Calculus - Chapter 9 Parametric Equations and Polar...

Chapter 9 Parametric Equations and Polar Coordinates
302
Chapter 9
Section 9.1
5.
– 4.7
– 3.1
4.7
3.1
2 2
2 2
2 2 2 2
2 2
3cos
3sin
9cos
9sin
9(sin cos )
9
x t
y t
x t
y t
x y t t
x y
= 

=


= 

=


+ = +
+ =
7.
– 4.7
– 5.1
4.7
1.1
2 2
2 2
2 2 2 2
2 2
1 2cos
2 2sin
1 2cos
2 2sin
( 1) 4cos
( 2) 4sin
( 1) ( 2) 4(sin cos )
( 1) ( 2) 4
x t
y t
x t
y t
x t
y t
x y t t
x y
= + 

= − +

− =


+ =


− =


+ = 

− + + = +
− + + =
9.
– 4.7
– 3.1
4.7
3.1
1 2
3
3
1 2
3
2
1
3
3 3
2 2
x t
y t
y
t
y
x
y x
y x
= − +


=

=
 
= − +
 
 
= +
= +
11.
– 4.7
0
4.7
6.2
2
2
2
1
2
1
( 1) 2
2 3
x t
y t
t x
y x
y x x
= + 

= +

= −
= − +
= − +
13.
– 4.7
– 3.1
4.7
3.1
2
2
2
1
2
2
1
1 1
2 4
x t
y t
y
t
y
x y

= −

=

=
 
= − = −
 
 

Chapter 9 Parametric Equations and Polar Coordinates
303
15.
– 2.7
0
6.7
6.2
2
2
2 2
1
1
1 ( 1)
2, 1
x t
y t
y x t t
y x x

= −


= +


− = + − −
= + ≥ −
17.
– 4.7
– 3.1
4.7
3.1
19.
– 9.4
– 3.1
9.4
9.3
21.
– 2.35
– 1.55
2.35
1.55
23.
– 2.35
– 1.55
2.35
1.55
25.
– 15
– 10
15
10
27.
– 12
– 8
12
8
29.
– 6
– 4
6
4
31.
– 2.35
– 1.55
2.35
1.55
33.
– 2.7
– 1.1
6.7
5.1
35. C. Since
2
( 1) x y + = , with x ±±²³±WKLV±LV±SDUW±RI
an upward-opening parabola.
37. B. x is bounded below by –1. y is bounded below
by –1 and above by 1. The graph will have x -
intercepts at t = n ±RU±
2
( ) 1. x n = π −
39.
A.
x
and
y
both oscillate between –1 and 1, but
with different periods.

Chapter 9 Parametric Equations and Polar Coordinates
304
41. Use 0 ± t ±²³±ZLWK± t = 0 corresponding to
(0, 1); then x = 0 + bt, y = 1 + dt. With t = 1,
x = 3 = 0 + b , y = 4 = 1 + d . So b = 3, d = 3, the
equations are x = 3 t , y = 1 + 3 t .
43. Use 0 ± t ±²³±ZLWK± t = 0 corresponding to
(–2, 4); then x = –2 + bt , y = 4 + dt . With t = 1,
x = 6 = –2 + b , y = 1 = 4 + d . So b = 8, d = –3, the
equations are x = –2 + 8 t , y = 4 – 3 t .
45.
2
, 1 x t y t = = + for 1 ± t ±´µ
47. Since x decreases from 2 to 0, let
2 2
2 , 2 (2 ) 4 2 x t y t t t = − = − − = − + − for
0 ± t ±´µ
49. Use cos , sin x a b t y c d t = + = + , with
0 ± t ±´ µ±7KH±FHQWHU±LV±DW±¶´³±²·³±WKHUHIRUH
a = 2 and c = 1. The radius is 3, therefore
b = d = 3. So the equations are x = 2 + 3 cos t ,
y = 1 + 3 sin t .
51. Solve
2
1
1 4
t s
t s
= +


− = −

s = t – 1, so
2 2
1 4 ( 1) 6 0 3 t t t t t − = − − ⇒ + − = ⇒ = − or
t = 2. So either t = –3 and s = –4, in which case
x = –3 and y = 8, or t = 2 and s = 1, in which case
x = 2 and y = 3. So the intersection points are (–3,
8) and (2, 3).
53. Solve
2
3 1
2
t s
t s
+ = +


= −

s = t + 2, so
2 2
2 ( 2) 0 0 t t t t t = − + ⇒ + = ⇒ =
or t = –1. So either t = 0 and s = 2, in which case x
= 3 and y = 0, or t = –1 and s = 1, in which case x
= 2 and y = 1. So the intersection points are (3, 0)
and (2, 1).
55. Solve 100 t = 500 – 500( t – 2);
100 t = 1500 –
15
500
6
t t ⇒ = . So at
5
2
t = , the
missiles have the same x -coordinate.
For the target missile, the y -coordinate is
2
15 15
80 16 100
6 6
   
− =
   
   
, and for the interceptor
missile, the y- coordinate is
2
15 15
208 2 16 2 100
6 6
   
− − − =
   
   
. Since the
y -coordinates are equal when x -coordinates are
equal, the interceptor missile hits its target.