paraeqns1 - Chapter 22 Parametric Equations Imagine a car is traveling along the highway and you look down at the situation from high above highway

# paraeqns1 - Chapter 22 Parametric Equations Imagine a car...

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This preview shows page 1 out of 8 pages. 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 © ¡ #  1P' 1) # ¥¥ © A X § ¥ £ ¡ B8! R¡ 2  § ¥ #  # ) &1) (1¤¡ £ ¡  ) # 2 ¡ 9 @ 6 ¥ ¡  9 ¦  2 &# 2 U  B!2 4E§ ¥ !¡ © A #1"' 1) h¥ #  B28! R¡ #10G1¤E§ ' # )¡ 9 @ [email protected] 9 @ 6 ¥ ¥ ¡ ¢ h¥ ¡ ¤E¦¥ ¡  £¡ § X # ' # ) § 1 Q1S¨¥ ¡ ¢ § ¢ ¢ X ¡ 9 ©  9 H &# 2 Y¡ ¥ U A ¦ A 2 ! 2 § ¦  ' 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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