9.1 PLANE CURVES AND PARAMETRIC EQUATIONS

Transcription

1 76 CHAPTER 9.. Parametric Equations and Polar Coordinates 9-9. PLANE CURVES AND PARAMETRIC EQUATIONS We often find it convenient to describe the location of a point (, )inthe plane in terms of a parameter. For instance, in tracking the movement of a satellite, we would naturall want to give its location in terms of time. In this wa, we not onl know the path it follows, but we also know when it passes through each point. Given an pair of functions (t) and (t) defined on the same domain D, the equations = (t), = (t) are called parametric equations. Notice that for each choice of t, the parametric equations specif a point (, ) = ((t), (t)) in the -plane. The collection of all such points is called the graph of the parametric equations. In the case where (t) and (t) are continuous functions and D is an interval of the real line, the graph is a curve in the -plane, referred to as a plane curve. The choice of the letter t to denote the independent variable (called the parameter) should make ou think of time, which is often what the parameter represents. For instance, we might represent the position ((t), (t)) of a moving object as a function of the time t. Infact, ou might recall that in section 5.5, we used a pair of equations of this tpe to describe two-dimensional projectile motion. In man applications, the parameter has an interpretation other than time; in others, it has no phsical meaning at all. In general, the parameter can be an quantit that is convenient for describing the relationship between and.ineample., we can simplif our discussion b eliminating the parameter. t EXAMPLE. Graphing a Plane Curve Sketch the plane curve defined b the parametric equations = 6 t, = t/, for t. Solution In the accompaning table, we list a number of values of the parameter t and the corresponding values of and. We have plotted these points and connected them with a smooth curve in Figure 9.. You might also notice that we can easil eliminate the parameter here, b solving for t in terms of.wehavet =, sothat = 6. The graph of this last equation is a parabola opening to the left. However, the plane curve we re looking for is the portion of this parabola corresponding to t. From the table, notice that this (, ) 8 6 (, ) (, ) (5, Q) (6, ) (5, Q) (, ) FIGURE 9. = 6 t, = t, t

2 9- SECTION 9... Plane Curves and Parametric Equations 77 corresponds to, so that the plane curve is the portion of the parabola indicated in Figure 9., where we have also indicated a number of points on the curve. You probabl noticed the small arrows drawn on top of the plane curve in Figure 9.. These indicate the orientation of the curve (i.e., the direction of increasing t). If t represents time and the curve represents the path of an object, the orientation indicates the direction followed b the object as it traverses the path, as in eample.. 6 t = t = t =.5 t = FIGURE 9. Path of projectile EXAMPLE. The Path of a Projectile Find the path of a projectile thrown horizontall with initial speed of ft/s from a height of 6 feet. Solution Following our discussion in section 5.5, the path is defined b the parametric equations = t, = 6 6t, for t, where t represents time (in seconds). This describes the plane curve shown in Figure 9.. Note that in this case, the orientation indicated in the graph gives the direction of motion. Although we could eliminate the parameter, as in eample., the parametric equations provide us with more information. It is important to recognize that while the corresponding - equation = 6 6 ( ) describes the path followed b the projectile, the parametric equations provide us with additional information, as the also tell us when the object is located at a given point and indicate the direction of motion. We indicate the location of the projectile at several times in Figure 9.. Graphing calculators and computer algebra sstems sketch a plane curve b plotting points corresponding to a large number of values of the parameter t and then connecting the plotted points with a curve. The appearance of the resulting graph depends greatl on the graphing window used and also on the particular choice of t-values. This can be seen in eample.. EXAMPLE. Parametric Equations Involving Sines and Cosines FIGURE 9.a = cos t, = sin t Sketch the plane curve defined b the parametric equations = cos t, = sin t, for (a) t π and (b) t π. (.) FIGURE 9.b = cos t, = sin t Solution (a) The default graph produced b most graphing calculators looks something like the curve shown in Figure 9.a (where we have added arrows indicating the orientation). With some thought, we can improve this sketch. First, notice that since = cos t, ranges between and. Similarl, ranges between and. Changing the graphing window to.. and.. produces the curve shown in Figure 9.b, which is an improvement over Figure 9.a. The curve still looks like an ellipse, but with some more thought we can identif it as a circle. Rather than eliminate the parameter b solving for t in terms of either or, instead notice from (.) that + = cos t + sin t = (cos t + sin t) =.

3 78 CHAPTER 9.. Parametric Equations and Polar Coordinates 9- So, the plane curve lies on the circle of radius centered at the origin. In fact, it s the whole circle, as we can see b recognizing what the parameter represents in this case. Recall from the definition of sine and cosine that if (, ) isapoint on the unit circle and θ is the angle from the positive -ais to the line segment joining (, ) and the origin, then we define cos θ = and sin θ =. Since we have = cos t and = sin t, the parameter t corresponds to the angle θ. Further, the curve is the entire circle of radius, traced out as the angle t ranges from to π. A square graphing window is one with the same scale on the - and -aes (though not necessaril the same and ranges). Such a square window gives us the circle seen in Figure 9.c. (b) Finall, what would change if the domain were limited to t π? Since we ve identified t as the angle as measured from the positive -ais, it should be clear that ou will now get the top half of the circle, as shown in Figure 9.d. FIGURE 9.c A circle FIGURE 9.d Top semicircle REMARK. To sketch a parametric graph on a CAS, ou ma need to write the equations in vector format. For instance, in the case of eample., instead of entering = cos t and = sin t, ou would enter the ordered pair of functions ( cos t, sin t). Simple modifications to the parametric equations in eample. will produce a variet of circles and ellipses. We eplore this in eample. and the eercises. EXAMPLE. More Circles and Ellipses Defined b Parametric Equations Identif the plane curves (a) = cos t, = sin t, (b) = + cos t, = + sin t and (c) = cos t, = sin t, all for t π. Solution A computer-generated sketch of (a) is shown in Figure 9.a. It s difficult to determine from the sketch whether the curve is an ellipse or simpl a distorted graph of a circle. You can rule out a circle, since the parametric equations produce -values between and and -values between and. To verif that this is an ellipse, observe that + 9 = cos t + 9 sin t 9 = cos t + sin t =. FIGURE 9.a = cos t, = sin t A computer-generated sketch of (b) is shown in Figure 9.b. You should verif that this is the circle ( ) + ( ) = 6. Finall, a computer sketch of (c) is shown in Figure 9.c. You should verif that this is the circle + = 9. So, what is the role of the in the argument of cosine and sine? If ou sketched this on a calculator, ou ma have noticed that the circle was completed long before the calculator finished graphing. Because of the, a complete circle corresponds to t π or t π. With the

4 9-5 SECTION 9... Plane Curves and Parametric Equations 79 REMARK. Look carefull at the plane curves in eamples. and. until ou can identif the roles of each of the constants in the equations = a + b cos ct, (t) = d + e sin ct. These interpretations are important in applications. 6 FIGURE 9.b = + cos t, = + sin t FIGURE 9.c = cos t, = sin t REMARK. There are infinitel man choices of parameters that produce a given curve. For instance, ou can verif that = + t, = + 5t, for t and = t, = + 5t, for t both produce the line segment from eample.5. We sa that each of these pairs of parametric equations is a different parameterization of the curve. domain t π, the circle is traced out twice. You might sa that the factor of in the argument doubles the speed with which the curve is traced. In eample.5, we see how to find parametric equations for a line segment. EXAMPLE.5 Parametric Equations for a Line Segment Find parametric equations for the line segment joining the points (, ) and (, 7). Solution For aline segment, notice that the parametric equations can be chosen to be linear functions. That is, = a + bt, = c + dt, for some constants a, b, c and d. (Eliminate the parameter t to see wh this generates a line.) The simplest wa to choose these constants is to have t = correspond to the starting point (, ). Note that if t =, the equations reduce to = a and = c.to start our segment at = and =, we set a = and c =. Now note that with t =, the equations are = a + b and = c + d. Toproduce the endpoint (, 7), we must have a + b = and c + d = 7. With a = and c =, solve to get b = and d = 5. We now have that = + t, = + 5t, for t is a pair of parametric equations describing the line segment. In general, for parametric equations of the form = a + bt, = c + dt, notice that ou can alwas choose a and c to be the - and -coordinates, respectivel, of the starting point (since = a, = b corresponds to t = ). Then b is the difference in -coordinates (endpoint minus starting point) and d is the difference in -coordinates. With these choices, the line segment is alwas sketched out for t. As we illustrate in eample.6, ever equation of the form = f () can be simpl epressed using parametric equations. EXAMPLE.6 Parametric Equations from an - Equation Find parametric equations for the portion of the parabola = from (, ) to (, 9).

5 7 CHAPTER 9.. Parametric Equations and Polar Coordinates 9-6 Solution An equation of the form = f () can be converted to parametric form simpl b defining t =. Here, this gives us = = t,sothat = t, = t, for t, 5 is a parametric representation of the curve. (Of course, ou can use the letter as the parameter instead of the letter t, ifou prefer.) Besides indicating an orientation, parametric representations of curves often also carr with them a built-in restriction on the portion of the curve included, as we see in eample.7. FIGURE 9.5a = ( + ) EXAMPLE.7 Parametric Representations of a Curve with a Subtle Difference FIGURE 9.5b = t, = t Sketch the plane curves (a) = t, = t and (b) = t, = t. Solution Since there is no restriction placed on t, wecan assume that t can be an real number. Eliminating the parameter in (a), we get t = +, so that the parametric equations in (a) correspond to the parabola = ( + ), shown in Figure 9.5a. Notice that the graph includes the entire parabola, since t and hence, = t can be an real number. (If our calculator sketch doesn t show both sides of the parabola, adjust the range of t-values in the plot.) The importance of this check is shown b (b). When we eliminate the parameter, we get t = + and so, = ( + ). This gives the same parabola as in (a). However, the initial computer sketch of the parametric equations shown in Figure 9.5b shows onl the right half of the parabola. To verif that this is correct, note that since = t, we have that for ever real number t. Therefore, the curve is onl the right half of the parabola = ( + ), as shown. Man plane curves described parametricall are unlike anthing ou ve seen so far in our stud of calculus. Man of these are difficult to draw b hand, but can be easil plotted with a graphing calculator or CAS. EXAMPLE.8 Some Unusual Plane Curves Sketch the plane curves (a) = t, = t t and (b) = t t, = t 5t +. FIGURE 9.6a = t, = t t Solution Asketch of (a) is shown in Figure 9.6a. From the vertical line test, this is not the graph of an function. Further, converting to an - equation here is mess and not particularl helpful. (Tr this to see wh.) However, eamine the parametric equations to see if important portions of the graph have been left out (e.g., is there supposed to be anthing to the left of =?). Here, = t for all t and = t t has no maimum or minimum (think about wh). It seems that most of the graph is indeed shown in Figure 9.6a. A computer sketch of (b) is shown in Figure 9.6b. Again, this is not a familiar graph. To get an idea of the scope of the graph, note that = t t has no maimum or minimum. To find the minimum of = t 5t +, note that critical numbers are at

6 9-7 SECTION 9... Plane Curves and Parametric Equations FIGURE 9.6b = t t, = t 5t + 5 FIGURE 9.7a Missile flight paths 5 FIGURE 9.7b Missile flight paths 5 t = and t =± with corresponding function values and 9, respectivel. You should conclude that 9,asindicated in Figure 9.6b. You should now have some idea of the fleibilit of parametric equations. Quite significantl, a large number of applications translate simpl into parametric equations. Bear in mind that parametric equations communicate more information than do the corresponding - equations. We illustrate this with eample.9. EXAMPLE.9 Intercepting a Missile in Flight Suppose that a missile is fired toward our location from 5 miles awa and follows a flight path given b the parametric equations = t, = 8t 6t, for t 5. Two minutes later, ou fire an interceptor missile from our location following the flight path = 5 (t ), = 8(t ) 6(t ), for t 7. Determine whether the interceptor missile hits its target. Solution In Figure 9.7a, we have plotted the flight paths for both missiles simultaneousl. The two paths clearl intersect, but this does not necessaril mean that the two missiles collide. For that to happen, the need to be at the same point at the same time.todetermine whether there are an values of t for which both paths are simultaneousl passing through the same point, we set the two -values equal: t = 5 (t ) and obtain one solution: t =. Note that this simpl sas that the two missiles have the same -coordinate when t =. Unfortunatel, the -coordinates are not the same here, since when t =, we have 8t 6t = 96 but 8(t ) 6(t ) = 6. You can see this graphicall b plotting the two paths simultaneousl for t onl, as we have done in Figure 9.7b. From the graph, ou can clearl see that the two missiles pass one another without colliding. So, b the time the interceptor missile intersects the flight path of the incoming missile, it is long gone! Another ver nice wa to observe this behavior is to plot the two sets of parametric equations on our graphing calculator in simultaneous plot mode. With this, ou can animate the flight paths and watch the missiles pass b one another. BEYOND FORMULAS When thinking of parametric equations, it is often helpful to think of t as representing time and the graph as representing the position of a moving particle. It is important to realize that the parameter can be anthing. For eample, in equations of circles and ellipses, the parameter ma represent the angle as ou rotate around the oval. Allowing the parameter to change from problem to problem gives us an incredible fleibilit to describe the relationship between and in the most convenient wa possible.

7 7 CHAPTER 9.. Parametric Equations and Polar Coordinates 9-8 EXERCISES 9. WRITING EXERCISES. Interpret in words the { roles of each of the constants in the = a + b parametric equations cos(ct) = a + b sin(ct).. An algorithm was given in eample.5 for finding parametric equations of a line segment. Discuss the advantages that this method has over the other methods presented in remark... As indicated in remark., a given curve can be described b numerous sets of parametric equations. Eplain wh several different equations can all be correct. (Hint: Emphasize the fact that t is a dumm variable.). In eample.9, ou saw that missiles don t collide even though their paths intersect. If ou wanted to determine the intersection point of the graphs, eplain wh ou would need to solve for values s and t (possibl different) such that t = 5 (s ) and 8t 6t = 8(s ) 6(s ). In eercises, sketch the plane curve defined b the given parametric equations and find a corresponding - equation for the curve. { { = cos t = + cos t.. = sin t = + sin t { { = + t = + t.. = t = t { { = + t = t 5. = t 6. + = t + { { = t 7. = t = t { = cos t = cos t. = t + { = sin t = cos t { = cos t,. Conjecture the difference between the graphs of = sin kt where k is an integer compared to when k is an irrational number. (Hint: Use eercises 5 and 6 and tr k =, k = and other values.) { = cos t. Compare the graphs of for k =, k =, = sin kt k =, k = and k = 5, and describe the role that k plas in the graph. { = cos t cos kt. Compare the graphs of = sin t sin kt for k =, k =, k = and k = 5, and describe the role that k plas in the graph.. Describe the role that r plas in the graph of { = r cos t and then describe how to sketch the graph of = r sin t { = t cos t. = t sin t In eercises 5, match the parametric equations with the corresponding plane curve displaed in Figures A F. Give reasons for our choices. { { = t = t 5. = t 6. = t { { = t = t = sin t = sin t 9. { = cos t = sin t. { = cos t = sin t In eercises, use our CAS or graphing calculator to sketch the plane curves defined b the given parametric equations. { { = t. t = t = t. t = t t { = t = t t { = cos t = sin 7t. 6. { = t = t t { = cos t = sin πt { = cos t + sin 5t = sin t + cos 5t 8. { = cos t + sin 6t = sin t + cos 6t 9. { = e t = e t. { = e t = e t FIGURE A

8 9-9 SECTION 9... Plane Curves and Parametric Equations 7 FIGURE B 6 8 FIGURE C 8 8 FIGURE D FIGURE F In eercises, find parametric equations describing the given curve.. The line segment from (, ) to (, ). The line segment from (, ) to (, ). The line segment from (, ) to (6, ). The line segment from (, ) to (, ) 5. The portion of the parabola = + from (, ) to (, 5) 6. The portion of the parabola = from (, ) to (, 7) 7. The portion of the parabola = from (, ) to (, ) 8. The portion of the parabola = + from (, ) to (, ) 9. The circle of radius centered at (, ), drawn counterclockwise. The circle of radius 5 centered at (, ), drawn counterclockwise In eercises, find all points of intersection of the two curves. { { = t = + s. = t and = s { { = t = + s. and = t + = s. { = t + = t and { = + s = s FIGURE E. { = t + = t + t and { = + s = s 5. Rework eample.9 with the interceptor missile following the flight path = 5 5(t ) and = 8(t ) 6(t ). 6. Rework eample.9 with the interceptor missile following the flight path = 5 t and = 8t 6t.

9 7 CHAPTER 9.. Parametric Equations and Polar Coordinates 9-7. In eample.9 and eercise 5, eplain wh the in the term t represents the time dela between the launches of the two missiles. For the equations in eample.9, find a value of the time dela such that the two missiles do collide. that the boat s wake will be described b the graphs of eercises 5 and 55. Graph the wake of a boat with speed.6v. 8. Eplain wh the missile path in eercise 6 must produce a collision (compare the -equations) but is unrealistic. Eercises 9 56 eplore the sound barrier problem discussed in the chapter introduction. Define unit to be the distance traveled b sound in second. 9. Suppose a sound wave is emitted from the origin at time. After t seconds (t > ), eplain wh the position in units of the sound wave is modeled b = t cos θ and = t sin θ, where the dumm parameter θ has range θ π. 5. Find parametric equations as in eercise 9 for the position at time t seconds (t > ) of a sound wave emitted at time c seconds from the point (a, b). 5. Suppose that a jet has speed.8 unit per second (i.e., Mach.8) with position function (t) =.8t and (t) =. To model the position at time t = 5 seconds of various sound waves emitted b the jet, do the following on one set of aes. (a) Graph the position after 5 seconds of the sound wave emitted from (, ); (b) graph the position after seconds of the sound wave emitted from (.8, ); (c) graph the position after seconds of the sound wave emitted from (.6, ); (d) graph the position after seconds of the sound wave emitted from (., ); (e) graph the position after second of the sound wave emitted from (., ); (f) mark the position of the jet at time t = Repeat eercise 5 for a jet with speed. unit per second (Mach ). You should notice that the sound waves all intersect at the jet s location. This is the sound barrier that must be broken. 5. Repeat eercise 5 for a jet with speed. units per second (Mach.). 5. In eercise 5, ou should find that the sound waves intersect each other. The intersections form the shock wave that we hear as a sonic boom. Theoreticall, the angle θ between the shock wave and the -ais satisfies the equation sin θ = l, where m is the Mach speed of the jet. Show that for m m =., the theoretical shock wave is formed b the lines (t) = 7.96t, (t) = t and (t) = 7.96t, (t) = t. Superimpose these lines onto the graph of eercise In eercise 5, the shock wave of a jet at Mach. is modeled b two lines. Argue that in three dimensions, the shock wave has circular cross sections. Describe the three-dimensional figure formed b revolving the lines in eercise 5 about the -ais. 56. If a pebble is dropped into water, a wave spreads out in an epanding circle. Let v be the speed of the propagation of the wave.if a boat moves through this water with speed.v,argue Eercises 57 6 show that a celestial object can incorrectl appear to be moving faster than the speed of light. 57. A bright object is at position (, D)attime, where D is a ver large positive number. The object moves toward the positive -ais with constant speed v at an angle θ from the vertical. Find parametric equations for the position of the object at time t. 58. For the object of eercise 57, let s(t) bethe distance from the object to the origin at time t. Then L(t) = s(t) gives the c amount of time it takes for light emitted b the object at time t to reach the origin. Show that L (t) = v t Dv cos θ. c s(t) 59. An observer stands at the origin and tracks the horizontal movement of the object in eercises 57 and 58. As computed in eercise 58, light received at time T was emitted b the object at time t, where T = t + L(t). Similarl, light received at time T + T was emitted at time t + dt, where tpicall dt T. The apparent -coordinate of the object at time T is a (T ) = (t). The apparent horizontal speed of the object at time T as measured b the observer is a (T + T ) a (T ) h(t ) = lim.tracing back to time t, T T (t +dt) (t) show that h(t) = lim = v sin θ dt T T (t) = v sin θ + L (t). cv sin θ 6. In eercise 59, show that h() = c v cos θ. 6. For the moving object of eercises 57 6, show that for a constant speed v, the maimum apparent horizontal speed h() occurs when the object moves at an angle with cos θ = v c. Find the maimum speed in terms of v and the contraction factor γ = v /c. 6. For the moving object of eercises 57 6, show that as v approaches c, the apparent horizontal speed can eceed c, causing the observer to measure an object moving faster than the speed of light! As v approaches c, show that the angle producing the

10 9- SECTION 9... Plane Curves and Parametric Equations 75 maimum apparent horizontal speed decreases to. Discuss wh this is paradoical. { { = cos t = cos t 6. Compare the graphs of and = sin t = sin t. Use the identities cos t = cos t sin t and sin t = cos t sin t to find - equations for each graph. { = cosh t 6. Sketch the graph of. Use the identit = sinh t cosh t sinh t = tofind an - equation for the graph. Eplain where the hperbolic in hperbolic sine and hperbolic cosine might come from. Assume that the central arm has length feet and rotates about its center. Also assume that the wheels have radius feet and rotate at the same speed as the central arm. Find parametric equations for the position of a rider and graph the rider s path. Adjust the speed of rotation of the wheels to improve the ride. { = 65. Sketch the graph of cos t cos t = sin t sin t. This heartshaped region is the largest feature of the Mandelbrot set, one of the most famous mathematical sets. Portions of the Mandelbrot set have been turned into colorful T-shirts and posters that ou ma have seen. Mandelbrot set Mandelbrot zoom To progress { further on a sketch of the Mandelbrot set, add the = + circle cos t = sin t to our initial sketch. 66. Determine parametric equations for the curves defined b n + n = r n for integers n. (Hint: Start with n =, + = r, then think of the general equation as ( n ) + ( n ) = r n.)sketch the graphs for n =, n = and n =, and predict what the curve will look like for large values of n. EXPLORATORY EXERCISES. Man carnivals have a version of the double Ferris wheel. A large central arm rotates clockwise. At each end of the central arm is a Ferris wheel that rotates clockwise around the arm.. The Fling Zucchini Circus Troupe has a human cannonball act, shooting a performer from a cannon into a speciall padded seat of a turning Ferris wheel. The Ferris wheel has a radius of feet and rotates counterclockwise at one revolution per minute. The special seat starts at ground level. Carefull { eplain wh parametric equations for the = cos( π seat are t π ) = + sin( π t. The cannon is located π ) feet left of the Ferris wheel with the muzzle feet above ground. The performer is launched 5 seconds after the wheel starts turning with an initial velocit of ft/s at an angle of π above the horizontal. Carefull eplain 5 wh { parametric equations for the human cannonball are = ( cos π )(t 5) 5 = 6(t 5) + ( sin π )(t 5) + (t 5). 5 Determine whether the act is safe or the Fling Zucchini comes down squash.. Rework eercise with initial velocit 5 ft/s, launch angle and a 7-second dela. How close does the Fling Zucchini get to the special seat? Given that a Ferris wheel seat actuall has height, width, and depth, do ou think that this is close enough? Repeat with (a) initial velocit 75 ft/s, launch angle 7 and 7.5-second dela; (b) initial velocit 8 ft/s, launch angle 5 and 8-second dela. Develop criteria for a safe and eciting human cannonball act. Consider each of the following: Should the launch velocit be large or small? Should the seat be high or low when the cannonball lands? Should the human have a positive or negative vertical velocit at landing? How close (verticall and horizontall) should the human need to get to the center of the seat? Based on our criteria, which of the launches in this eercise is the best? Find an initial velocit, launch angle and launch dela that is better.

11 76 CHAPTER 9.. Parametric Equations and Polar Coordinates 9-9. CALCULUS AND PARAMETRIC EQUATIONS FIGURE 9.8a The Scrambler FIGURE 9.8b Path of a Scrambler rider REMARK. Be careful with how ou interpret equation (.). The primes on the right side of the equation refer to derivatives with respect to the parameter t. We recommend that ou (at least initiall) use the Leibniz notation, which also gives ou a simple wa to accuratel remember the chain rule. The Scrambler is a popular carnival ride consisting of two sets of rotating arms (see Figure 9.8a). Suppose that the inner arms have length and rotate counterclockwise. In this case, we can describe the location ( i, i )ofthe end of one of the inner arms b the parametric equations i = cos t, i = sin t. Atthe end of each inner arm, a set of outer arms rotate clockwise at roughl twice the speed. If the outer arms have length, parametric equations describing the outer arm rotation are o = sin t, o = cos t. Here, the reversal of sine and cosine terms indicates that the rotation is clockwise and the factor of inside the sine and cosine terms indicates that the speed of the rotation is double that of the inner arms. The position of a person riding the Scrambler is the sum of the two component motions; that is, = cos t + sin t, = sin t + cos t. The graph of these parametric equations is shown in Figure 9.8b. Passengers on the Scrambler feel like the rapidl accelerate to the outside of the ride, momentaril stop, then change direction and accelerate to a different point on the outside of the ride. Figure 9.8b suggests that this is an accurate description of the ride, but we need to develop the calculus of parametric equations to determine whether the riders actuall come to a complete stop. Our initial aim is to find a wa to determine the slopes of tangent lines to curves that are defined parametricall. First, recall that for a differentiable function = f (), the slope of the tangent line at the point = a is given b f (a). Written in Leibniz notation, the slope is d (a). In the case of the Scrambler ride, both and are functions of the parameter t. d Notice that if = (t) and = (t) both have derivatives that are continuous at t = c, the chain rule gives us As long as d (c), we then have dt d dt d d (a) = = d d d dt. d dt (c) = d dt (c) where a = f (c). In the case where (c) = (c) =, we define d (a) = lim d t c d dt d dt (c) (c), (.) (t) = lim t c (t), (.) provided the limit eists.

12 9- SECTION 9... Calculus and Parametric Equations 77 CAUTION Look carefull at (.) and convince ourself that d d d dt. d dt Equating these two epressions is a common error. You should be careful to avoid this trap. t d t w FIGURE 9.9 Tangent lines to the Scrambler path t We can use (.) to calculate second (as well as higher order) derivatives. Notice that if we replace b d d,weget ( ) d EXAMPLE. d d = d d ( ) d = d d dt d dt d Slopes of Tangent Lines to the Path of the Scrambler. (.) Find the slope of the tangent line to the path of the Scrambler = cos t + sin t, = sin t + cos t at (a) t = ; (b) t = π and (c) the point (, ). Solution (a) First, note that d d = sin t + cos t and = cos t sin t. dt dt From (.), the slope of the tangent line at t = isthen d d d = dt () cos sin = t= d dt () sin + cos =. (b) The slope of the tangent line at t = π is d ( π ) d cos π d = dt t=π/ d ( π ) = sin π sin π dt + cos π =. (c) To determine the slope at the point (, ), we must first determine a value of t that corresponds to the point. In this case, notice that t = π/gives = and =. Here, we have ( ) d π = d ( ) π = dt dt and consequentl, we must use (.) to compute d. Since the limit has the d indeterminate form,weusel Hôpital s Rule, to get ( ) d π cos t sin t = lim d t π/ sin t + cos t = lim sin t cos t t π/ cos t sin t, which does not eist, since the limit in the numerator is 6 and the limit in the denominator is. This sas that the slope of the tangent line at t = π/isundefined. In Figure 9.9, we have drawn in the tangent lines at t =,π/ and π/. Notice that the tangent line at the point (, ) is vertical. For the passenger on the Scrambler of eample., notice that the slope of the tangent line indicates the direction of motion and does not correspond to speed, which we discuss shortl. Finding slopes of tangent lines can help us identif man points of interest.

13 78 CHAPTER 9.. Parametric Equations and Polar Coordinates 9- EXAMPLE. Finding Vertical and Horizontal Tangent Lines Identif all points at which the plane curve = cos t, = sin t has a horizontal or vertical tangent line. Solution Asketch of the curve is shown in Figure 9.. There appear to be two locations (the top and bottom of the bow) with horizontal tangent lines and one point (the far right edge of the bow) with a vertical tangent line. Recall that horizontal tangent lines occur where d d =. From (.), we then have d d = (t) =, which can occur (t) onl when = (t) = cos t, FIGURE 9. = cos t, = sin t provided that (t) = sin t for the same value of t. Since cos θ = onl when θ is an odd multiple of π,wehave that (t) = cos t =, onl when t = π, π, 5π,...and so, t = π 6, π 6, 5π,... The corresponding points on the curve 6 are then ( ( π ) ( π )) (, = cos π 6 6, sin π ) ( ) =,, ( ( ) ( )) ( π π, = cos π, sin π ) = (, ), 6 6 ( ( ) ( )) ( 7π 7π, = cos 7π 6 6, sin 7π ) ( ) =, and ( ( ) 9π, 6 ( )) ( 9π = cos π, sin 9π 6 ) = (, ). Note that t = 5π π and t = reproduce the first and third points, respectivel, and 6 6 so on. The points (, ) and (, ) are on the top and bottom of the bow, respectivel, where there clearl are horizontal tangents. The points (, ) and (, ) should not seem quite right, though. These points are on the etreme ends of the bow and certainl don t look like the have vertical or horizontal tangents. In fact, the don t. Notice that at both t = π and t = π,wehave (t) = (t) = and so, the slope must be computed as a limit using (.). We leave it as an eercise to show that the slopes at t = π and t = π are 9 and 9, respectivel. To find points where there is a vertical tangent, we need to see where (t) = but (t). Setting = (t) = sin t, wegetsin t =, which occurs if t =,π, π,...or t =, π,π,... The corresponding points are ((), ()) = (cos, sin ) = (, ), ((π), (π)) = (cos π, sin π) = (, ) and the points corresponding to t = π and t = π, which we have alread discussed (where (t) =, also). Since (t) = cos t, for t = ort = π, there is a vertical tangent line onl at the point (, ). Theorem. generalizes what we observed in eample..

14 9-5 SECTION 9... Calculus and Parametric Equations 79 THEOREM. Suppose that (t) and (t) are continuous. Then for the curve defined b the parametric equations = (t) and = (t), (i) if (c) = and (c), there is a horizontal tangent line at the point ((c), (c)); (ii) if (c) = and (c), there is a vertical tangent line at the point ((c), (c)). PROOF The proof depends on the calculation of derivatives for parametric curves and is left as an eercise. ['(t)] ['(t)] '(t) Recall that our introductor question about the Scrambler was whether or not the rider ever comes to a complete stop. To answer this question, we will need to be able to compute velocities. Recall that if the position of an object moving along a straight line is given b the differentiable function f (t), the object s velocit is given b f (t). The situation with parametric equations is completel analogous. If the position is given b ((t), (t)), for differentiable functions (t) and (t), then the horizontal component of velocit is given b (t) and the vertical component of velocit is given b (t) (see Figure 9.). We define the speed to be [ (t)] + [ (t)]. From this, note that the speed is if and onl if (t) = (t) =. In this event, there is no horizontal or vertical motion. '(t) FIGURE 9. Horizontal and vertical components of velocit and speed EXAMPLE. Velocit of the Scrambler For the path of the Scrambler = cos t + sin t, = sin t + cos t, find the horizontal and vertical components of velocit and speed at times t = and t = π, and indicate the direction of motion. Also determine all times at which the speed is zero. Solution Here, the horizontal component of velocit is d = sin t + cos t and dt the vertical component is d = cos t sin t. Att =, the horizontal and vertical dt components of velocit both equal and the speed is + = 8. The rider is located at the point ((), ()) = (, ) and is moving to the right [since () > ] and up [since () > ]. At t = π, the velocit has components (horizontal) and (vertical) and the speed is 6 + =. At this time, the rider is located at the point (, ) and is moving to the left [since ( ) π < ]. In general, the speed of the rider at time t is given b (d ) s(t) = + dt ( ) d = ( sin t + cos t) dt + ( cos t sin t) = sin t 8 sin t cos t + cos t + cos t 8 cos t sin t + sin t = 8 8 sin t cos t 8 cos t sin t = 8 8 sin t, using the identities sin t + cos t =, cos t + sin t = and sin t cos t + sin t cos t = sin t. So, the speed is whenever sin t =.

15 7 CHAPTER 9.. Parametric Equations and Polar Coordinates 9-6 This occurs when t = π, 5π, 9π,...,or t = π 6, 5π 6, 9π,... The corresponding points 6 on the curve are ( ( ) ( π 6, π )) ( 6 = ( (, ), 5π ) ( 6, 5π )) ( 6 =, ) and ( ( 9π ) ( 6, 9π )) 6 = (, ). You can easil verif that these points are the three tips of the path seen in Figure 9.8b. We just showed that riders in the Scrambler of Figure 9.8b actuall come to a brief stop at the outside of each loop. As ou will eplore in the eercises, for similar Scrambler paths, the riders slow down but have a positive speed at the outside of each loop. This is true of the Scrambler at most carnivals, for which a more complicated path makes up for the lack of stopping. Notice that the Scrambler path shown in Figure 9.8b begins and ends at the same point and so, encloses an area. An interesting question is to determine the area enclosed b such a curve. Computing areas in parametric equations is a straightforward etension of our original development of integration. Recall that for a continuous function f defined on [a, b], where f () on[a, b], the area under the curve = f () for a b is given b A = b a f () d = b a d. Now, suppose that this same curve is described parametricall b = (t) and = (t), where the curve is traversed eactl once for c t d. Wecan then compute the area b making the substitution = (t). It then follows that d = (t) dt and so, the area is given b A = b a }{{} (t) d }{{} (t)dt = d c (t) (t) dt, where ou should notice that we have also changed the limits of integration to match the new variable of integration. We generalize this result in Theorem.. THEOREM. (Area Enclosed b a Curve Defined Parametricall) Suppose that the parametric equations = (t) and = (t), with c t d, describe a curve that is traced out clockwise eactl once, as t increases from c to d and where the curve does not intersect itself, ecept that the initial and terminal points are the same [i.e., (c) = (d) and (c) = (d)]. Then the enclosed area is given b A = d c (t) (t) dt = d c (t) (t) dt. (.) If the curve is traced out counterclockwise, then the enclosed area is given b A = d c (t) (t) dt = d c (t) (t) dt. (.5) PROOF This result is a special case of Green s Theorem, which we will develop in section.. The new area formulas given in Theorem. turn out to be quite useful. As we see in eample., we can use these to find the area enclosed b a parametric curve.

16 9-7 SECTION 9... Calculus and Parametric Equations 7 EXAMPLE. Finding the Area Enclosed b a Curve Find the area enclosed b the path of the Scrambler = cos t + sin t, = sin t + cos t. Solution Notice that the curve is traced out counterclockwise once for t π. From (.5), the area is then A = = π π (t) (t) dt = π ( cos t + sin t)( cos t sin t) dt ( cos t cos t sin t sin t) dt = π, where we evaluated the integral using a CAS. In eample.5, we use Theorem. to derive a formula for the area enclosed b an ellipse. Pa particular attention to how much easier this is to do with parametric equations than it is to do with the original - equation. EXAMPLE.5 Finding the Area Enclosed b an Ellipse Find the area enclosed b the ellipse a + = (for constants a, b > ). b Solution One wa to compute the area is to solve the equation for to obtain =±b and then integrate: a [ ( a A = b )] a b d. a a You can evaluate this integral b trigonometric substitution or b using a CAS, but a simpler, more elegant wa to compute the area is to use parametric equations. Notice that the ellipse is described parametricall b = a cos t, = b sin t, for t π. The ellipse is then traced out counterclockwise eactl once for t π, sothat the area is given b (.5) to be π π π A = (t) (t) dt = (b sin t)( a sin t) dt = ab sin tdt= abπ, where this last integral can be evaluated b using the half-angle formula: sin t = ( cos t). We leave the details of this calculation as an eercise. BEYOND FORMULAS The formulas in this section are not new, but are simpl modifications of the wellestablished rules for differentiation and integration. If ou think of them this wa, the are not complicated memorization eercises, but instead are old standards epressed in a slightl different wa.

17 7 CHAPTER 9.. Parametric Equations and Polar Coordinates 9-8 EXERCISES 9. WRITING EXERCISES. In the derivation of parametric equations for the Scrambler, we{ used the fact that reversing the sine and cosine functions = sin t to causes the circle to be traced out clockwise. = cos t Eplain wh this is so b starting at t = and following the graph as t increases to π.. Eplain wh Theorem. makes sense. (Hint: If (c) =, what does that sa about the change in -coordinates on the graph? Wh do ou also need (c) toguarantee a horizontal tangent?). Imagine an object with position given b (t) and (t). If a right triangle has a horizontal leg of length (t) and a vertical leg of length (t), what would the length of the hpotenuse represent? Eplain wh this makes sense.. Eplain wh the sign (±) of d (t) (t) dt in Theorem. c is different for curves traced out clockwise and counterclockwise. In eercises 6, find the slopes of the tangent lines to the given curves at the indicated points.. { = t = t t (a) t =, (b) t =, (c) (, ). { = t t = t 5t + (a) t =, (b) t =, (c) (, ). { = cos t = sin t (a) t = π, (b) t = π, (c) (, ). { = cos t = sin t (a) t = π, (b) t = π, (c) (, ) 5. { = cos t = sin t ( ) (a) t = π, (b) t = π, (c) 6. { = cos t = sin t (a) t = π, (b) t = π, (c) (, ) In eercises 9, identif all points at which the curve has (a) a horizontal tangent and (b) a vertical tangent. { { = cos t = cos t 9.. = sin t = sin 7t { { = t. = t = t. t = t t { { = cos t + sin t = cos t + sin t.. = sin t + cos t = sin t + cos t In eercises 5, parametric equations for the position of an object are given. Find the object s velocit and speed at the given times and describe its motion. 5. { = cos t = sin t (a) t =, (b) t = π 6. { = cos t = sin t (a) t =, (b) t = π 7. { = t = t 6t (a) t =, (b) t = 8. { = t + 5 = + t 6t (a) t =, (b) t = 9. { = cos t + sin 5t = sin t + cos 5t (a) t =, (b) t = π. { = cos t + sin t = sin t + cos t (a) t =, (b) t = π In eercises 8, find the area enclosed b the given curve. { { = cos t = 6 cos t.. = sin t = sin t { =. cos t { cos t = cos t + cos t = sin t sin t. = sin t + sin t { = cos t 5. = sin t, π t π In eercises 7 and 8, sketch the graph and find the slope of the curve at the given point. { = t 7. = t at (, ) t { = t 8. t = t 5t at (, ) { = t sin t = t cos t, π t π { = t t = t, t { = t t = t, t

18 9-9 SECTION 9... Calculus and Parametric Equations 7 In eercises 9 and, find the speed of the object each time it crosses the -ais. { { = cos 9. t + cos t = 6 cos t + 5 cos t. = ( cos t) sin t = 6 sin t 5 sin t. A{ modification of the Scrambler in eample. is = cos t + sin 5t.Ineample., the ratio of the speed = sin t + cos 5t of the outer arms to the speed of the inner arms is -to-. What is the ratio in this version of the Scrambler? Sketch a graph showing the motion of this new Scrambler.. Compute the speed of the Scrambler in eercise. Using trigonometric identities as in eample., show that the speed is at a minimum when sin 8t = but that the speed is never zero. Show that the minimum speed is reached at the outer points of the path.. Find parametric equations for a Scrambler that is the same as in eample. ecept that the outer arms rotate three times as fast as the inner arms. Sketch a graph of its motion and determine its minimum and maimum speeds.. Find parametric equations for a Scrambler that is the same as in eample. ecept that the inner arms have length. Sketch a graph of its motion and determine its minimum and maimum speeds. { = sin t 5. Suppose an object follows the path. Show that = cos t its speed is constant. Show that, at an time t, the tangent line is perpendicular to a line connecting the origin and the object. 6. A Ferris wheel has height feet and completes one revolution in minutes at a constant speed. Compute the speed of a rider in the Ferris wheel. 7. Suppose ou are standing at the origin watching an object that has position ((t), (t)) at time t. Show that, ( from ) our (t) perspective, the object is moving clockwise if < ( ) (t) (t) and is moving counterclockwise if >. (t) 8. In the Ptolemaic model of planetar motion, the earth was at the center of the solar sstem and the sun and planets orbited the earth. Circular orbits, which were preferred for aesthetic reasons, could not account for the actual motion of the planets as viewed from the earth. Ptolem modified the circles into epiccloids, which are circles on circles similar to the Scrambler of eample.. Suppose that a planet s motion is given b { = cos 6πt + cos πt. Using the result of eercise = sin 6πt + sin πt 7, find the intervals in which the planet rotates clockwise and the intervals in which the planet rotates counterclockwise. 9. Find parametric equations for the path traced out b a specific point on a circle of radius r rolling from left to right at a constant speed v>r. Assume that the point starts at (r, r) at time t =. (Hint: First, find parametric equations for the center of the circle. Then, add on parametric equations for the point going around the center of the circle.) Find the minimum and maimum speeds of the point and the locations where each occurs. Graph the curve for v = and r =. This curve is called a ccloid.. Find parametric equations for the path traced out b a specific point inside the circle as the circle rolls from left to right. (Hint: If r is the radius of the circle, let d < r be the distance from the point to the center.) Find the minimum and maimum speeds of the point and the locations where each occurs. Graph the curve for v =, r = and d =. This curve is called a trochoid.. A hpoccloid is the path traced out b a point on a smaller circle of radius b that is rolling inside a larger circle of radius a > b. Find parametric equations for the hpoccloid and graph it for a = 5 and b =. Find an equation in terms of the parameter t for the slope of the tangent line to the hpoccloid and determine one point at which the tangent line is vertical. What interesting simplification occurs if a = b? Figure for eercise Figure for eercise. An epiccloid is the path traced out b a point on a smaller circle of radius b that is rolling outside a larger circle of radius a > b. Find parametric equations for the epiccloid and graph it for a = 8 and b = 5. Find an equation in terms of the parameter t for the slope of the tangent line to the epiccloid and determine one point at which the slope is vertical. What interesting simplification occurs if a = b?. Suppose that = cos t and = sin t.atthe point (, ), show that d d ( ) d dt (π/6). d (π/6) dt. For = at and = b for nonzero constants a and b, determine whether there are an values of t such that d d ((t)) = d dt (t). d (t) dt

19 7 CHAPTER 9.. Parametric Equations and Polar Coordinates 9- EXPLORATORY EXERCISES. B varing the speed of the outer { arms, the Scrambler of eample. can be generalized to for some = cos t + sin kt = sin t + cos kt positive constant k. Show that the minimum speed for an such Scrambler is reached at the outside of a loop. Show that the onl value of k that actuall produces a speed of is k =. B varing the lengths { of the arms, ou can further generalize = r cos t + sin kt the Scrambler to for positive constants = r sin t + cos kt r > and k.sketch the paths for several such Scramblers and determine the relationship between r and k needed to produce a speed of.. Bézier curves are essential in almost all areas of modern engineering design. (For instance, Bézier curves were used for sketching man of the figures for this book.) One version of a Bézier curve starts with control points at (a, a ), (b, b ), (c, c ) and (d, d ). The Bézier curve passes through the points (a, a ) and (d, d ). The tangent line at = a passes through (b, b ) and the tangent line at = d passes through (c, c ). Show that these criteria are met, for t, with = (a + b c d)t + (d b + c a)t + (b a)t + a = ( a + b c d )t + ( d b + c a )t + ( b a )t + a Use this formula to find and graph the Bézier curve with control points (, ), (, ), (, ) and (, ). Eplore the effect of moving the middle control points, for eample, moving them up to (, ) and (, ), respectivel. 9. ARC LENGTH AND SURFACE AREA IN PARAMETRIC EQUATIONS ((a), (a)) FIGURE 9.a The plane curve C ((b), (b)) In this section, we investigate arc length and surface area for curves defined parametricall. Along the wa, we eplore one of the most famous and interesting curves in mathematics. Let C be the curve defined b the parametric equations = (t) and = (t), for a t b (see Figure 9.a), where,, and are continuous on the interval [a, b]. We further assume that the curve does not intersect itself, ecept possibl at a finite number of points. Our goal is to compute the length of the curve (the arc length). As we have done countless times now, we begin b constructing an approimation. First, we divide the t-interval [a, b] into n subintervals of equal length, t: a = t < t < t < < t n = b, where t i t i = t = b a, for each i =,,,...,n. For each subinterval [t i, t i ], n we approimate the arc length s i of the portion of the curve joining the point ((t i ), (t i )) to the point ((t i ), (t i )) with the length of the line segment joining these points. This approimation is shown in Figure 9.b for the case where n =. We have s i d{((t i ), (t i )), ((t i ), (t i ))} = [(t i ) (t i )] + [(t i ) (t i )]. FIGURE 9.b Approimate arc length Recall that from the Mean Value Theorem (see section.9 and make sure ou know wh we can appl it here), we have that and (t i ) (t i ) = (c i )(t i t i ) = (c i ) t (t i ) (t i ) = (d i )(t i t i ) = (d i ) t,

20 7 CHAPTER 9.. Parametric Equations and Polar Coordinates 9- EXPLORATORY EXERCISES. B varing the speed of the outer { arms, the Scrambler of eample. can be generalized to for some = cos t + sin kt = sin t + cos kt positive constant k. Show that the minimum speed for an such Scrambler is reached at the outside of a loop. Show that the onl value of k that actuall produces a speed of is k =. B varing the lengths { of the arms, ou can further generalize = r cos t + sin kt the Scrambler to for positive constants = r sin t + cos kt r > and k.sketch the paths for several such Scramblers and determine the relationship between r and k needed to produce a speed of.. Bézier curves are essential in almost all areas of modern engineering design. (For instance, Bézier curves were used for sketching man of the figures for this book.) One version of a Bézier curve starts with control points at (a, a ), (b, b ), (c, c ) and (d, d ). The Bézier curve passes through the points (a, a ) and (d, d ). The tangent line at = a passes through (b, b ) and the tangent line at = d passes through (c, c ). Show that these criteria are met, for t, with = (a + b c d)t + (d b + c a)t + (b a)t + a = ( a + b c d )t + ( d b + c a )t + ( b a )t + a Use this formula to find and graph the Bézier curve with control points (, ), (, ), (, ) and (, ). Eplore the effect of moving the middle control points, for eample, moving them up to (, ) and (, ), respectivel. 9. ARC LENGTH AND SURFACE AREA IN PARAMETRIC EQUATIONS ((a), (a)) FIGURE 9.a The plane curve C ((b), (b)) In this section, we investigate arc length and surface area for curves defined parametricall. Along the wa, we eplore one of the most famous and interesting curves in mathematics. Let C be the curve defined b the parametric equations = (t) and = (t), for a t b (see Figure 9.a), where,, and are continuous on the interval [a, b]. We further assume that the curve does not intersect itself, ecept possibl at a finite number of points. Our goal is to compute the length of the curve (the arc length). As we have done countless times now, we begin b constructing an approimation. First, we divide the t-interval [a, b] into n subintervals of equal length, t: a = t < t < t < < t n = b, where t i t i = t = b a, for each i =,,,...,n. For each subinterval [t i, t i ], n we approimate the arc length s i of the portion of the curve joining the point ((t i ), (t i )) to the point ((t i ), (t i )) with the length of the line segment joining these points. This approimation is shown in Figure 9.b for the case where n =. We have s i d{((t i ), (t i )), ((t i ), (t i ))} = [(t i ) (t i )] + [(t i ) (t i )]. FIGURE 9.b Approimate arc length Recall that from the Mean Value Theorem (see section.9 and make sure ou know wh we can appl it here), we have that and (t i ) (t i ) = (c i )(t i t i ) = (c i ) t (t i ) (t i ) = (d i )(t i t i ) = (d i ) t,

21 9- SECTION 9... Arc Length and Surface Area in Parametric Equations 75 where c i and d i are some points in the interval (t i, t i ). This gives us s i [(t i ) (t i )] + [(t i ) (t i )] = [ (c i ) t] + [ (d i ) t] = [ (c i )] + [ (d i )] t. Notice that if t is small, then c i and d i are close together. So, we can make the further approimation s i [ (c i )] + [ (c i )] t, for each i =,,...,n. The total arc length is then approimatel n s [ (c i )] + [ (c i )] t. i= Taking the limit as n then gives us the eact arc length, which ou should recognize as an integral: n b s = lim [ (c i )] + [ (c i )] t = [ (t)] + [ (t)] dt. n i= We summarize this discussion in Theorem.. a THEOREM. (Arc Length for a Curve Defined Parametricall) For the curve defined parametricall b = (t), = (t), a t b, if and are continuous on [a, b] and the curve does not intersect itself (ecept possibl at a finite number of points), then the arc length s of the curve is given b b b (d ) ( ) s = [ d (t)] + [ (t)] dt = + dt. (.) dt dt a a In eample., we illustrate the use of (.) to find the arc length of the Scrambler curve from eample.. EXAMPLE. Finding the Arc Length of a Plane Curve Find the arc length of the Scrambler curve = cos t + sin t, = sin t + cos t, for t π. Also, find the average speed of the Scrambler over this interval. Solution The curve is shown in Figure 9.. First, note that,, and are all continuous on the interval [, π]. From (.), we then have FIGURE 9. = cos t + sin t, = sin t + cos t, t π s = = = b a π π (d ) + dt ( ) d π dt = ( sin t + cos t) + ( cos t sin t) dt dt sin t 8 sin t cos t + cos t + cos t 8 cos t sin t + sin tdt π 8 8 sin t cos t 8 cos t sin t dt= 8 8 sin t dt 6,

22 76 CHAPTER 9.. Parametric Equations and Polar Coordinates 9- since sin t + cos t =, cos t + sin t = and sin t cos t + sin t cos t = sin t and where we have approimated the last integral numericall. To find the average speed over the given interval, we simpl divide the arc length (i.e., the distance traveled), b the total time, π, toobtain s ave 6 π.56. We want emphasize that Theorem. allows the curve to intersect itself at a finite number of points, but a curve cannot intersect itself over an entire interval of values of the parameter t. Tosee wh this requirement is needed, notice that the parametric equations = cos t, = sin t, for t π, describe the circle of radius centered at the origin. However, the circle is traversed twice as t ranges from to π. Ifou were to appl (.) to this curve, ou d obtain π (d ) + dt ( ) d π dt = ( sin t) + cos dt tdt= π, which corresponds to twice the arc length (circumference) of the circle. As ou can see, if a curve intersects itself over an entire interval of values of t, the arc length of such a portion of the curve is counted twice b (.). EXAMPLE. Finding the Arc Length of a Complicated Plane Curve Find the arc length of the plane curve = cos 5t, = sin 7t, for t π. Solution This unusual curve (an eample of a Lissajous curve) issketched in Figure 9.. We leave it as an eercise to verif that the hpotheses of Theorem. are met. From (.), we then have that π (d ) ( ) d π s = + dt = ( 5 sin 5t) + (7 cos 7t) dt dt dt 6.5, where we have approimated the integral numericall. This is a long curve to be confined within the rectangle,! The arc length formula (.) should seem familiar to ou. Parametric equations for a curve = f () are = t, = f (t) and from (.), the arc length of this curve for a b is then FIGURE 9. A Lissajous curve s = b a (d ) + dt ( ) d dt = dt b a + [ f (t)] dt, which is the arc length formula derived in section 5.. Thus, the formula developed in section 5. is a special case of (.). Observe that the speed of the Scrambler calculated in eample. and the length of the Scrambler curve found in eample. both depend on the same (d ) ( ) d quantit: +. Observe that if the parameter t represents time, then dt dt (d ) ( ) d + represents speed and from Theorem., the arc length (i.e., the dis- dt dt tance traveled) is the integral of the speed with respect to time.

23 9- SECTION 9... Arc Length and Surface Area in Parametric Equations 77 A FIGURE 9.5 Downhill skier B We can use our notion of arc length to address a famous problem called the brachistochrone problem. We state this problem in the contet of a downhill skier. Consider a ski slope consisting of a tilted plane, where a skier wishes to get from a point A at the top of the slope to a point B down the slope (but not directl beneath A)inthe least time possible (see Figure 9.5). Suppose the path taken b the skier is given b the parametric equations = (u) and = (u), u, where and determine the position of the skier in the plane of the ski slope. (For simplicit, we orient the positive -ais so that it points down. Also, we name the parameter u since u will, in general, not represent time.) To derive a formula for the time required to get from point A to point B, start with the simple formula d = r t relating the distance to the time and the rate. As seen in the derivation of the arc length formula (.), for a small section of the curve, the distance is approimatel [ (u)] + [ (u)]. The rate is harder to identif since we aren t given position as a function of time. For simplicit, we assume that the onl effect of friction is to keep the skier on the path and that (t). In this case, using the principle of conservation (u) of energ, it can be shown that the skier s speed is given b for some constant k. k Putting the pieces together, the total time from point A to point B is given b [ Time = k (u)] + [ (u)] du. (.) (u) Your first thought might be that the shortest path from point A to point B is along a straight line. If ou re thinking of short in terms of distance, ou re right, of course. However, if ou think of short in terms of time (how most skiers would think of it), this is not true. In eample., we show that the fastest path from point A to point B is, in fact, not along a straight line, b ehibiting a faster path. EXAMPLE. Skiing a Curved Path that Is Faster Than Skiing a Straight Line If point A in our skiing eample is (, ) and point B is (π, ), show that the ccloid defined b = πu sin πu, = cos πu is faster than the line segment connecting the points. Eplain the result in phsical terms. Solution First, note that the line segment connecting the points is given b = πu, = u, for u. Further, both curves meet the endpoint requirements that ((), ()) = (, ) and ((), ()) = (π, ). For the ccloid, we have from (.) that [ Time = k (u)] + [ (u)] du (u) (π π cos πu) = k + (π sin πu) du cos πu = k cos πu π cos πu du = k π.

24 78 CHAPTER 9.. Parametric Equations and Polar Coordinates 9- Similarl, for the line segment, we have that [ Time = k (u)] + [ (u)] du (u) π + = k du u FIGURE 9.6 Two skiing paths = k π +. Notice that the ccloid route is faster since π< π +. The two paths are shown in Figure 9.6. Observe that the ccloid is ver steep at the beginning, which would allow a skier to go faster following the ccloid than following the straight line. As it turns out, the greater speed of the ccloid more than compensates for the longer distance of the ccloid path. HISTORICAL NOTES Jacob Bernoulli (65 75) and Johann Bernoulli (667 78) Swiss mathematicians who were instrumental in the development of the calculus. Jacob was the first of several generations of Bernoullis to make important contributions to mathematics. He was active in probabilit, series and the calculus of variations and introduced the term integral. Johann followed his brother into mathematics while also earning a doctorate in medicine. Johann first stated l Hôpital s Rule, one of man results over which he fought bitterl (usuall with his brother, but, after Jacob s death, also with his own son Daniel) to receive credit. Both brothers were sensitive, irritable, egotistical (Johann had his tombstone inscribed, The Archimedes of his age ) and quick to criticize others. Their competitive spirit accelerated the development of calculus. We will ask ou to construct some skiing paths of our own in the eercises. However, it has been proved that the ccloid is the plane curve with the shortest time (which is what the Greek root words for brachistochrone mean). In addition, we will give ou an opportunit to discover another remarkable propert of the ccloid, relating to another famous problem, the tautochrone problem. Both problems have an interesting histor focused on brothers Jacob and Johann Bernoulli, who solved the problem in 697 (along with Newton, Leibniz and l Hôpital) and argued incessantl about who deserved credit. Much as we did in section 5., we can use our arc length formula to find a formula for the surface area of a surface of revolution. Recall that if the curve = f () for c d is revolved about the -ais (see Figure 9.7), the surface area is given b Surface Area = d c π f () + [ f ()] }{{} d. }{{} radius arc length Let C be the curve defined b the parametric equations = (t) and = (t) with a t b, where,, and are continuous and where the curve does not intersect itself for a t b. Weleave it as an eercise to derive the corresponding formula for parametric equations: Surface Area = b a π (t) [ (t)] }{{} + [ (t)] dt. }{{} radius arc length More generall, we have that if the curve is revolved about the line = c, the surface area is given b Surface Area = b a π (t) c [ (t)] }{{} + [ (t)] dt. (.) }{{} radius arc length Likewise, if we revolve the curve about the line = d, the surface area is given b Surface Area = b a π (t) d [ (t)] }{{} + [ (t)] dt. (.) }{{} radius arc length

25 9-5 SECTION 9... Arc Length and Surface Area in Parametric Equations 79 Look carefull at what all of the surface area formulas have in common. That is, in each case, the surface area is given b f() Surface Area = SURFACE AREA b a π(radius)(arc length) dt. (.5) c d Look carefull at the graph of the curve and the ais about which ou are revolving, to see how to fill in the blanks in (.5). As we observed in section 5., it is ver important that ou draw a picture here. Circular cross sections FIGURE 9.7 Surface of revolution EXAMPLE. Finding Surface Area with Parametric Equations Find the surface area of the surface formed b revolving the half-ellipse 9 + =,, about the -ais (see Figure 9.8). Solution It would trul be a mess to set up the integral for = f () = /9. (Think about this!) Instead, notice that ou can represent the curve b the parametric equations = cos t, = sin t, for t π. From (.), the surface area is then given b FIGURE 9.8 = 9 Surface Area = π = π π ( sin t) ( sin t) + ( cos t) }{{} dt }{{} radius arc length π sin t 9 sin t + cos tdt = π 9 5 sin ( 5/) + 5 where we used a CAS to evaluate the integral. 67.7, FIGURE 9.9 = sin t, = cos t EXAMPLE.5 Revolving about a Line Other Than a Coordinate Ais Find the surface area of the surface formed b revolving the curve = sin t, = cos t, for t π/, about the line =. Solution Asketch of the curve is shown in Figure 9.9. Since the -values on the curve are all less than, the radius of the solid of revolution is = sin t and so, from (.), the surface area is given b π/ Surface Area = π ( sin t) [ cos t] + [ sin t] }{{} dt., }{{} radius arc length where we have approimated the value of the integral numericall. In eample.6, we model a phsical process with parametric equations. Since the modeling process is itself of great importance, be sure that ou understand all of the steps. See if ou can find an alternative approach to this problem.

26 7 CHAPTER 9.. Parametric Equations and Polar Coordinates 9-6 FIGURE 9. Ladder sliding down a wall EXAMPLE.6 Arc Length for a Falling Ladder An 8-foot-tall ladder stands verticall against a wall. The bottom of the ladder is pulled along the floor, with the top remaining in contact with the wall, until the ladder rests flat on the floor. Find the distance traveled b the midpoint of the ladder. Solution We first find parametric equations for the position of the midpoint of the ladder. We orient the - and -aes as shown in Figure 9.. Let denote the distance from the wall to the bottom of the ladder and let denote the distance from the floor to the top of the ladder. Since the ladder is 8 feet long, observe that + = 6. Defining the parameter t =,wehave = 6 t. The midpoint of the ladder has coordinates (, ) and so, parametric equations for the midpoint are { (t) = t (t) =. 6 t When the ladder stands verticall against the wall, we have = and when it lies flat on the floor, = 8. So, t 8. From (.), the arc length is then given b 8 ( ) ( ) t 8 ( s = + dt = + t ) dt 6 t 6 t = 6 t dt = (t/8) dt. Substituting u = t 8 gives us du = dt or dt = 8 du.for the limits of integration, note 8 that when t =, u = and when t = 8, u =. The arc length is then 8 s = (t/8) dt = u 8 du = sin u u= u= ( π ) = = π. Since this is a rare arc length integral that can be evaluated eactl, ou might be suspicious that there is an easier wa to find the arc length. We eplore this in the eercises. EXERCISES 9. WRITING EXERCISES. In the derivation preceding Theorem., we justified the equation g(t i ) g(t i ) = g (c i ) t. Thinking of g(t)asposition and g (t)asvelocit, eplain wh this makes sense.. The curve in eample. was a long curve contained within a small rectangle. What would ou guess would be the maimum length for a curve contained in such a rectangle? Briefl eplain.. In eample., we noted that the steeper initial slope of the ccloid would allow the skier to build up more speed than the straight-line path. The ccloid takes this idea to the limit b having a vertical tangent line at the origin. Eplain wh, despite the vertical tangent line, it is still phsicall possible for the skier to sta on this slope. (Hint: How do the two dimensions of the path relate to the three dimensions of the ski slope?). The tautochrone problem discussed in eplorator eercise involves starting on the same curve at two different places and

27 9-7 SECTION 9... Arc Length and Surface Area in Parametric Equations 7 comparing the times required to reach the end. For the ccloid, compare the speed of a skier starting at the origin versus one starting halfwa to the bottom. Eplain wh it is not clear whether starting halfwa down would get ou to the bottom faster. In eercises, find the arc length of the curve; approimate numericall, if needed.. { { = cos t = cos t. = sin t = + sin t. { = t t = t, t. { { = t t = cos t = t, t 5. t = sin t 6. { = cos 7t = sin t 7. { = t cos t = t sin t, t 8. { = t cos t = t sin t, t 9. { = sin t cos t = sin t sin t, t π/. { = sin t cos t = sin t sin t, t π/. { = sin t = sin πt, t π. { = sin t = sin t, t π In eercises 6, show that the curve starts at the origin at t and reaches the point (π, ) at t. Then use the time formula (.) to determine how long it would take a skier to take the given path. In eercises 6, compute the surface area of the surface obtained b revolving the given curve about the indicated ais.. { = t = t t, t, about the -ais. { = t = t t, t, about the -ais. { = t = t t, t, about the -ais. { = t = t t, t, about = 5. { = t t = t, t, about the -ais 6. { = t t = t, t, about = 7. An 8-foot-tall ladder stands verticall against a wall. The top of the ladder is pulled directl awa from the wall, with the bottom remaining in contact with the wall, until the ladder rests on the floor. Find parametric equations for the position of the midpoint of the ladder. Find the distance traveled b the midpoint of the ladder. 8. The answer in eercise 7 equals the circumference of a quarter-circle of radius. Discuss whether this is a coincidence or not. Compare this value to the arc length in eample.6. Discuss whether or not this is a coincidence. 9. The figure shown here is called Cornu s spiral. It is defined b the parametric equations = t cos πs ds and = t sin πs ds. Each of these integrals is important in the stud of Fresnel diffraction. Find the arc length of the spiral for (a) π t π and (b) general a t b. Use this result to discuss the rate at which the spiralling occurs { { = πt = πt =. t = t { { = π(cos πt ) = πt.6 sin πt = t + 7 sin πt 6. = t +. sin πt In eercises 7, find the slope at the origin and the arc length for the curve in the indicated eercise. Compare to the ccloid from eample.. 7. eercise 8. eercise 9. eercise 5. eercise 6. Accloid is the curve traced out b a point on a circle as the circle rolls along the -ais. Suppose the circle has radius, the point we are following starts at (, 8) and the circle rolls from left to right. Find parametric equations for the ccloid and find the arc length as the circle completes one rotation.

28 7 CHAPTER 9.. Parametric Equations and Polar Coordinates 9-8 EXPLORATORY EXERCISES. For the brachistochrone problem, two criteria for the fastest curve are: () steep slope at the origin and () concave down (note in Figure 9.6 that the positive -ais points downward). Eplain wh these criteria make sense and identif other criteria. Then find parametric equations for a curve (different from the ccloid or those of eercises 6) that meet all the criteria. Use the formula of eample. to find out how fast our curve is. You can t beat the ccloid, but get as close as ou can!. The tautochrone problem is another surprising problem that was studied and solved b the same seventeenth-centur mathematicians as the brachistochrone problem. (See Journe Through Genius b William Dunham for a description of this interesting piece of histor, featuring the brilliant et combative Bernoulli brothers.) Recall that the ccloid of eample. runs from (, ) to (π, ). It takes the skier k π = π/g seconds to ski the path. How long would it take the skier starting partwa down the path, for instance, at (π/, )? Find the slope of the ccloid at this point and compare it to the slope at (, ). Eplain wh the skier would build up less speed starting at this new point. Graph the speed function for the ccloid with u and eplain wh the farther down the slope ou start, the less speed ou ll have. To see how speed and distance balance, use the time formula T = π cos πu du g a cos πa cos πu for the time it takes to ski the ccloid starting at the point (πa sin πa, cos πa), < a <. What is the remarkable propert that the ccloid has? 9. POLAR COORDINATES (, ) You ve probabl heard the cliche about how difficult it is to tr to fit a round peg into a square hole. In some sense, we have faced this problem on several occasions so far in our stud of calculus. For instance, if we were to use an integral to calculate the area of the circle + = 9, we would have A = [ 9 ( )] 9 d = 9 d. (.) FIGURE 9. Rectangular coordinates u r FIGURE 9. Polar coordinates (r, u) Note that ou can evaluate this integral b making the trigonometric substitution = sin θ. (It s a good thing that we alread know a simple formula for the area of a circle!) A better plan might be to use parametric equations, such as = cos t, = sin t, for t π, to describe the circle. In section 9., we saw that the area is given b π (t) (t) dt = = 9 π π ( cos t)( cos t) dt cos tdt. This is certainl better than the integral in (.), but it still requires some effort to evaluate this. The basic problem is that circles do not translate well into the usual - coordinate sstem. We often refer to this sstem as a sstem of rectangular coordinates, because a point is described in terms of the horizontal and vertical distances from the origin (see Figure 9.). An alternative description of a point in the -plane consists of specifing the distance r from the point to the origin and an angle θ (in radians) measured from the positive -ais counterclockwise to the ra connecting the point and the origin (see Figure 9.). We describe the point b the ordered pair (r,θ) and refer to r and θ as polar coordinates for the point.

29 7 CHAPTER 9.. Parametric Equations and Polar Coordinates 9-8 EXPLORATORY EXERCISES. For the brachistochrone problem, two criteria for the fastest curve are: () steep slope at the origin and () concave down (note in Figure 9.6 that the positive -ais points downward). Eplain wh these criteria make sense and identif other criteria. Then find parametric equations for a curve (different from the ccloid or those of eercises 6) that meet all the criteria. Use the formula of eample. to find out how fast our curve is. You can t beat the ccloid, but get as close as ou can!. The tautochrone problem is another surprising problem that was studied and solved b the same seventeenth-centur mathematicians as the brachistochrone problem. (See Journe Through Genius b William Dunham for a description of this interesting piece of histor, featuring the brilliant et combative Bernoulli brothers.) Recall that the ccloid of eample. runs from (, ) to (π, ). It takes the skier k π = π/g seconds to ski the path. How long would it take the skier starting partwa down the path, for instance, at (π/, )? Find the slope of the ccloid at this point and compare it to the slope at (, ). Eplain wh the skier would build up less speed starting at this new point. Graph the speed function for the ccloid with u and eplain wh the farther down the slope ou start, the less speed ou ll have. To see how speed and distance balance, use the time formula T = π cos πu du g a cos πa cos πu for the time it takes to ski the ccloid starting at the point (πa sin πa, cos πa), < a <. What is the remarkable propert that the ccloid has? 9. POLAR COORDINATES (, ) You ve probabl heard the cliche about how difficult it is to tr to fit a round peg into a square hole. In some sense, we have faced this problem on several occasions so far in our stud of calculus. For instance, if we were to use an integral to calculate the area of the circle + = 9, we would have A = [ 9 ( )] 9 d = 9 d. (.) FIGURE 9. Rectangular coordinates u r FIGURE 9. Polar coordinates (r, u) Note that ou can evaluate this integral b making the trigonometric substitution = sin θ. (It s a good thing that we alread know a simple formula for the area of a circle!) A better plan might be to use parametric equations, such as = cos t, = sin t, for t π, to describe the circle. In section 9., we saw that the area is given b π (t) (t) dt = = 9 π π ( cos t)( cos t) dt cos tdt. This is certainl better than the integral in (.), but it still requires some effort to evaluate this. The basic problem is that circles do not translate well into the usual - coordinate sstem. We often refer to this sstem as a sstem of rectangular coordinates, because a point is described in terms of the horizontal and vertical distances from the origin (see Figure 9.). An alternative description of a point in the -plane consists of specifing the distance r from the point to the origin and an angle θ (in radians) measured from the positive -ais counterclockwise to the ra connecting the point and the origin (see Figure 9.). We describe the point b the ordered pair (r,θ) and refer to r and θ as polar coordinates for the point.

30 9-9 SECTION 9... Polar Coordinates 7 EXAMPLE. Converting from Polar to Rectangular Coordinates Plot the points with the indicated polar coordinates and determine the corresponding rectangular coordinates (, ) for: (a) (, ), (b) (, π ), (c) (, π ) and (d) (,π). Solution (a) Notice that the angle θ = locates the point on the positive -ais. At a distance of r = units from the origin, this corresponds to the point (, ) in rectangular coordinates (see Figure 9.a). (b) The angle θ = π locates points on the positive -ais. At a distance of r = units from the origin, this corresponds to the point (, ) in rectangular coordinates (see Figure 9.b). (c) The angle is the same as in (b), but a negative value of r indicates that the point is located units in the opposite direction, at the point (, ) in rectangular coordinates (see Figure 9.b). (d) The angle θ = π corresponds to the negative -ais. The distance of r = units from the origin gives us the point (, ) in rectangular coordinates (see Figure 9.c). (, q) (, ) q (, p) p (, q) FIGURE 9.a The point (, ) in polar coordinates FIGURE 9.b ( The points, π ) ( and, π ) in polar coordinates FIGURE 9.c The point (,π)in polar coordinates FIGURE 9.a Polar coordinates for the point (, ) d EXAMPLE. Converting from Rectangular to Polar Coordinates Find a polar coordinate representation of the rectangular point (, ). Solution From Figure 9.a, notice that the point lies on the line =, which makes an angle of π with the positive -ais. From the distance formula, we get that r = + =. This sas that we can write the point as (, π )inpolar coordinates. Referring to Figure 9.b (on the following page), notice that we can specif the same point b using a negative value of r, r =, with the angle 5π. (Think about this some.) Notice further, that the angle 9π = π + π corresponds to the

31 7 CHAPTER 9.. Parametric Equations and Polar Coordinates 9- h, d p FIGURE 9.b An alternative polar representation of (, ) FIGURE 9.c Another polar representation of the point (, ) REMARK. As we see in eample., each point (, ) inthe plane has infinitel man polar coordinate representations. For a given angle θ, the angles θ ± π, θ ± π and so on, all correspond to the same ra. For convenience, we use the notation θ + nπ (for an integer n) torepresent all of these possible angles. (r, u) same ra shown in Figure 9.a (see Figure 9.c). In fact, all of the polar points (, π + nπ) and (, 5π + nπ) for an integer n correspond to the same point in the -plane. Referring to Figure 9.5, notice that it is a simple matter to find the rectangular coordinates (, ) of a point specified in polar coordinates as (r,θ). From the usual definitions for sin θ and cos θ, weget = r cos θ and = r sin θ. (.) From equations (.), notice that for a point (, ) inthe plane, + = r cos θ + r sin θ = r (cos θ + sin θ) = r and for, = r sin θ r cos θ = sin θ = tan θ. cos θ That is, ever polar coordinate representation (r,θ)ofthe point (, ), where must satisf r = + and tan θ =. (.) r r sin u Notice that since there s more than one choice of r and θ, wecannot actuall solve equations (.) to produce formulas for r and θ. Inparticular, while ou might be tempted to write θ = tan ( ) u, this is not the onl possible choice. Remember that for (r,θ)tobe a polar representation of the point (, ),θ can be an angle for which tan θ =, while tan ( ) ( gives ou an angle θ in the interval π, ) r cos u π. Finding polar coordinates for a given point is tpicall a process involving some graphing and some thought. FIGURE 9.5 Converting from polar to rectangular coordinates EXAMPLE. Converting from Rectangular to Polar Coordinates Find all polar coordinate representations for the rectangular points (a) (, ) and (b) (, ). Solution (a) With = and =, we have from (.) that r = + = + =,

32 9- SECTION 9... Polar Coordinates 75 REMARK. Notice that for an point (, ) specified in rectangular coordinates ( ), we can alwas write the point in polar coordinates using either of the polar angles tan ( ) or tan ( ) + π. You can determine which angle corresponds to r = + and which corresponds to r = + b looking at the quadrant in which the point lies. so that r =±. Also, tan θ = =. One angle is then θ = tan ( ).98 radian. To determine which choice of r corresponds to this angle, note that (, ) is located in the first quadrant (see Figure 9.6a). Since.98 radian also puts ou in the first quadrant, this angle corresponds to the positive value of r, sothat (, tan ( )) is one polar representation of the point. The negative choice of r corresponds to an angle one half-circle (i.e., π radians) awa (see Figure 9.6b), so that another representation is (, tan ( ) ) + π.ever other polar representation is found b adding multiples of π to the two angles used above. That is, ever polar representation of the point (, ) must have the form (, tan ( ) + nπ ) or (, tan ( ) + π + nπ ), for some integer choice of n. (, ) (, ) u tan ( ) u tan ( ) p u tan ( ) FIGURE 9.6a The point (, ) FIGURE 9.6b Negative value of r (, ) u tan ( W) p (b) For the point (, ), we have = and =. From (.), we have r = + = ( ) + =, so that r =±. Further, tan θ = =, FIGURE 9.7 The point (, ) u tan ( W) so that the most obvious choice for the polar angle is θ = tan ( )., which lies in the fourth quadrant. Since the point (, ) is in the second quadrant, this choice of the angle corresponds to the negative value of r (see Figure 9.7). The positive value of r then corresponds to the angle θ = tan ( ) + π. Observe that all polar coordinate representations must then be of the form (, tan ( ) + nπ)or (, tan ( ) + π + nπ), for some integer choice of n. Observe that the conversion from polar coordinates to rectangular coordinates is completel straightforward, as in eample.. EXAMPLE. Converting from Polar to Rectangular Coordinates Find the rectangular coordinates for the polar points (a) (, π 6 ) and (b) (, ). Solution For (a), we have from (.) that = r cos θ = cos π 6 =

33 76 CHAPTER 9.. Parametric Equations and Polar Coordinates 9- and = r sin θ = sin π 6 =. The rectangular point is then (, ).For (b), we have and = r cos θ = cos.98 = r sin θ = sin.8. r The rectangular point is ( cos, sin ), which is located at approimatel (.98,.8). The graph of a polar equation r = f (θ) isthe set of all points (, ) for which = r cos θ, = r sin θ and r = f (θ). In other words, the graph of a polar equation is a graph in the -plane of all those points whose polar coordinates satisf the given equation. We begin b sketching two ver simple (and familiar) graphs. The ke to drawing the graph of a polar equation is to alwas keep in mind what the polar coordinates represent. FIGURE 9.8a The circle r = EXAMPLE.5 Some Simple Graphs in Polar Coordinates Sketch the graphs of (a) r = and (b) θ = π/. u Solution For (a), notice that = r = + and so, we want all points whose distance from the origin is (with an polar angle θ). Of course, this is the definition of a circle of radius with center at the origin (see Figure 9.8a). For (b), notice that θ = π/ specifies all points with a polar angle of π/ from the positive -ais (at an distance r from the origin). Including negative values for r, this defines a line with slope tan π/ = (see Figure 9.8b). It turns out that man familiar curves have simple polar equations. 6 FIGURE 9.8b The line θ = π 6 6 EXAMPLE.6 Converting an Equation from Rectangular to Polar Coordinates Find the polar equation(s) corresponding to the hperbola = 9 (see Figure 9.9). Solution From (.), we have 9 = = r cos θ r sin θ = r (cos θ sin θ) = r cos θ. Solving for r,weget r = 9 = 9 sec θ, cos θ so that r =± sec θ. 6 FIGURE 9.9 = 9 Notice that in order to keep sec θ >, we can restrict θ to lie in the interval π < θ < π,sothat π <θ< π. Observe that with this range of values of θ, the hperbola is drawn eactl once, where r = sec θ corresponds to the right branch of the hperbola and r = sec θ corresponds to the left branch.

34 9- SECTION 9... Polar Coordinates 77 q p w p FIGURE 9.a = sin plotted in rectangular coordinates FIGURE 9.b The circle r = sin θ EXAMPLE.7 A Surprisingl Simple Polar Graph Sketch the graph of the polar equation r = sin θ. Solution For reference, we first sketch a graph of the sine function in rectangular coordinates on the interval [, π] (see Figure 9.a). Notice that on the interval θ π, sin θ increases from to its maimum value of. This corresponds to a polar arc in the first quadrant from the origin (r = ) to unit up on the -ais. Then, on the interval π θ π, sin θ decreases from to. This corresponds to an arc in the second quadrant, from unit up on the -ais back to the origin. Net, on the interval π θ π, sin θ decreases from to its minimum value of. Since the values of r are negative, remember that this means that the points plotted are in the opposite quadrant (i.e., the first quadrant). Notice that this traces out the same curve in the first quadrant as we ve alread drawn for θ π. Likewise, taking θ in the interval π θ π retraces the portion of the curve in the second quadrant. Since sin θ is periodic of period π, taking further values of θ simpl retraces portions of the curve that we have alread drawn. A sketch of the polar graph is shown in Figure 9.b. We now verif that this curve is actuall a circle. Notice that if we multipl the equation r = sin θ through b r, weget r = r sin θ. You should immediatel recognize from (.) and (.) that = r sin θ and r = +. This gives us the rectangular equation + = or = +. Completing the square, we get = + ( + ) or, adding to both sides, ( ) ( = +. ) This is the rectangular equation for the circle of radius centered at the point (, ), which is what we see in Figure 9.b. The graphs of man polar equations are not the graphs of an functions of the form = f (), as in eample.8. EXAMPLE.8 An Archimedian Spiral FIGURE 9. The spiral r = θ,θ Sketch the graph of the polar equation r = θ, for θ. Solution Notice that here, as θ increases, so too does r. That is, as the polar angle increases, the distance from the origin also increases accordingl. This produces the spiral (an eample of an Archimedian spiral) seen in Figure 9.. The graphs shown in eamples.9,. and. are all in the general class known as limaçons. This class of graphs is defined b r = a ± b sin θ or r = a ± b cos θ,for positive constants a and b. Ifa = b, the graphs are called cardioids.

35 78 CHAPTER 9.. Parametric Equations and Polar Coordinates 9-5 q p w p FIGURE 9. = + cos in rectangular coordinates 5 EXAMPLE.9 A Limaçon Sketch the graph of the polar equation r = + cos θ. Solution We begin b sketching the graph of = + cos in rectangular coordinates on the interval [, π], to use as a reference (see Figure 9.). Notice that in this case, we have r = + cos θ>for all values of θ. Further, the maimum value of r is 5 (corresponding to when cos θ = atθ =, π, etc.) and the minimum value of r is (corresponding to when cos θ = atθ = π, π, etc.). In this case, the polar graph is traced out with θ π. Wesummarize the intervals of increase and decrease for r in the following table. Interval cos θ r cos θ [ ], π Decreases from to Decreases from 5 to [ π,π] Decreases from to Decreases from to [ ] π, π Increases from to Increases from to [ π, π] Increases from to Increases from to 5 In Figures 9.a 9.d, we show how the sketch progresses through each interval indicated in the table, with the completed figure (called a limaçon) shown in Figure 9.d. FIGURE 9.a θ π FIGURE 9.b θ π FIGURE 9.c θ π FIGURE 9.d θ π EXAMPLE. The Graph of a Cardioid Sketch the graph of the polar equation r = sin θ. Solution As we have done several times now, we first sketch a graph of = sin in rectangular coordinates, on the interval [, π], as in Figure 9.. We summarize the intervals of increase and decrease in the following table.

36 9-5 SECTION 9... Polar Coordinates 79 q p w p FIGURE 9. = sin in rectangular coordinates Interval sin θ r sin θ [ ], π Increases from to Decreases from to [ π,π] Decreases from to Increases from to [ ] π, π Decreases from to Increases from to [ π, π] Increases from to Decreases from to Again, we sketch the graph in stages, corresponding to each of the intervals indicated in the table, as seen in Figures 9.5a 9.5d. 5 FIGURE 9.5a θ π 5 FIGURE 9.5b θ π 5 5 FIGURE 9.5c θ π FIGURE 9.5d θ π The completed graph appears in Figure 9.5d and is sketched out for θ π.you can see wh this figure is called a cardioid ( heartlike ). EXAMPLE. ALimaçon with a Loop Sketch the graph of the polar equation r = sin θ. Solution We again begin b sketching a graph of = sin in rectangular coordinates, as in Figure 9.6. We summarize the intervals of increase and decrease in the following table.

37 75 CHAPTER 9.. Parametric Equations and Polar Coordinates 9-6 q p w p FIGURE 9.6 = sin in rectangular coordinates Interval sin θ r sin θ [ ], π Increases from to Decreases from to [ π,π] Decreases from to Increases from to [ ] π, π Decreases from to Increases from to [ π, π] Increases from to Decreases from to Notice that since r assumes both positive and negative values in this case, we need to eercise a bit more caution, as negative values for r cause us to draw that portion of the graph in the opposite quadrant. Observe that r = when sin θ =, that is, when sin θ =. This will occur when θ = π 6 and when θ = 5π.For this reason, we epand 6 the above table, to include more intervals and where we also indicate the quadrant where the graph is to be drawn, as follows: Interval sin θ r sin θ Quadrant [ ], π Increases from to Decreases from to First 6 [ π, ] π Increases from to Decreases from to Third 6 [ π, ] 5π Decreases from to Increases from to Fourth 6 [ 5π,π] Decreases from to Increases from to Second 6 [ ] π, π Decreases from to Increases from to Third [ π, π] Increases from to Decreases from to Fourth We sketch the graph in stages in Figures 9.7a 9.7f, corresponding to each of the intervals indicated in the table. FIGURE 9.7a θ π 6 FIGURE 9.7b θ π FIGURE 9.7c θ 5π 6 The completed graph appears in Figure 9.7f and is sketched out for θ π.you should observe from this the importance of determining where r =, as well as where r is increasing and decreasing.

38 9-7 SECTION 9... Polar Coordinates 75 FIGURE 9.7d θ π FIGURE 9.7e θ π FIGURE 9.7f θ π EXAMPLE. AFour-Leaf Rose Sketch the graph of the polar equation r = sin θ. Solution As usual, we will first draw a graph of = sin in rectangular coordinates on the interval [, π], as seen in Figure 9.8. Notice that the period of sin θ is onl π. We summarize the intervals on which the function is increasing and decreasing in the following table. q p w FIGURE 9.8 = sin in rectangular coordinates p Interval r sin θ Quadrant [ ], π Increases from to First [ π, ] π Decreases from to First [ π, ] π Decreases from to Fourth [ π,π] Increases from to Fourth [ ] π, 5π [ 5π, ] π [ π, ] 7π Increases from to Third Decreases from to Third Decreases from to Second [ 7π, π] Increases from to Second We sketch the graph in stages in Figures 9.9a 9.9h, each one corresponding to the intervals indicated in the table, where we have also indicated the lines =±,asa guide. This is an interesting curve known as a four-leaf rose. Notice again the significance of the points corresponding to r =, or sin θ =. Also, notice that r reaches a maimum of when θ = π, 5π,...or θ = π, 5π,...and r reaches a minimum of when θ = π, 7π,...or θ = π, 7π,...Again, ou must keep in mind that when the value of r is negative, this causes us to draw the graph in the opposite quadrant.

39 75 CHAPTER 9.. Parametric Equations and Polar Coordinates 9-8 FIGURE 9.9a θ π FIGURE 9.9b θ π FIGURE 9.9c θ π FIGURE 9.9d θ π FIGURE 9.9e θ 5π FIGURE 9.9f θ π FIGURE 9.9g θ 7π FIGURE 9.9h θ π Note that in eample., even though the period of the function sin θ is π, ittook θ-values ranging from to π to sketch the entire curve r = sin θ.bcontrast, the period of the function sin θ is π, but the circle r = sin θ was completed with θ π. To determine the range of values of θ that produces a graph, ou need to carefull identif important points as we did in eample.. The Trace feature found on graphing calculators can be ver helpful for getting an idea of the θ-range, but remember that such Trace values are onl approimate. You will eplore a variet of other interesting graphs in the eercises. BEYOND FORMULAS The graphics in Figures 9.5, 9.7 and 9.9 provide a good visual model of how to think of polar graphs. Most polar graphs r = f (θ) can be sketched as a sequence of connected arcs, where the arcs start and stop at places where r = or where a new quadrant is entered. B breaking the larger graph into small arcs, ou can use the properties of f to quickl determine where each arc starts and stops.

40 9-9 SECTION 9... Polar Coordinates 75 EXERCISES 9. WRITING EXERCISES. Suppose a point has polar representation (r,θ). Eplain wh another polar representation of the same point is ( r,θ + π).. After working with rectangular coordinates for so long, the idea of polar representations ma seem slightl awkward. However, polar representations are entirel natural in man settings. For instance, if ou were on a ship at sea and another ship was approaching ou, eplain whether ou would use a polar representation (distance and bearing) or a rectangular representation (distance east-west and distance north-south).. In eample.7, the graph (a circle) of r = sin θ is completel traced out with θ π. Eplain wh graphing r = sin θ with π θ π would produce the same full circle.. Two possible advantages of introducing a new coordinate sstem are making previous problems easier to solve and allowing new problems to be solved. Give two eamples of graphs for which the polar equation is simpler than the rectangular equation. Give two eamples of polar graphs for which ou have not seen a rectangular equation. In eercises 6, plot the given polar points (r,θ) and find their rectangular representation.. (, ). (,π). (,π) ( ) ( )., π 5. (, π) 6. 5, π In eercises 7, find all polar coordinate representations of the given rectangular point. 7. (, ) 8. (, ) 9. (, ). (, ). (, ). (, 5) In eercises 8, find rectangular coordinates for the given polar point. ( ) ( )., π., π 5. (, ) ( ) ( ) 6., π 7. 8, π 8. (, ) In eercises 9 6, sketch the graph of the polar equation and find a corresponding - equation. 9. r =. r =. θ = π/6. θ = π/. r = cos θ. r = cos θ 5. r = sin θ 6. r = sin θ In eercises 7 5, sketch the graph and identif all values of θ where r and a range of values of θ that produces one cop of the graph. 7. r = cos θ 8. r = cos θ 9. r = sin θ. r = sin θ. r = + sin θ. r = cos θ. r = sin θ. r = + cos θ 5. r = + sin θ 6. r = 6 cos θ 7. r = θ 8. r = eθ/ 9. r = cos(θ π/). r = sin(θ π). r = cos θ + sin θ. r = cos θ + sin θ. r = tan θ. r = θ/ θ + 5. r = + cos θ 6. r = sin θ 7. r = 8. r = + sin θ sin θ 9. r = 5. r = + cos θ cos θ 5. Graph r = cos θ sin θ and eplain wh there is no curve to the left of the -ais. 5. Graph r = θ cos θ for π θ π. Eplain wh this is called the Garfield curve. GARFIELD c 5 Paws, Inc. Reprinted with permission of UNIVERSAL PRESS SYNDICATE. All rights reserved. 5. Based on our graphs in eercises and, conjecture the graph of r = a cos θ for an positive constant a. 5. Based on our graphs in eercises 5 and 6, conjecture the graph of r = a sin θ for an positive constant a. 55. Based on the graphs in eercises 7 and 8 and others (tr r = cos θ and r = cos 5θ), conjecture the graph of r = cos nθ for an positive integer n. 56. Based on the graphs in eercises 9 and and others (tr r = sin θ and r = sin 5θ), conjecture the graph of r = sin nθ for an positive integer n. In eercises 57 6, find a polar equation corresponding to the given rectangular equation. 57. = = = = 6. = 6. =

41 75 CHAPTER 9.. Parametric Equations and Polar Coordinates 9-6. Sketch the graph of r = cos θ first for θ π, then for θ π, then for θ π,..., and finall for θ π. Discuss an patterns that ou find and predict what will happen for larger domains. 6. Sketch the graph of r = cos πθ first for θ, then for θ, then for θ,... and finall for θ. Discuss an patterns that ou find and predict what will happen for larger domains. 65. One situation where polar coordinates appl directl to sports is in making a golf putt. The two factors that the golfer tries to control are distance (determined b speed) and direction (usuall called the line ). Suppose a putter is d feet from the hole, which has radius h = 6. Show that the path of the ball will intersect the hole if the angle A in the figure satisfies sin (h/d) < A < sin (h/d). (r, A) (d, ) (, ) A 66. The distance r that the golf ball in eercise 65 travels also needs to be controlled. The ball must reach the front of the hole. In rectangular coordinates, the hole has equation ( d) + = h, so the left side of the hole is = d h. Show that this converts in polar coordinates to r = d cos θ d cos θ (d h ). (Hint: Substitute for and, isolate the square root term, square both sides, combine r terms and use the quadratic formula.) 67. The golf putt in eercises 65 and 66 will not go in the hole if it is hit too hard. Suppose that the putt would go r = d + c feet if it did not go in the hole (c > ). For a putt hit toward the center of the hole, define b to be the largest value of c such that the putt goes in (i.e., if the ball is hit more than b feet past the hole, it is hit too hard). Eperimental evidence (see Dave Pelz s Putt Like the Pros) shows ( that at other angles A, the distance r [ ] ) A must be less than d + b. The results of sin (h/d) eercises 65 and 66 define limits for the angle A and distance r of a successful putt. Identif the functions r (A) and r (A) such that r (A) < r < r (A) and constants A and A such that A < A < A. 68. Take the general result of eercise 67 and appl it to a putt of d = 5 feet with a value of b = feet. Visualize this b graphing the region 5 cos θ 5 cos θ (5 /6) ( [ ] ) θ < r < 5 + sin (/9) with sin (/9) <θ<sin (/9). A good choice of graphing windows is.8 9 and.5.5. EXPLORATORY EXERCISES. In this eercise, ou will eplore the roles of the constants a, b and c in the graph of r = af(bθ + c). To start, sketch r = sin θ followed b r = sin θ and r = sin θ. What does the constant a affect? Then sketch r = sin(θ + π/) and r = sin(θ π/). What does the constant c affect? Now for the tough one. Sketch r = sin θ and r = sin θ. What does the constant b seem to affect? Test all of our hpotheses on the base function r = + cos θ and several functions of our choice.. The polar curve r = ae bθ is sometimes called an equiangular curve. To see wh, sketch the curve and then show that dr = br. Asomewhat complicated geometric argument dθ shows that dr = r cot α, where α is the angle between the dθ tangent line and the line connecting the point on the curve to the origin. Comparing equations, conclude that the angle α is constant (hence equiangular ). To illustrate this propert, compute α for the points at θ = and θ = π for r = e θ. This tpe of spiral shows up often in nature, possibl because the equal-angle propert can be easil achieved. Spirals can be found among shellfish (the picture shown here is of an ammonite fossil from about 5 million ears ago) and the florets of the common dais. Other eamples, including the connection to sunflowers, the Fibonacci sequence and the musical scale, can be found in H. E. Huntle s The Divine Proportion.