**Unformatted text preview: **Chapter 22
Parametric Equations
Imagine a car is traveling along the highway and you look down at the
situation from high above:
highway curve (static) eplacements
-axis
-axis car moving point (dynamic) -axis Figure 22.1: The dynamic motion of a car on a static highway. We can adopt at least two different viewpoints: We can focus on the
entire highway all at once, which is modeled by a curve in the plane;
this is a “static viewpoint”. We could study the movement of the car
along the highway, which is modeled by a point moving along the curve;
this is a “dynamic viewpoint”. The ideas in this chapter are “dynamic”,
involving motion along a curve in the plane; in contrast, our previous
work has tended to involve the “static” study of a curve in the plane. We
will combine our understanding of linear functions, quadratic functions
and circular functions to explore a variety of dynamic problems. 303 CHAPTER 22. PARAMETRIC EQUATIONS 304 22.1 Parametric Equations
Gatorade
(200,300) (−20,260) -axis
(400,50) Solution. As a ﬁrst step, we can model the lines along
which both Tim and Michael will travel: (a) Tim and Michael
running toward
refreshments. Michael’s line of travel:
Tim’s line of travel: 6 dist(Tim, ) feet A 9
¢@ A ©
U
$B!2 #
§
G#1¤¡T' H ¥C% G#%$4¡ E§
)
)
# ! D
©
) C%%S§
2 )
# D
GF¡ E§
feet H CB)2
¥ #¡ ' H¥ )
QFPIFC% dist(Mike, ) 22.2: Visualizing
moving points. A 9
¢@ Figure H¥
RCB)2 (b) Modeling Tim and
Michael as points moving
on a path. 6 Michael’s line of travel 6
) ¡
8% 7§ (0, 0) A 9
¢@ (400, 50) -axis ¥
CB)2 § ¥ £¡
3¦¤¢ Tim’s line of travel -axis P=(125, 187.5) -axis It is an easy matter to determine where these two lines
cross: Set
and solve for , getting
, so
the lines intersect at
.
Unfortunately, we have NOT yet determined if the runners collide. The difﬁculty is that we have found where
the two lines of travel cross, but we have not worried
about the individual locations of Michael and Tim along
the lines of travel. In fact, if we compute the distance from
the starting point of each person to , we ﬁnd: ¥ £
54¡ (−20, 260) £ (200, 300) ments and §
0£ michael tim -axis )
% -axis Example 22.1.1. After a vigorous soccer match, Tim and
Michael decide to have a glass of their favorite refreshment.
They each run in a straight line along the indicated paths
at a speed of 10 ft/sec. Will Tim and Michael collide?
2 # ) ' ¥ # # !
&10(&%$" ¤¡ ¦¤¡
£ § ¥ £ PowerPunch
© § ¥ £¡
£ ¨¦¤¢ ments 6 Since these distances are different and both runners have the same
speed, Tim and Michael do not collide! 22.2 Motivation: Keeping track of a bug ements Imagine a bug is located on your desktop. How can you
best study its motion as time passes?
P
-axis
Let’s denote the location of the bug when you ﬁrst ob-axis
served it by . If we let represent time elapsed since ﬁrst
Figure 22.3: A bug on your
spotting the bug (say in units of seconds), then we can let
desktop.
be the new location of the bug at time . When
,
which is the instant you ﬁrst spot the bug, the location
is the initial location. For example, the path followed by the bug
might look something like the dashed path in the next Figure; we have
indicated the bug’s explicit position at four future times:
.
-axis X #
P§ X X V 6 X 6
¥ ¡ 6 gX b d X
hcfecb H SIaX
X b ` § ¥ #
¦&Y¡ 6 W 22.3. EXAMPLES OF PARAMETRIZED CURVES 305
¡ How can we describe the curve in Figure 22.4? To
PSfrag replacements
start, lets deﬁne a couple of new functions. Given a time
, we have the point
in the plane, so we can deﬁne: P(0) -axis P(t )
3 2 -axis Figure 22.4: A bug’s path. X 6
¥ ¡
X 6
¥ ¡
X 6
¥ ¡ is described as X ¢
X 6¥ ¡
¥ ¡ X X X
¥ ¡ ¢ X ¢ ¥ X ¡ ¢
§
X
£ §
S¨¥ ¡ £ § In other words, the point at time
at time P(t2) P(t )
4
P(t )
1 X 6
¥ ¡ -coordinate of
-coordinate of -axis X
¥ ¡ £ § £
X
X
X 6
2 ¥ ¥ ¡ ¢ h¥ ¤7C¥ ¡
¡ £¡ § We usually call
and
the coordinate functions of
.
Also, it is common to call the pair of functions
and
the
parametric equations for the curve. Anytime we describe a curve using
parametric equations, we usually call it a parametrized curve.
Given parametric equations
and
, the domain will be
the set of values we are allowed to plug in. Notice, we are using the
same set of -values to plug into both of the equations. Describing the
curve in Figure 22.4 amounts to ﬁnding the parametric equations
and
. In other words, we typically want to come up with “formulas”
for the functions
and
. Depending on the situation, this can be
easy or very hard. § ¢ X
¥ ¡ £ § £ X
¥ ¡ ¢ § ¢ ¢ X
¥ T§ £
¡ £ X X X
¥ £
¡ X
¥ ¡ ¢ X
¥ ¡ ¢ X
¥ ¡ £ 22.3 Examples of Parametrized Curves
We have already worked with some interesting examples of parametric
equations.
PSfrag replacements
Example 22.3.1. A bug begins at the location (1,0) on the
-axis
unit circle and moves counterclockwise with an angular -axis
bug starts moving
speed of
rad/sec. What are the parametric equaat 2 rad/sec
-axis
tions for the motion of the bug during the ﬁrst 5 seconds?
Indicate, via “snapshots”, the location of the bug at 1 secFigure 22.5: A circular path.
ond time intervals. £ X
T§ X £ § X ¤ P(2)
P(4) -axis X
2 ¥ ¤¡
X
¥ 4¡
cos
sin P(3) P(5) Figure 22.6: Six snapshots. X
¥ ¥ ¤¥ ¡ to the domain
cos
sin ¦
§) (# X
¥ ¤¡
¡ X
§¨¥ ¡ ¢ § ¢
X
§ ¥ S§ £
¡ £
If we restrict
is given by
snapshots: P(1)
P(0) , then the location of the bug at time
. We locate the bug via six one-second X § PSfrag replacements
Solution. We can use Fact 14.2.2 to ﬁnd the angle swept
out after seconds:
radians. The parametric -axis
-axis
equations are now easy to describe: X 6
§ ¥ X¡ CHAPTER 22. PARAMETRIC EQUATIONS 306 When modeling motion along a curve in the plane, we would typically
be given the curve and try to ﬁnd the parametric equations. We can turn
this around: Given a pair of functions
and
, let X
¥ ¡ ¢ § ¢ X
¥ S§ £
¡ £ X
X
X 6
h¥ ¥ ¡ ¢ h¥ ¤78¥ ¡
¡ £¡ § (22.1)
¢ £ which assigns to each input a point in the -plane. As ranges over
a given domain of allowed values, we will obtain a collection of points
in the plane. We refer to this as the graph of the parametric equations
. Thus, we have now described a process which allows us to
obtain a picture in the plane given a pair of equations in a common
single variable . Again, we call curves that arise in this way parametrized
curves. The terminology comes from the fact we are describing the curve
using an auxiliary variable , which is called the describing “parameter”.
In applications, often represents time. X X X ¡
¢¥ X ¢ X
¡ h¥ ¡ £
X X -axis Example 22.3.2. The graph of the parametric equations
and
on the domain
is
pictured; it is a line segment. As we let increase from
to , we can observe the motion of the corresponding points
on the curve.
G ¤
¥G
' X X
§ ¥ ¡ ¢ X X © X
§ ¥ £
¡
V Figure 22.7: Observing the
motion of . X -axis ¤ P(−1)=(−3,0)
P(−2)=(−6,−1) £ 22.4 Function graphs
It is important to realize that the graph of every function can be thought
of as a parametrized curve. Here is the reason why: Given a function
, recall the graph consists of points
, where runs over the
allowed domain values. If we deﬁne £ ¥ ¦¤¢ 5¤¡
¥ £¡ £ ¥ £¡
5¤¢ X
¡
h¥ ¢ X
X
¥¥ ¡ ¡ X
X
§ ¥ ¥ ¡ ¢ h¥ ¤¡
¡ £ X § ¢ X
§¨¥ ¡ ¢ § ¢
X
§ ¥ S§ £
¡ £ X 6
§ ¥ ¡ then plotting the points
gives us the graph
of . We gain one important thing with this new viewpoint: Letting
increase in the domain, we now have the ability to dynamically view
a point
moving along the function graph. See how this works in
Example 22.4.1. § £ X 6
¥ ¡ ¤ £ ¤
§¦G X £
H P§ ¢ on the domain
.
Example 22.4.1. Consider the function
As a parametrized curve, we would view the graph of
as all points
of the form
, where
. If increases from
to , the
corresponding points
move along the curve as pictured:
¥ ¡
§ ¥ H ¡ H £ § ¢ X 6
§
I¥ ¡ ,
G ¥
8! 8GR¡ ¤ X X 6
¥ ¡ X X
¥ H ¡ ¤
¦G § ¥ H CGR¡ 8GR¡
¥ § CGR¡
¥
6 X 6
§ ¥ ¡ Solution. For example,
etc. W P(2)=(6,3)
P(1)=(3,2) -axis X ements , 22.4. FUNCTION GRAPHS 307 W P(−2) -axis P(2)
P(1.75) P(−1.75) eplacements V −2 2 P(−1) -axis P(0) −2 static graph y=x 2 -axis P(1)
V W -axis 2 -axis motion along curve Figure 22.8: Visualizing dynamic motion along a static curve. PSfrag replacements W V Not every parametrized curve is the graph of a function. For example, consider these possible curves in the
plane: The second curve from the left is the graph of a
function; the other curves violate the vertical line test. -axis -axis -axis Figure 22.9: Some curves
that are not functions. 22.4.1 A useful trick
There is an approach to understanding a parametrized curve which is
sometimes useful: Begin with the equation
. Solve the equation
for in terms of the single variable ; i.e., obtain
. Then
substitute
into the other equation
, leading to an equation
involving only the variables and . If we were given the allowed values,
we can use the equation
to determine the allowed values, which
will be the domain of values for the function
. This may be a
function with which we are familiar or can plot using available software. ¥ £
5¤¡ § X X § £ £ X
¥ ¡ ¢ § ¢ £ ¥ £
¦¤¡ £ ¢ X
¥ ¡ £ § £
¥ ¦¤¡ ¡ ¢ § ¢
¥ £
H 100 W X
¥ £
¡ §X X
¥ ¡ T§ £
£ X £ Example 22.4.2. Start with the parametrized curve given
by the equations
and
, when
. Find a function
whose PSfrag gives this
graph replacements
parametrized curve. -axis 80 60 X
§ ¥ ¡ ¢ X § ¢ ¢
¥5£¤¢ ¡
X§
X
) ' § ¥ ¡ P§ £
£ # ¤ ¤ X # 40 Solution. Following the suggestion, we begin by solving
-axis
for , giving
. Plugging this into -axis
the second equation gives
.
Figure 22.10: Finding the
is on the parametrized curve if and
Conclude that
path equation.
only if the equation
is
satisﬁed. This is a quadratic function, so the graph will be an upward
opening parabola with vertex (5,0).
Since the domain is
, we get a new inequality for the
domain:
. Solving this, we get
, so
20 10 V H
¢
§ ¢¥
£
H £C) ¤¡ ` ¡ §
X
H
¥
C) 4¡ ` §
£ −10 20 ¥ )
£¡
H C ¤f¢ ` £ ¥
£
H C) ¤¡
¥ ¡
&# ¤ )
S £ ¤ # ¡ g ¢ ` ¡ g H
§ H ¢ C0 4¡ `
¥ )
£
# ¤ ¡ § ¦¤¢7§ ¢
¥ £¡ X # ¤ # ) X ¤ ¥ ¢ ¤¡
£ ' ¥ )
CS ¤¡
£ ¢ `H ¡ X ¤ # X § £ CHAPTER 22. PARAMETRIC EQUATIONS 308 )
%c¤ £ ) ¤ . This means the graph of the parametrized curve is the graph
of the function
, with the domain of values
. Here is
a plot of the graph of
; the thick portion is the parametrized curve
we are studying.
¦ )
§% 8) ¥ £¡
5¤¢ § ¢ g
¥
£
H C) 4¡ ¢ ` £ ¥ ¡ § ¢ 22.5 Circular motion
We can describe the motion of an object around a circle
using parametric equations. This will involve the trigonometric functions. The general setup to imagine is pictured: An object moving around a circle of radius centered at a point
in the
-plane. The path traced
out is the circle. However, the location of the object at
time will depend on a number of things: ¥ ¡ ¢ @¢¤¡
¡ £ £ The starting location £ The angular speed £ The radius V Figure 22.11: Circular motion. of the object;
of the object; £ -axis 6 P(t)=(x(t),y(t))
= location at time t -axis ¥ ¡ ¢ @¤¤¡
¡ £ r
(x c,yc ) X ements ¢ £ W P = starting
location -axis and the center . We will build up to the general solution by considering two cases, the
ﬁrst being a special case of the second. 22.5.1 Standard circular motion
As a ﬁrst case to consider, assume that the center of the
circle is
and the starting location
, as pictured below. If the angular speed is , then the angle
swept out in time will be
; this requires that the
time units in
agree with the time units of ! We denote by
the -coordinates of the object
at time . At time , we can compute the coordinates of
using the circular functions: X £ X § X ¤ ¢ £ X £ X
2¥
¡
§
¨¥ ¡ ¢ § ¢
X £ X
¥
¡
§ ¥ S§ £
¡ £
X
X
X 6
¥ ¥ ¡ ¢ h¥ ¤78¥ ¡
¡ £¡ §
X
X
X
X
X 6
¥ ¥ ¡ ¢ ¥ ¤¡ £ § ¥ ¡
¡ £ P = (r,0)
= P(0)
= starting
location £ cos
sin § Figure 22.12: Standard circular motion. ¥
&# ¡ This parametrizes motion starting at
. Using the shifting technology of Chapter 9, we are led to a general description of this type of
circular motion, which involves a circle of radius centered at a point
; we refer to this situation as standard circular motion. 6 ¥ ¡ ¢ ¥¤¡
¡ £ Important Fact 22.5.1 (Standard circular motion). Assume an object
is moving around a circle of radius centered at
with a constant
angular speed of . Assume the object begins at
. Then
the location of the object at time is given by the parametric equations:
cos
and
sin
. ¥ ¡ ¢ ' ¤¤¡
¡ £ 6
§
¥ ¡ ¢ @¢¤¡
¡ £
¥ X £ ¡ X
'
¡ ¢ ¨¥ ¡X ¢ § ¢
§ £ ¥ X £ ¡ V P(t)=(x(t),y(t)) -axis ¤ -axis ω
ωt 6 W ements ¥ #
&# (F¡ 2ω § P(1) P(2) ¥
&# ¡ -axis X
' ¥S¨¥ S§ £
¡ £ § ¡ £ 22.5. CIRCULAR MOTION 309 Example 22.5.2. Cosmo the dog is tied to a 20 foot long
tether, as in Figure 14.1. Assume Cosmo starts at the locaPSfrag replacements
tion “ ” in the Figure and maintains a tight tether, moving
around the circle at a constant angular speed
ra- -axis
dians/second. Parametrize Cosmos motion and determine -axis
where the dog is located after 3 seconds and after 3 min- -axis
utes. Cosmo
S motors
around Q the ¢¡ P 20 feet R circle § £ Figure 22.13: Cosmo on a running on a circular path.
Solution. Impose a coordinate system so that the pivot
point of the tether is
. Since
,
Cosmo is walking counterclockwise around the circle. By
(4.1.6), the location of Cosmo after
seconds is
cos
sin
. After 3 seconds, Cosmo is located at
cos
sin
. After 3 minutes = 180 seconds,
the location of the dog will be
cos
sin
,
which is the original starting point. # © 6
§ ¥
X¡
X
§ ¥ ¥ ¡ ¢ h¥ ¡ ¤¡
£ ¥ # ¡
&# (1¤7§ ¢ ¢ ¨
¡ £ ¢ ` ¥ ¥ #
&# (F¡ £ ¡ X 6
§
"¥ ¡ ¥
@£ ¢ ¤¤¡
£ £ § X #
1 ¢ ¨
¡ £ ¢ ©` ¢ ¢ 9
@ 6
#1 ¡ 8&
§ ¥
¦# 9
§ ¡ @
¥
C # 2 &U R¡
2 ¡ § ¢ ¢ X¡ ¡ ¢¡ #
1
#
1 ¡ ¢d ¢ ¢ X¡ ¢¡ ¢d ¡ ¡ #1
#
1 ¡
¡ 22.5.2 General circular motion
The circular motion of an object can begin at any location
-axis
PSfrag replacements
rel -axis
on the circle. To handle the general case, we follow an
P=starting position
earlier idea and introduce an auxiliaryrelative coordinate
θo
system : The rel rel -coordinates are obtained by drawing
rel -axis
lines parallel to the -axis and passing through
. -axis
We are using the subscript “rel” to stand for “relative”.
initial angle
(xc ,yc )
and
This new relative coordinate system has origin
-axis
allows us to deﬁne the initial angle , which indexes the
Figure 22.14: Initial angle
starting location , as pictured below:
and auxiliary axis.
Assume the object starts at and is moving at a constant angular speed around the pictured circle of radius
. Then after time has elapsed, the location of the object is indexed by
sweeping out an angle
, starting from . In other words, the location
after units of time is going to be determined by the central standard
angle
with initial side the positive rel -axis. This means that if
is the location of the object at time , W W V 6 ¥ ¡ ¢ ¡ ¤¡
£ ¥ ¡ ¢ ¡ 4¡
£ V ¢ ¢ £ £ ¤ £ £ X £ ¤ £ X X £
2¥
' £
X £
'
I£
¥ £ 6 6 ¡
¤
¤ ¡ cos
sin X
X
'¡ ¢ ¨¥ ¡ ¢ § ¢
§ X
' ¡¥£S§ ¥ S§ £
¡ £
X
X
X 6
¥ ¥ ¢ ¡ £ ¡¤ §
X¡ £ ¥ ¤E8¥ ¡
'
G£ X Notice, the case of standard circular motion is just the scenario when
and these parametric equations collapse to those of Fact 22.5.1. #
§ £ ¤ Important Fact 22.5.3 (General circular motion). Assume an object
is moving around a circle of radius centered at
with a constant
angular speed of . Assume the object begins at the location with initial 6 ¥ ¡ ¢ ¢¤¡
¡ £ £ CHAPTER 22. PARAMETRIC EQUATIONS
, as in Figure 22.14. The object location at time
cos
and
sin X £ ¤ Example 22.5.4. A rider jumps on a merry-go-round of radius 20 feet at the pictured location. The ride rotates at the
constant angular speed of
radians/second. The
center of the platform is located 50 feet East and 50 feet
North of the ticket booth for the ride. What are the parametric equations describing the location of the rider? Where is
the rider after 18 seconds have elapsed? How far from the
ticket booth is the rider after 18 seconds have elapsed? # ¡ § £ Figure 22.15: A rider jumps
on a merry-go-round. is given by:
. ¡ x−axis (east) ¤ ¤ 50 ft X £ £ X
' ¥S¨¥ S§ £
¡ £ § ¡ £
50 ft Solution. In this example, since
, the rotation is
clockwise. Since the angular speed is given in radians,
we need to convert the initial angle to radians as well:
radians. Impose a coordinate system so that the center of the ride is
and its radius is 20 feet. By Fact 22.5.3, the parametric equations for the rider are given by
cos
and
sin
. The location after 18 seconds will be
¢ X ¡ 9
@ ¡ ) # 2 9 © ¡ #
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§ ¥ £
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¥
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#
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9
@
[email protected]
9
@ 6
¥ ¥ ¡ ¢ h¥ ¡ ¤E¦¥ ¡
£¡ §
X
# ' # ) §
1 Q1S¨¥ ¡ ¢ § ¢
¢ X ¡ 9 © 9
H &# 2 Y¡
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U A ¦ A
2
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§
¦
' H ¥ 2! 4¡ E§
! D
feet to the origin is ¢ The distance from sin § £ cos 9 ©
) # 2 b Ticket Booth ' £ ¡ P=P(0) -axis
-axis ¥ -axis X
'
¨¡ ¢ ¨¥ ¡ ¢ § ¢
§ ements o
218
−π/7 rad/sec ' £ ¡ y−axis (north) X angle ¥ 310 £ ...

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