6.3 Parametric Equations and Motion


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1 SECTION 6.3 Parametric Equations and Motion 475 What ou ll learn about Parametric Equations Parametric Curves Eliminating the Parameter Lines and Line Segments Simulating Motion with a Grapher... and wh These topics can be used to model the path of an object such as a baseball or a golf ball. t = 0, = 420 t = 4, = 164 t = 5, = 20 t = 1, = 404 t = 2, = 356 t = 3, = 276 [0, 5] b [ 10, 500] FIGURE 6.23 The position of the rock at 0, 1, 2, 3, 4, and 5 seconds. 6.3 Parametric Equations and Motion Parametric Equations Imagine that a rock is dropped from a 420ft tower. The rock s height in feet above the ground t seconds later (ignoring air resistance) is modeled b = 16t as we saw in Section 2.1. Figure 6.23 shows a coordinate sstem imposed on the scene so that the line of the rock s fall is on the vertical line = 2.5. The rock s original position and its position after each of the first 5 seconds are the points (2.5, 420), (2.5, 404), (2.5, 356), (2.5, 276), (2.5, 164), (2.5, 20), which are described b the pair of equations = 2.5, = 16t , when t = 0, 1, 2, 3, 4, 5. These two equations are an eample of parametric equations with parameter t. As is often the case, the parameter t represents time. Parametric Curves In this section we stud the graphs of parametric equations and investigate motion of objects that can be modeled with parametric equations. DEFINITION Parametric Curve, Parametric Equations The graph of the ordered pairs 1, 2 where = ƒ1t2, = g1t2 are functions defined on an interval I of tvalues is a parametric curve. The equations are parametric equations for the curve, the variable t is a parameter, and I is the parameter interval. When we give parametric equations and a parameter interval for a curve, we have parametrized the curve. A parametrization of a curve consists of the parametric equations and the interval of tvalues. Sometimes parametric equations are used b companies in their design plans. It is then easier for the compan to make larger and smaller objects efficientl b just changing the parameter t. Graphs of parametric equations can be obtained using parametric mode on a grapher. EXAMPLE 1 Graphing Parametric Equations For the given parameter interval, graph the parametric equations = t 22, = 3t. 3 t 12 t 3 (c) 3 t 3 SOLUTION In each case, set Tmin equal to the left endpoint of the interval and Tma equal to the right endpoint of the interval. Figure 6.24 shows a graph of the parametric equations for each parameter interval. The corresponding relations are different because the parameter intervals are different. Now tr Eercise 7.
2 476 CHAPTER 6 Applications of Trigonometr [ 10, 10] b [ 10, 10] [ 10, 10] b [ 10, 10] [ 10, 10] b [ 10, 10] (c) FIGURE 6.24 Three different relations defined parametricall. (Eample 1) Eliminating the Parameter When a curve is defined parametricall it is sometimes possible to eliminate the parameter and obtain a rectangular equation in and that represents the curve. This often helps us identif the graph of the parametric curve as illustrated in Eample 2. EXAMPLE 2 Eliminating the Parameter Eliminate the parameter and identif the graph of the parametric curve = 12t, = 2  t,  q 6 t 6 q. SOLUTION We solve the first equation for t: = 12t 2t = 1  t = Then we substitute this epression for t into the second equation: = 2  t [ 10, 5] b [ 5, 5] FIGURE 6.25 The graph of = (Eample 2) = = The graph of the equation = is a line with slope 0.5 and intercept 1.5 (Figure 6.25). Now tr Eercise 11. EXPLORATION 1 Graphing the Curve of Eample 2 Parametricall 1. Use the parametric mode of our grapher to reproduce the graph in Figure Use 2 for Tmin and 5.5 for Tma. 2. Prove that the point 117, 102 is on the graph of = Find the corresponding value of t that produces this point. 3. Repeat part 2 for the point 23, Assume that 1a, b2 is on the graph of = Find the corresponding value of t that produces this point. 5. How do ou have to choose Tmin and Tma so that the graph in Figure 6.25 fills the window?
3 SECTION 6.3 Parametric Equations and Motion 477 If we do not specif a parameter interval for the parametric equations = ƒ1t2, = g1t2, it is understood that the parameter t can take on all values that produce real numbers for and. We use this agreement in Eample 3. EXAMPLE 3 Eliminating the Parameter Eliminate the parameter and identif the graph of the parametric curve = t 22, = 3t. Parabolas The inverse of a parabola that opens up or down is a parabola that opens left or right. We will investigate these curves in more detail in Chapter 8. SOLUTION Here t can be an real number. We solve the second equation for t, obtaining t = /3, and substitute this value for into the first equation. = t 22 = a 2 3 b  2 = = Figure 6.24c shows what the graph of these parametric equations looks like. In Chapter 8 we will call this a parabola that opens to the right. Interchanging and, we can identif this graph as the inverse of the graph of the parabola 2 = Now tr Eercise 15. EXAMPLE 4 Eliminating the Parameter Eliminate the parameter and identif the graph of the parametric curve = 2 cos t, = 2 sin t, 0 t 2p. [ 4.7, 4.7] b [ 3.1, 3.1] FIGURE 6.26 The graph of the circle of Eample 4. SOLUTION The graph of the parametric equations in the square viewing window of Figure 6.26 suggests that the graph is a circle of radius 2 centered at the origin. We confirm this result algebraicall = 4 cos 2 t + 4 sin 2 t = 41cos 2 t + sin 2 t2 = 4112 = 4 cos 2 t + sin 2 t = 1 The graph of = 4 is a circle of radius 2 centered at the origin. Increasing the length of the interval 0 t 2p will cause the grapher to trace all or part of the circle more than once. Decreasing the length of the interval will cause the grapher to onl draw a portion of the complete circle. Tr it! Now tr Eercise 23. In Eercise 65, ou will find parametric equations for an circle in the plane. Lines and Line Segments We can use vectors to help us find parametric equations for a line as illustrated in Eample 5.
4 478 CHAPTER 6 Applications of Trigonometr A( 2, 3) O 1 B(3, 6) P(, ) FIGURE 6.27 Eample 5 uses vectors to construct a parametrization of the line through A and B. EXAMPLE 5 Finding Parametric Equations for a Line Find a parametrization of the line through the points A = 12, 32 and B = 13, 62. SOLUTION Let P1, 2 be an arbitrar point on the line through A and B. As ou! can see from Figure 6.27, the vector is the tailtohead vector sum of and. You can also see that AP! OP! is a scalar multiple of AB! OA! AP. If we let the scalar be t, we have OP! OP! = OA! = OA! + AP!! + t # AB 8, 9 = 82, 39 + t , , 9 = 82, 39 + t85, 39 8, 9 = t, 3 + 3t9 This vector equation is equivalent to the parametric equations = t and = 3 + 3t. Together with the parameter interval 1 q, q2, these equations define the line. We can confirm our work numericall as follows: If t = 0, then = 2 and = 3, which gives the point A. Similarl, if t = 1, then = 3 and = 6, which gives the point B. Now tr Eercise 27. The fact that t = 0 ields point A and t = 1 ields point B in Eample 5 is no accident, as a little reflection on Figure 6.27 and the vector equation OP! = OA!! + t # AB should suggest. We use this fact in Eample 6. EXAMPLE 6 Finding Parametric Equations for a Line Segment Find a parametrization of the line segment with endpoints A = 12, 32 and B = 13, 62. SOLUTION In Eample 5 we found parametric equations for the line through A and B: = t, = 3 + 3t We also saw in Eample 5 that t = 0 produces the point A and t = 1 produces the point B. A parametrization of the line segment is given b = t, = 3 + 3t, 0 t 1. As t varies between 0 and 1 we pick up ever point on the line segment between A and B. Now tr Eercise 29. Simulating Motion with a Grapher Eample 7 illustrates several was to simulate motion along a horizontal line using parametric equations. We use the variable t for the parameter to represent time. EXAMPLE 7 Simulating Horizontal Motion Gar walks along a horizontal line (think of it as a number line) with the coordinate of his position (in meters) given b s = 0.11t 320t t where 0 t 12. Use parametric equations and a grapher to simulate his motion. Estimate the times when Gar changes direction.
5 SECTION 6.3 Parametric Equations and Motion 479 T=0 X=8.5 Y=5 Start, t = 0 SOLUTION We arbitraril choose the horizontal line = 5 to displa this motion. The graph C 1 of the parametric equations, simulates the motion. His position at an time t is given b the point 1 1 1t2, 52. Using TRACE in Figure 6.28 we see that when t = 0, Gar is 8.5 m to the right of the ais at the point (8.5, 5), and that he initiall moves left. Five seconds later he is 9 m to the left of the ais at the point (9, 5). And after 8 seconds he is onl 2.7 m to the left of the ais. Gar must have changed direction during the walk. The motion of the trace cursor simulates Gar s motion. A variation in 1t2, C 1 : 1 = 0.1(t 320t t  85), 1 = 5, 0 t 12, C 2 : 2 = 0.11t 320t t  852, 2 = t, 0 t 12, can be used to help visualize where Gar changes direction. The graph C 2 shown in Figure 6.29 suggests that Gar reverses his direction at 3.9 seconds and again at 9.5 seconds after beginning his walk. Now tr Eercise 37. T=5 X= 9 Y=5 5 sec later, t = 5 C 1 C 1 T=8 X= 2.7 Y=5 3 sec after that, t = 8 (c) FIGURE 6.28 Three views of the graph C 1 : 1 = 0.11t 320t t  852, 1 = 5, 0 t 312, in the 124 b 310, 104 viewing window. (Eample 7) Grapher Note The equation 2 = t is tpicall used in the parametric equations for the graph C 2 in Figure We have chosen 2 = t to get two curves in Figure 6.29 that do not overlap. Also notice that the coordinates of C 1 are constant 1 1 = 52, and that the coordinates of C 2 var with time t 1 2 = t2. C 2 T=3.9 X= Y= 3.9 [ 12, 12] b [ 15, 15] [ 12, 12] b [ 15, 15] Eample 8 solves a projectilemotion problem. Parametric equations are used in two was: to find a graph of the modeling equation and to simulate the motion of the projectile. C 2 T=9.5 X= Y= 9.5 FIGURE 6.29 Two views of the graph C 1 : 1 = 0.1(t 320t t  85), and the graph C 2 : 2 = 0.11t 320t 2 1 = 5, 0 t t  852, 2 = t, 0 t 12 in the 312, 124 b 315, 154 viewing window. (Eample 7) EXAMPLE 8 Simulating Projectile Motion A distress flare is shot straight up from a ship s bridge 75 ft above the water with an initial velocit of 76 ft/sec. Graph the flare s height against time, give the height of the flare above water at each time, and simulate the flare s motion for each length of time. 1 sec 2 sec (c) 4 sec (d) 5 sec SOLUTION An equation that models the flare s height above the water t seconds after launch is = 16t t A graph of the flare s height against time can be found using the parametric equations 1 = t, 1 = 16t t (continued)
6 480 CHAPTER 6 Applications of Trigonometr To simulate the flare s flight straight up and its fall to the water, use the parametric equations 2 = 5.5, 2 = 16t t (We chose 2 = 5.5 so that the two graphs would not intersect.) Figure 6.30 shows the two graphs in simultaneous graphing mode for 0 t 1, 0 t 2, (c) 0 t 4, and (d) 0 t 5. We can read that the height of the flare above the water after 1 sec is 135 ft, after 2 sec is 163 ft, after 4 sec is 123 ft, and after 5 sec is 55 ft. Now tr Eercise 39. T=1 X=5.5 Y=135 T=2 X=5.5 Y=163 T=4 X=5.5 Y=123 T=5 X=5.5 Y=55 [0, 6] b [0, 200] [0, 6] b [0, 200] [0, 6] b [0, 200] (c) [0, 6] b [0, 200] (d) FIGURE 6.30 Simultaneous graphing of (height against time) and 2 = 5.5, 2 = 16t 2 1 = t, 1 = 16t t t + 75 (the actual path of the flare). (Eample 8) 0 v 0 v 0 sin v 0 cos FIGURE 6.31 Throwing a baseball. In Eample 8 we modeled the motion of a projectile that was launched straight up. Now we investigate the motion of objects, ignoring air friction, that are launched at angles other than 90 with the horizontal. Suppose that a baseball is thrown from a point 0 feet above ground level with an initial speed of v 0 ft/sec at an angle u with the horizontal (Figure 6.31). The initial velocit can be represented b the vector The path of the object is modeled b the parametric equations The component is simpl v = 8v 0 cos u, v 0 sin u9. = 1v 0 cos u2t, = 16t 2 + 1v 0 sin u2t + 0. distance = 1component of initial velocit2 * time. The component is the familiar vertical projectilemotion equation using the component of initial velocit. [0, 450] b [0, 80] FIGURE 6.32 The fence and path of the baseball in Eample 9. See Eploration 2 for was to draw the wall. EXAMPLE 9 Hitting a Baseball Kevin hits a baseball at 3 ft above the ground with an initial speed of 150 ft/sec at an angle of 18 with the horizontal. Will the ball clear a 20ft wall that is 400 ft awa? SOLUTION The path of the ball is modeled b the parametric equations = 1150 cos 18 2t, = 16t sin 18 2t + 3. A little eperimentation will show that the ball will reach the fence in less than 3 sec. Figure 6.32 shows a graph of the path of the ball using the parameter interval 0 t 3 and the 20ft wall. The ball does not clear the wall. Now tr Eercise 43.
7 SECTION 6.3 Parametric Equations and Motion 481 EXPLORATION 2 Etending Eample 9 1. If our grapher has a line segment feature, draw the fence in Eample Describe the graph of the parametric equations = 400, = 201t/32, 0 t Repeat Eample 9 for the angles 19, 20, 21, and ft 30 ft FIGURE 6.33 The Ferris wheel of Eample 10. A In Eample 10 we see how to write parametric equations for position on a moving Ferris wheel, using time t as the parameter. EXAMPLE 10 Riding on a Ferris Wheel Jane is riding on a Ferris wheel with a radius of 30 ft. As we view it in Figure 6.33, the wheel is turning counterclockwise at the rate of one revolution ever 10 sec. Assume that the lowest point of the Ferris wheel (6 o clock) is 10 ft above the ground and that Jane is at the point marked A (3 o clock) at time t = 0. Find parametric equations to model Jane s path and use them to find Jane s position 22 sec into the ride. 40 P 30 θ A FIGURE 6.34 A model for the Ferris wheel of Eample 10. SOLUTION Figure 6.34 shows a circle with center 10, 402 and radius 30 that models the Ferris wheel. The parametric equations for this circle in terms of the parameter u, the central angle of the circle determined b the arc AP, are = 30 cos u, = sin u, 0 u 2p. To take into account the rate at which the wheel is turning we must describe u as a function of time t in seconds. The wheel is turning at the rate of 2p radians ever 10 sec, or 2p/10 = p/5 rad/sec. So, u = 1p/52t. Thus, parametric equations that model Jane s path are given b = 30 cos a p 5 tb, = sin a p 5 tb, t Ú 0. We substitute t = 22 into the parametric equations to find Jane s position at that time: = 30 cos a p 5 # 22b = sin a p 5 # 22b L 9.27 L After riding for 22 sec, Jane is approimatel 68.5 ft above the ground and approimatel 9.3 ft to the right of the ais, using the coordinate sstem of Figure Now tr Eercise 51.
8 482 CHAPTER 6 Applications of Trigonometr QUICK REVIEW 6.3 (For help, go to Sections P.2, P.4, 1.3, 4.1, and 6.1.) In Eercises 1 and 2, find the component form of the vectors OA!, OB!, and (c) AB! where O is the origin. 1. A = 13, 2), B = 14, A = 11, 3), B = 14, 32 In Eercises 3 and 4, write an equation in pointslope form for the line through the two points , 22, 14, , 32, 14, 32 In Eercises 5 and 6, find and graph the two functions defined implicitl b each given relation = = 5 In Eercises 7 and 8, write an equation for the circle with given center and radius , 02, , 52, 3 In Eercises 9 and 10, a wheel with radius r spins at the given rate. Find the angular velocit in radians per second. 9. r = 13 in., 600 rpm 10. r = 12 in., 700 rpm SECTION 6.3 EXERCISES Eercise numbers with a gra background indicate problems that the authors have designed to be solved without a calculator. In Eercises 1 4, match the parametric equations with their graph. Identif the viewing window that seems to have been used. (c) 1. = 4 cos 3 t, = 2 sin 3 t 2. = 3 cos t, = sin 2t (d) 3. = 2 cos t + 2 cos 2 t, = 2 sin t + sin 2t 4. = sin t  t cos t, = cos t + t sin t In Eercises 5 and 6, complete the table for the parametric equations and plot the corresponding points. 5. = t + 2, = 1 + 3/t t = cos t, = sin t t 0 /2 pp 3p/2 2p In Eercises 7 10, graph the parametric equations = 3  t 2, = 2t, in the specified parameter interval. Use the standard viewing window t t t t 4 In Eercises 11 26, use an algebraic method to eliminate the parameter and identif the graph of the parametric curve. Use a grapher to support our answer. 11. = 1 + t, = t 12. = 23t, = 5 + t 13. = 2t  3, = 94t, 3 t = 53t, = 2 + t, 15. = t 2, = t + 1 [Hint: Eliminate t and solve for in terms of.] 16. = t, = t = t, = t 32t = 2t 21, = t [Hint: Eliminate t and solve for in terms of.] 19. = 4  t 2, = t [Hint: Eliminate t and solve for in terms of.] 20. = 0.5t, = 2t 33, 2 t = t  3, = 2/t, 5 t = t + 2, = 4/t, t Ú = 5 cos t, = 5 sin t 24. = 4 cos t, = 4 sin t 1 t = 2 sin t, = 2 cos t, 0 t 3p/2 26. = 3 cos t, = 3 sin t, 0 t p In Eercises find a parametrization for the curve. 27. The line through the points 12, 52 and 14, The line through the points 3, 321 and 15, The line segment with endpoints 13, 42 and 16, 32
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