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10
PARAMETRIC EQUATIONS AND POLAR COORDINATES
10.1
1.
2.
3.
Curves Defined by Parametric Equations
−1
0
1
2
0
1
0
−3
−3
0
1
0
=1−
2
=2 −
,
2
−1 ≤ ≤ 2
,
−2
−1
0
1
2
−10
−2
0
2
10
6 = 3+ ,
3 =
−
−
2
0
2
−
−
2+1
0
2+1
−1 0 1 0 = + sin , = cos , − ≤ ≤
−2 2
−2
5 39 −2
−
=
5.
= 2 − 1, (a)
−1
0
−1
1
+2
=
1 2
−1
1 −1
1 72 −1
2 14 + , =
4.
AL
3 6 22 + 2, −2 ≤ ≤ 2
+1
1 37
+1
−1
1
1 37 − , −2 ≤ ≤ 2 +1
−4
−2
0
2
4
−9
−5
−1
3
7
−1
0
1
2
3
1 72
2 −2
+2
2 14 2
−2
5 39
=2 −1
(b)
=
864
6.
¤
1 2
+1 =
= 3 + 2,
(b)
1 2
1 2
+
⇒
+1 1 2
+1 =
1 2
= 1 4
1 4
+
+ 12 , so
+1
⇒
=
1 4
+
5 4
= 2 +3 −4
−2
0
2
4
−10
−4
2
8
14
−5
−1
3
7
11
⇒
=3 +2
=2 +3 =2
=
2
− 3,
(b)
−2
3 = −
1 3
2 3
+3 =
−1
1
3
6
−2
−2
6
−1
1
3
5
⇒
2 3
= −
4 3
−
1 3
2 3,
so
⇒
+3
=
2 3
+
5 3
≤3
−3
= +2
− 2, so
=
=
2
− 3 = ( − 2)2 − 3 =
=
2
− 4 + 1, −1 ≤
= 1 − cos ,
= sin ,
⇒
−3 ≤
= + 2,
(a)
8.
2 =
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
(a)
7.
⇒
2
− 4 +4 − 3
⇒
≤5
0≤
≤2
(a) 0
(b)
3
2
2
0
1
0
−1
0
0
1
2
1
0
= 1 − cos
[or
= sin , 2
2
− 1 = − cos ]
+ ( − 1)2 = (sin )2 + (− cos )2
⇒
⇒ 2
+ ( − 1)2 = 1.
As varies from 0 to 2 , the circle with center (0 1) and radius 1 is traced out.
9.
=
√
,
= 1−
(a) 0
1
2
3
4
0
1
1 414
1 732
2
1
0
−1
−2
−3
AL
(b)
=
√
⇒
⇒
2
=
= 1− = 1−
2
. Since ≥ 0,
= 1−
So the curve is the right half of the parabola
2
≥ 0.
.
SECTION 10.1
10.
=
2
,
3
=
(a)
(b)
11. (a)
−2
−1
0
1
2
4
1
0
1
4
−8
−1
0
1
8
⇒
3
=
1
= sin
=
,
+
2
= cos
=
1 2
1 2
,− ≤
+ cos2
≤ 0 and 0 ≤
≤
2
3
=
1 2
2
≤ 1. For 0
≤
≥ 0.
≤ 0, we have ≤1
, we have 0
2
1
= cos2
≤
+ sin2
= 1
ALE (b)
. ⇒
4
2
+
1
= 1 ⇒
2
4
22
= 1, which is an equation of an ellipse with
-intercepts ± 12 and -intercepts ±2. For 0 ≤ ≥ 0 and 0 ≤
and 2
13. (a)
∈ R,
2
+
2)2
≥
∈ R,
(b)
2
1 2
.
.
= 1. For − ≤
= 2 sin , 0 ≤
cos ,
(2 )2 +
(1
2 3
=
≥ 0. The graph is a semicircle.
and 1
12. (a)
1
2
=
2
= sin2
−1 ≤
⇒
3
2 2
CURVES DEFINED BY PARAMETRIC EQUATIONS
= sin
≤ 2. For
2
≤
≤
2, we have ≥ − 21
, we have 0
≥ 0. So the graph is the top half of the ellipse.
= csc , 0
.
= csc =
1
=
1
.
(b)
¤
865
2
For 0
, we have 0
2
1 and
the portion of the hyperbola
14. (a)
866
¤
−2
= ( )−2 =
=1
−2
with
2
=1
1. Thus, the curve is
for
1.
0 since
=
⇒
= ln
(b)
√
=
, so
+1
= sinh ,
2
= ( )2 =
2
.
(b)
−
⇒
2
2
−
2
= cosh2
− sinh2 = 1.
(b)
= 1.
= sec , −
= sec2
−
≤ 0, we have
2
and 1
2
⇒
1 + tan2
0
(b)
2.
1+
=
≥ 0 and
2
⇒
=
≥ 1. For 0
2
− 1. For
=
2 to 0, the point (
approaches (0 1) along the parabola. As increases from 0 to point (
AL
2, we have
. Thus, the curve is the portion of the parabola
in the first quadrant. As increases from −
2
−1
) 2, the
) retreats from (0 1) along the parabola.
−5 = 5 + 2 cos −5 2
(b)
= +1
= cosh
= tan2 ,
18. (a)
2
= cosh ≥ 1, we have the upper branch of the hyperbola
Since 2
⇒
=
⇒ = 2 − 1. √ √ 2 − 2. The curve is the part of = − 1 = ( 2 − 1) − 1 = √ the hyperbola 2 − 2 = 2 with ≥ 2 and ≥ 0. =
17. (a)
19.
.
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
15. (a)
16. (a)
=
sin
2
+
,
= 3 + 2 sin −3 2
⇒
cos
=
2
−3 , sin
=
2
2
.
2
cos ( ) + sin ( ) = 1 ⇒
2
= 1. The motion of the particle takes place on a circle centered at (5 3) with a radius 2. As goes
−5 2
from 1 to 2, the particle starts at the point (3 3) and moves counterclockwise along the circle
2
+
−3 2
2
= 1 to
(7 3) [one-half of a circle]. −1 20.
= 2 + sin ,
= 1 + 3 cos
⇒
sin =
− 2, cos =
3
2
.
sin
2
2
+ cos
= 1
⇒
( − 2) +
−1 3
2
= 1.
The motion of the particle takes place on an ellipse centered at (2 1). As goes from
2 to 2 , the particle starts at the point
(3 1) and moves counterclockwise three-fourths of the way around the ellipse to (2 4).
SECTION 10.1 2
21.
= 5 sin ,
= 2 cos
⇒
sin =
5
, cos =
2
.
sin
CURVES DEFINED BY PARAMETRIC EQUATIONS 2
2
+ cos
= 1
⇒
867
2
+
5
¤
= 1. The motion of the
2
particle takes place on an ellipse centered at (0 0). As goes from − to 5 , the particle starts at the point (0 −2) and moves clockwise around the ellipse 3 times. 22.
= cos2 = 1 − sin2
= 1−
2
. The motion of the particle takes place on the parabola
= 1−
2
. As goes from −2 to
− , the particle starts at the point (0 1), moves to (1 0), and goes back to (0 1). As goes from − to 0, the particle moves to (−1 0) and goes back to (0 1). The particle repeats this motion as goes from 0 to 2 . 23. We must have 1 ≤
≤ 4 and 2 ≤
≤ 3. So the graph of the curve must be contained in the rectangle [1 4] by [2 3].
24. (a) From the first graph, we have 1 ≤
≤ 2. From the second graph, we have −1 ≤
≤ 1 The only choice that satisfies
either of those conditions is III. (b) From the first graph, the values of of
cycle through the values from −2 to 2 six times. Choice I satisfies these conditions.
(c) From the first graph, the values of 0≤
cycle through the values from −2 to 2 four times. From the second graph, the values
cycle through the values from −2 to 2 three times. From the second graph, we have
≤ 2. Choice IV satisfies these conditions.
(d) From the first graph, the values of
cycle through the values from −2 to 2 two times. From the second graph, the values of
do the same thing. Choice II satisfies these conditions. 25. When = −1, (
) = (1 1). As increases to 0,
As increases from 0 to 1,
increases from 0 to 1 and
−1. As increases beyond 1, decrease. For
−1,
and
and
both decrease to 0.
decreases from 0 to
continues to increase and
continues to
are both positive and decreasing. We could
achieve greater accuracy by estimating - and -values for selected values of from the given graphs and plotting the corresponding points. 26. When = −1, (
while
) = (0 0). As increases to 0,
increases from 0 to 1,
first decreases to −1 and then increases to 0. As increases from 0 to 1,
decreases from 1 to 0, while
y 1
first increases to 1 and then decreases to 0. We
could achieve greater accuracy by estimating - and -values for selected values of from the given graphs and plotting the corresponding points.
0
t=_1, 1 (0, 0)
t=0 (1, 0) 1
x
_1
27. When = −1, (
) = (0 1). As increases to 0,
increases from 0 to 1 and
decreases from 1 to 0. As increases from 0 to 1, the curve is retraced in the opposite direction with
decreasing from 1 to 0 and
increasing from 0 to 1.
We could achieve greater accuracy by estimating - and -values for selected values of from the given graphs and plotting the corresponding points.
868
¤
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
28. (a)
=
4
− +1 =(
+ 1) −
4
0 [think of the graphs of
=
4
+ 1 and
= ] and
=
2
≥ 0, so these equations
are matched with graph V. √
≥ 0.
2
(b)
=
=
(c)
= sin 2 has period 2
− 2 = ( − 2) is negative for 0 2=
2, so these equations are matched with graph I.
. Note that
( + 2 ) = sin[ + 2 + sin 2( + 2 )] = sin( + 2 + sin 2 ) = sin( + sin 2 ) = ( ), so cycles through the values −1 to 1 twice as
These equations match graph II since (d)
= cos 5 has period 2
5 and
= sin 2 has period , so
takes on the values −1 to 1. Note that when = 0, ( (e)
= + sin 4 ,
=
2
+ cos 3 .
, so the graph will look like the graph of (f )
=
29. Use
30. Use
sin 2 , 4+ 2
= and
1
= ,
1
=
cos 2 . 4+ 2
= − 2 sin
=
3
As → ∞,
= and
2
=
3
−4,
2
will take on the values −1 to 1, and then 1 to −1, before
2
become the dominant terms in the expressions for
and
, but with oscillations. These equations are matched with graph IV. both approach 0. These equations are matched with graph III.
with a -interval of [−
− 4 and
cycles through those values once.
) = (1 0). These equations are matched with graph VI
As becomes large, and 2
has period 2 .
].
= with a -interval of
[−3 3]. There are 9 points of intersection; (0 0) is fairly obvious. The point in quadrant I is approximately (2 2 2 2), and by symmetry, the point in quadrant III is approximately (−2 2 −2 2). The other six points are approximately (∓1 9 ±0 5), (∓1 7 ±1 7), and (∓0 5 ±1 9).
31. (a)
=
1
through
+(
2
2( 2
−
1) 2)
,
=
1
+(
2
−
when = 1. For 0
1)
,0≤ 1,
≤ 1. Clearly the curve passes through is strictly between 2
−
1
1
and
2
and
1( 1
1)
when = 0 and
is strictly between
1
and
2.
For
every value of , 1( 1
1)
and
and 2( 2
Finally, any point (
satisfy the relation
−
1
= 2
−
( −
1 ),
which is the equation of the line through
1
2 ).
) on that line satisfies
− 2 −
1 1
=
− 2 −
1 1
; if we call that common value , then the given
parametric equations yield the point (
); and any (
) on the line between
1( 1
in [0 1]. So the given parametric equations exactly specify the line segment from (b)
= −2 + [3 − (−2)] = −2 + 5 and
= 7 + (−1 − 7) = 7 − 8 for 0 ≤
SECTION 10.1
32. For the side of the triangle from
to
, use (
1)
1
= (1 1) and (
and
1( 1
1)
2( 2
to
2)
2( 2
yields a value of 2 ).
≤ 1.
¤
CURVES DEFINED BY PARAMETRIC EQUATIONS 2)
2
1)
869
= (4 2).
Hence, the equations are = = Graphing
1
+( +(
to
− − 2 2
= 1 + (4 − 1) = 1 + 3 , ) = 1 + (2 − 1) = 1 + . 1 1)
= 1 + with 0 ≤
= 1 + 3 and
triangle from and
1
≤ 1 gives us the side of the
. Similarly, for the side
we use
= 4 − 3 and
= 2 + 3 , and for the side
we use
=1
= 1 +4 .
33. The circle
2
+ ( − 1)2 = 4 has center (0 1) and radius 2, so by Example 4 it can be represented by
= 1 + 2 sin , 0 ≤
≤ 2 . This representation gives us the circle with a counterclockwise orientation starting at (2 1).
(a) To get a clockwise orientation, we could change the equations to
= 1 − 2 sin , 0 ≤
= 2 cos ,
(b) To get three times around in the counterclockwise direction, we use the original equations the domain expanded to 0 ≤
≤2 .
= 2 cos ,
= 1 + 2 sin with
≤6 .
(c) To start at (0 3) using the original equations, we must have = 2 cos ,
= 2 cos ,
= 1 + 2 sin ,
2
≤
≤
2
1
= 0; that is, 2 cos = 0. Hence, =
2
. So we use
.
Alternatively, if we want to start at 0, we could change the equations of the curve. For example, we could use = −2 sin , 34. (a) Let
2
2
= 1 + 2 cos , 0 ≤
= sin2
and
= cos with 0 ≤ 2
2
+
2
2
2
2
≤
.
= cos2 to obtain
= sin and
≤ 2 as possible parametric equations for the ellipse
= 1.
(b) The equations are
= 3 sin and
= cos for ∈ {1 2 4 8}.
(c) As increases, the ellipse stretches vertically.
35. Big circle: It’s centered at (2 2) with a radius of 2, so by Example 4, parametric equations are
= 2 + 2 cos
= 2 + 2 sin
0≤
≤2
Small circles: They are centered at (1 3) and (3 3) with a radius of 0 1. By Example 4, parametric equations are
and
(left)
= 1 + 0 1 cos
= 3 + 0 1 sin
0≤
≤2
(right)
= 3 + 0 1 cos
= 3 + 0 1 sin
0≤
≤2
Semicircle: It’s the lower half of a circle centered at (2 2) with radius 1. By Example 4, parametric equations are = 2 + 1 cos
= 2 + 1 sin
≤
≤2
To get all four graphs on the same screen with a typical graphing calculator, we need to change the last -interval to[0 2 ] in order to match the others. We can do this by changing to 0 5 . This change gives us the upper half. There are several ways to get the lower half—one is to change the “+” to a “−” in the -assignment, giving us
= 2 + 1 cos(0 5 )
= 2 − 1 sin(0 5 )
0≤
≤2
870
¤
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
36. If you are using a calculator or computer that can overlay graphs (using multiple -intervals), the following is appropriate.
Left side:
= 1 and
goes from 1 5 to 4, so use =
15≤
≤4
=
15≤
≤4
=15
1≤
≤ 10
=1 Right side:
= 10 and
goes from 1 5 to 4, so use = 10
Bottom:
goes from 1 to 10 and
= 1 5, so use =
Handle: It starts at (10 4) and ends at (13 7), so use = 10 +
0≤
= 4+
≤3
Left wheel: It’s centered at (3 1), has a radius of 1, and appears to go about 30◦ above the horizontal, so use = 3 + 1 cos
= 1 + 1 sin
6
≤
≤
13 6
6
≤
≤
13 6
Right wheel: Similar to the left wheel with center (8 1), so use = 8 + 1 cos
= 1 + 1 sin
If you are using a calculator or computer that cannot overlay graphs (using one -interval), the following is appropriate. We’ll start by picking the -interval [0 2 5] since it easily matches the -values for the two sides. We now need to find parametric equations for all graphs with 0 ≤ Left side:
= 1 and
≤ 2 5.
goes from 1 5 to 4, so use =1
Right side:
= 10 and
0≤
≤25
=15+
0≤
≤25
goes from 1 5 to 4, so use = 10
Bottom:
=15+
goes from 1 to 10 and
= 1 5, so use = 1 +36
0≤
=15
≤25
To get the x-assignment, think of creating a linear function such that when = 0, = 10. We can use the point-slope form of a line with ( −1 =
10 − 1 ( − 0) 25−0
⇒
1
1)
= (0 1) and (
2
= 1 and when = 2 5, 2)
= (2 5 10).
= 1 +36 .
Handle: It starts at (10 4) and ends at (13 7), so use = 10 + 1 2 (
1
1)
= (0 10) and (
2
2)
= (2 5 13) gives us
0≤
= 4 +1 2 − 10 =
≤ 25
13 − 10 ( − 0) 25−0
⇒
= 10 + 1 2 .
7− 4 (
1
1 ) = (0 4) and (
2
2 ) = (2 5 7) gives us − 4 =
25 −0
( − 0)
⇒
= 4 +12 .
SECTION 10.1
CURVES DEFINED BY PARAMETRIC EQUATIONS
¤
871
Left wheel: It’s centered at (3 1), has a radius of 1, and appears to go about 30◦ above the horizontal, so use = 3 + 1 cos
(
1)
1
= 0
and (
2)
2
+
15
gives us −
5
=
6
= 1 + 1 sin
6
2
5 6
=
+
15
−
13 6 5
6 2
5 6
5 6
( − 0)
0≤
≤ 25
⇒
=
−0
+ 6
. 15
Right wheel: Similar to the left wheel with center (8 1), so use = 8 + 1 cos 37. (a)
=
3
⇒
1 3
=
, so
We get the entire curve
=
+
15
2 3
= 1 + 1 sin
6
=
2
2 3
traversed in a left to
=
.
(b)
right direction.
(c)
= = If
−3 −2
=( =(
0, then
−
)3
−
)2 = (
and
−
[so
=
1 3 2
) =
1 3
=
2 3
0 and
+
= , so
=
−2
=
−2
= cos ,
= sec2
=
cos2
=
1
2.
1 6
≤ 25 , so
=
curve
=
2 3
. 0, then
and
0, the curve never quite
=1
2
traversed in a
Since sec ≥ 1, we only get the
4
=
4 6
=
2 3
≥ 0, we only get the right half of the
6
.
AL
. We get the entire curve
1
=
=
left-to-right direction.
(b)
6
⇒
6
reaches the origin. 38. (a)
0≤
5
Since
],
are both larger than 1. If
are between 0 and 1. Since
15
.
2
≥ 1. We get the first quadrant portion of
parts of the curve
=1
with
the curve when
0, that is, cos
portion of the curve when
(c)
=
,
=
−2
0, and we get the second quadrant
0, that is, cos
= ( )−2 =
−2
0.
. Since
only get the first quadrant portion of the curve
¤
872
are both positive, we 2
=1
.
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
39. The case
and
is illustrated.
2
has coordinates (
has coordinates (
+ cos( − )) = (
[since cos( − ) = cos cos
+ sin
) as in Example 7, (1 − cos ))
= − cos
sin
, so
has
− sin( − ) (1 − cos )) = ( ( − sin ) (1 − cos ))
coordinates (
[since sin( − ) = sin parametric equations
− cos sin
cos
= ( − sin ),
= sin
+ sin( − )
−
2,
. As in Example 7,
+ cos( − )) = (
diagram) has coordinates (
. Again we have the
= (1 − cos ).
40. The first two diagrams depict the case
(
−2
and
cos ). That i
−
has coordinates (
cos ), so a typical point
has coordinates (
), where
−
=
). Now
(in the second
of the trochoid has coordinates
sin
and
=
−
cos . When
= , these equations agree with those of the cycloid.
41. It is apparent that
= =
= cos and cos and
sin2
=
=(
2
+ cos2
has coordinates ( cos
42.
( sec
0). It follows tha
= (2 cot
43.
Then ∠
=
=
= sin . Thus, the parametric equations are
⇒
= = 1 =
2
. From the diagram,
we rearrange: sin cos2
2
+
=(
2
sin ). Since
2
is a right angle and ∠
2
) . Adding the two
. Thus, we have an ellipse.
has coordinates ( sec is
is a right triangle and
sin ). Thus, the parametric equations are
= 2 cot . Le
= , so
,∆
= 2 sin
has coordinates
= sec ,
= sin .
= (0 2 ). and
(2 sin ) sin ). Thus, the -coordinate of
= 2 sin2 .
44. (a) Let
AL
⇒
=
is perpendicular to
2 ), so the -coordinate of
= ((2 sin ) cos is
=
= sin . To eliminate
) and cos
equations: sin2
and
be the angle of inclination of segment
cos . Then
=
2
.
Le
= (2 0). Then by use of right triangle
we see that
= 2 cos . (b)
Now
=
=2
= 1 cos
− − cos

=2
1 − cos2 cos
=2
sin2 cos
= 2 sin
tan
So
has coordinates
= 2 sin
· cos = 2 sin2
tan
and
= 2 sin
tan
SECTION 10.1
45. (a)
· sin
= 2 sin2
tan .
CURVES DEFINED BY PARAMETRIC EQUATIONS
¤
873
There are 2 points of intersection: (−3 0) and approximately (−2 1 1 4).
(b) A collision point occurs when
=
1
2
and
1
=
2
for the same . So solve the equations:
3 sin = −3 + cos
(1)
2 cos = 1 + sin
(2)
From (2), sin = 2 cos − 1. Substituting into (1), we get 3(2 cos − 1) = −3 + cos cos = 0
⇒
=
occurs when =
2
3 2
or
2
. We check that =
satisfies (1) and (2) but =
2
2
⇒
5 cos = 0 ( )
does not. So the only collision point
, and this gives the point (−3 0). [We could check our work by graphing
functions of and, on another plot, 3 2
pairs of graphs intersect is =
1
and
2
⇒
1
and
2
together as
as functions of . If we do so, we see that the only value of for which both
.]
(c) The circle is centered at (3 1) instead of (−3 1). There are still 2 intersection points: (3 0) and (2 1 1 4), but there are ⇒
no collision points, since ( ) in part (b) becomes 5 cos = 6 = 30◦ and
46. (a) If
0
=
1 2 (9
8)
2
= 250 − 4 9 2 . √
250
49
≈ 51 s. Then
The formula for
= 250
2
with equality when =
= 0 when = 0 (when the gun is fired) and again when
3
49
≈ 22,092 m, so the bullet hits the ground about 22 km from the gun.
−
= −4 9
250
2
−
125
250
+
49
125 2
cos )
⇒
1252
= −4 9
49
s, so the maximum height attained is
125 2
−
125 2
+
49
125 2
≤
49
1252 49
≈ 3189 m.
49
As
OT 0
+
49
49
=(
√ 3 and
250
49
(c)
1.
is quadratic in . To find the maximum -value, we will complete the square:
= −4 9
(b)
6 5
= (500 cos 30◦ ) = 250
= 500 m s, then the equations become
= (500 sin 30◦ ) −
cos =
= 0
cos
(0◦
90◦ ) increases up to 45◦ , the projectile attains a
greater height and a greater range. As
increases past 45◦ , the
projectile attains a greater height, but its range decreases.
. 2
=(
0
sin ) −
1 2
2
⇒
=(
0
sin
) 0
cos
−
2
0
cos
= (tan ) −
2
2
0 2
cos2
,
which is the equation of a parabola (quadratic in ).
874
¤
47.
=
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 2
=
3
−
. We use a graphing device to produce the graphs for various values of with − ≤
the members of the family are symmetric about the -axis. For cusp at (0 0) and for
48.
=2
−4
3
=−
0 the graph crosses itself at
2
≤
. Note that all
0, the graph does not cross itself, but for = 0 it has a
= , so the loop grows larger as increases.
+ 3 4 . We use a graphing device to produce the graphs for various values of with − ≤
Note that all the members of the family are symmetric about the -axis. When of even degree, but when = 0 there is a corner at the origin, and when
= + cos
0, the graph crosses itself at the origin, and has
0. From the first figure, we see that
= + sin
curves roughly follow the line
= , and they start having loops when
is between 1 4 and 1 6. The loops increase in size as
increases.
While not required, the following is a solution to determine the exact values for which the curve has a loop, that is, we seek the values of ( + cos
for which there exist parameter values and
+ sin ) = ( + cos
such that
and
+ sin ). In the diagram at the left, and
denotes the point (
the point ( + cos
Since
=
angles,
=∠
=2 − + =
2
4
−
(1).
=
4
=∠ −
)
+ sin ) = ( + cos
= , the triangle and
.
0, the graph resembles that of a polynomial
two cusps below the -axis. The size of the “swallowtail” increases as increases.
49.
≤
the point (
),
+ sin ).
is isosceles. Therefore its base are equal. Since
the relation
=
= −
implies that
4
and
SECTION 10.1
Since cos
= distance(( =
− =
1 2
=
( − )
√ 2 cos
−
) ( √ 2
2( − )2 =
− =
√ 2 cos , that is,
(2). Now cos − 4 = sin 2 − − 4 = sin 4 − , √ − = 2 sin 34 − (20 ). Subtracting (20 ) from (1) and
4
so we can rewrite (2) as
−
dividing by 2, we obtain = 4
Since
√ 2
−
sin
2
implicitly assumed that 0 by + 2 merely increases √ 3 2 4 − (3), we get = sin
4
−
, or
3
− =
4
4
−
−
sin √ 2
4
it follows from (20 ) that sin
0 and
875
√ 2 ( − ), we see that
)) =
, so
¤
CURVES DEFINED BY PARAMETRIC EQUATIONS
(3).
4
0. Thus from (3) we see that
4.
[We have
by the way we drew our diagram, but we lost no generality by doing so since replacing and
by 2 . The curve’s basic shape repeats every time we change by 2 .] Solving for
. Write
=
3
4
. Then −
=
√ 2 sin
, where
0. Now sin
for
0, so
√ 2.
in
→ 0+ , that is, as →
As
50. Consider the curves
− 4
= sin + sin
→
,
,
radius 2 centered at the origin. For
√ 2 .
= cos + cos
, where
is a positive integer. For
= 1, we get a circle of − 1 loops as
1, we get a curve lying on or inside that circle that traces out
ranges from 0 to 2 . 2
Note:
+
2
= (sin + sin
)2 + (cos + cos
= sin2 + 2 sin sin
+ sin2
= (sin2 + cos2 ) + (sin2 = 1 + 1 + 2 cos( − with equality for
)2 + cos2 + 2 cos cos
+ cos2
+ cos2
) + 2(cos cos
+ sin sin
)
) = 2 + 2 cos((1 − ) ) ≤ 4 = 22 ,
= 1. This shows that each curve lies on or inside the curve for
= 1, which is a circle of radius 2 centered
at the origin.
NOT = 1
= 2
= 3
51. Note that all the Lissajous figures are symmetric about the -axis. The parameters
- and -directions respectively. For
876
¤
= =
and simply stretch the graph in the
= 1 the graph is simply a circle with radius 1. For
= 2 the graph crosses
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
itself at the origin and there are loops above and below the -axis. In general, the figures have all of which are on the -axis, and a total of
= =1 52.
=5
= cos ,
= sin − sin
=2 .
If = 1, then
graphs are shown for = 2 3 4 and 5.
− 1 points of intersection,
closed loops.
=3
= 0, and the curve is simply the line segment from (−1 0) to (1 0). The
F R
It is easy to see that all the curves lie in the rectangle [−1 1] by [−2 2]. When is an integer, ( + 2 ) = ( ) and ( + 2 ) = ( ), so the curve is closed. When is a positive integer greater than 1, the curve intersects the x-axis + 1 times and has loops (one of which degenerates to a tangency at the origin when is an odd integer of the form 4 + 1). √ As increases, the curve’s loops become thinner, but stay in the region bounded by the semicircles = ± 1 + 1 − and the line segments from (−1 −1) to (−1 1) and from (1 −1) to (1 1). This is true because
2
= sin − sin
≤ sin + sin ≤
√ 1−
2
+ 1. This curve appears to fill the entire region when is very large, as
shown in the figure for = 1000.
LABORATORY PROJECT
RUNNING CIRCLES AROUND CIRCLES
¤
When is a fraction, we get a variety of shapes with multiple loops, but always within the same region. For some fractional values, such as = 2 359, the curve again appears to fill the region.
S LE
LABORATORY PROJECT Running Circles Around Circles 1. The cente
Arc
of the smaller circle has coordinates (( − )cos
on circ
has length
( − )sin ).
since it is equal in length to arc
(the smaller circle rolls without slipping against the larger.) Thus, ∠
and 2. With
=
and ∠
− , so
=
has coordinates
= ( − )cos + cos(∠
) = ( − )cos + cos
= ( − )sin − sin(∠
) = ( − )sin − sin
= 1 and
− −
a positive integer greater than 2, we obtain a hypocycloid of
cusps. Shown in the figure is the graph for
= 4. Let
= 4 and = 1. Using the
sum identities to expand cos 3 and sin 3 , we obtain = 3 cos + cos 3 = 3 cos + 4 cos3 and
.
− 3 cos
= 3 sin − sin 3 = 3 sin − 3 sin − 4 sin3
= 4 cos3 = 4 sin3 .
877
3. The graphs at the right are obtained with 1
1
1
1
2
3
4
10
= , , , and
with −2 ≤
= 1 and
≤ 2 . We
conclude that as the denominator increases, the graph gets smaller, but maintains the basic shape shown.
[continued]
¤
878
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
Letting
So if
= 2 and
= 3, 5, and 7 with −2 ≤
is held constant and
obtain a hypocycloid of graphs have
4. If
varies, we get a graph with
cusps. As
= 32 , 54 , and
≤ 2 gives us the following:
cusps (assuming
=
+ 1, we
increases, we must expand the range of in order to get a closed curve. The following
11 10 .
S
= 1, the equations for the hypocycloid are = ( − 1) cos + cos (( − 1) )
which is a hypocycloid of
is in lowest form). When
= ( − 1) sin − sin (( − 1) )
cusps (from Problem 2). In general, if
1, we get a figure with cusps on the “outside ring” and if
1, the cusps are on the “inside ring”. In any case, as the values of get larger, we get a figure that looks more and more like a washer. If we were to graph the hypocycloid for all values of , every point on the washer would eventually be arbitrarily close to a point on the curve.
= 5. The cente
Arc and ∠
√ 2,
−10 ≤
= − 2,
≤ 10
of the smaller circle has coordinates (( + ) cos
has length =
−
(as in Problem 1), so that ∠ −
Thus, the coordinates of
=
since ∠
+
−
=
( + ) sin ).
,∠
=
−
,
= .
are
= ( + ) cos + cos
−
+
= ( + ) cos − cos
+ and
0≤
= ( + ) sin − sin
−
+
+ = ( + ) sin − sin
.
≤ 446
LABORATORY PROJECT
6. Let
= 1 and the equations become = ( + 1) cos − cos(( + 1) )
If
RUNNING CIRCLES AROUND CIRCLES
= 1, we have a cardioid. If
= ( + 1) sin − sin(( + 1) )
is a positive
integer greater than 1, we get the graph of an “ -leafed clover”, with cusps that are
units
from the origin. (Some of the pairs of figures are not to scale.) = 3, −2 ≤ If
=
with
≤2
= 10, −2 ≤
≤2
= 1, we obtain a figure that
does not increase in size and requires −
≤
≤
to be a closed curve traced
exactly once.
S LE
= 1 , −4 ≤
R
≤4
= 1 , −7 ≤
≤ 7
¤
879
4
Next, we keep
constant and let
7
vary. As
increases, so does the size of the figure. There is an -pointed star in the middle.
Now if
= 25 , −5 ≤
≤5
= 75 , −5 ≤
≤ 5
= 43 , −3 ≤
≤3
= 76 , −6 ≤
≤ 6
√ 2, 0 ≤
≤ 200
= − 2, 0 ≤
+ 1 we obtain figures similar to the
=
previous ones, but the size of the figure does not increase.
If
is irrational, we get washers that increase in
size as
increases.
=
¤
880
10.2 1.
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
Calculus with Parametric Curves ,
=
=
√ 1+
1+
⇒
=
=
3.
=
,
3
+ 1,
=
4
+ ;
(1 + )−1
2
= √
1
,
=
(1 + )(1) − (1) (1 + )2
2 1+
, and
(1 + )2
(1 + ) 1 3 2 = √ = (1 + ) . 2 2 1+
⇒
= 1 + cos ,
= −1.
=4
3
=
+
=
( + 1), and
= 3 2 , and
+ 1,
=
=
=
4
=
3
3
4.
1
=
2
1 (2 1 + )
= + sin
1 2
1 (1 + )2
2.
=
√
=
and
≤ 446
+1 2
1 + cos . ( + 1)
hen = −1, ( .W
) = (0 0)
= −3 3 = −1, so an equation of the tangent to the curve at the point corresponding to = −1 is
− 0 = −1( − 0), or √ = , = 2−2;
=− . = 4.
= 2 − 2,
1
= 2
√
, and
=
√ √ = (2 − 2)2 = 4( − 1) . When = 4,
(
) = (2 8) and
= 4(3)(2) = 24, so an equation of the tangent to the curve at the point corresponding to = 4 is
− 8 = 24( − 2), or 5.
= cos ,
= sin ;
When = , (
=
sin
= .
,
2
=
=−
is
−0 =
− (− )], or
;
= 0. 2
=
= cos + sin ,
0) and
) = (−
corresponding to = 6.
= 24 − 40.
=2
2
,
=
cos + sin . − sin + cos
=
2
+
( cos
.
) + (sin
)
=
( cos
+ sin
), and
2
= ( cos
+ sin
)
=
cos
. When = 0, (
+ sin
of the tangent to the curve at the point corresponding to = 0 is − 1 = 7. (a)
=
(−1) = , so an equation of the tangent to the curve at the point
=
2
= (− sin ) + cos , and
= 1 + ln ,
=
2
+ 2; (1 3).
=2
=
1
and
2
) = (0 1) and
( − 0), or
=
=
2
=2
2
=
, so an equation
+ 1.
= 2 2 . At (1 3),
1 = 1 + ln = 1 or (b)
ln = 0
⇒
= 1 and
−1
⇒
=
⇒
= 1 + ln
= 1+
0
√
= ,
ln =
2(1)−2
=
2
−1
, so
= 1+ =4
√
= 2
⇒
=
2
−1 2
+2 =(
) +2 =
2 −2
· 2 = 2, so an equation of the tangent is − 3 = 2( − 1), or ; (2 ).
=
2
2, ·
or
= 2, so an equation of the tangent is
− 3 = 2( − 1),
= 2 + 1.
At (1 3), 8. (a)
⇒
√
= 1
⇒
= 1 and
= 2
1 , and √
=
=
+ 2, and
0
=
2 −2
· 2.
= 2 + 1. 2
1 2
2
√
= 4 , so an equation of the tangent is
=4
−
3 2
2
. At (2 ),
= 4 ( − 2),
−7 .
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