1.
y g( t) 4t 3 3 4 t . One possible parameterization is: x 2 3 t,
26. For simplicity, we assume that x and y are linear functions of t and that the point ( x, y ) starts
at ( 1, 2) for t 0 and passes through (0, 0) at t 1. Then x f ( t ), where f (0) 1 and f (1) 0.
0 ( 1)
Since slope x
t 1 0
y
Since slope 0 2
t 1 0
y 2 2 t, t 0.
1,
2.
x f ( t) 1 t ( 1) 1 t . Also, y g( t), where g(0) 2 and g(1) 0.
y g( t) 2t 2 2 2 t . One possible parameterization is: x 1 t,
27. Since we only want the top half of a circle, y 0, so let x 2cos t, y 2 sin t ,0 t 4
28. Since we want x to stay between 3 and 3,let x 3 sin t , then
2
y 9sin t, 0 t
2 2
y (3 sin t) 9sin t , thus x 3sin t,
29.
2 2 2
dy dy
2 2 0 x dy
x y a x y ; let x
.
dx dx y
t dx y
t x yt Substitution yields
2 2 2 2
y t y a y a and x at , t
2
1 t
1 t
30. In terms of , parametric equations for the circle are x a cos , y a sin , 0 2 . Since s , the arc
a
length parametrizations are: x a cos s ,
a
y a sin s
a
, and 0 s 2 0 s
a
2 a is the interval for s.
31. Drop a vertical line from the point ( x, y ) to the x -axis, then is an angle in a right triangle, and from
y
trigonometry we know that tan y x tan . The equation of the line through (0, 2) and (4, 0)
x
is given by y 1 x 2. Thus x tan 1 x 2 x 4 and
2
2 2 tan 1
4 tan
y where
2 tan 1
0 .
2
32. Drop a vertical line from the point ( x, y ) to the x -axis, then is an angle in a right triangle, and from
y
trigonometry we know that tan y x tan . Since
x
2
x cot y cot where 0 .
2
2 2
y x y x ( x tan ) x
2 2
33. The equation of the circle is given by ( x 2) y 1. Drop a vertical line from the point ( x, y ) on the circle
to the x -axis, then is an angle in a right triangle. So that we can start at (1, 0) and rotate in a clockwise
direction, let x 2 cos , y sin , 0 2 .
34. Drop a vertical line from the point ( x, y ) to the x -axis, then is an angle in a right triangle, whose height is y
and whose base is x 2. By trigonometry we have
y
tan y ( x 2) tan . The equation of the circle
x 2
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