Section 10.7 Parametric Equations


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1 299 Section 10.7 Parametric Equations Objective 1: Defining and Graphing Parametric Equations. Recall when we defined the x (rcos(θ), rsin(θ)) and ycoordinates on a circle of radius r as a function of θ. The value of x and y depended on r the value of θ. Notice that the graph of the equation of the θ circle, x 2 + y 2 = r 2, is not a function since you can find a vertical line that passes through more than one point of the curve, But, by defining x = rcos(θ) and y = rsin(θ) on the interval [0, 2π), each coordinate is a function of θ. This means we can view each coordinate as a separate curve. x = rcos(θ) is the curve of the cosine function with amplitude r and y = r sin(θ) is the curve of the sine function with amplitude r. When we graph the points (x, y) on the graph to obtain the circle of radius r, the variable θ no longer is part of the equation for the plane curve. We call the variable θ in this case a parameter and the equations x = rcos(θ) and y = rsin(θ) parametric equations. Definition Let x = f(t) and y = g(t) be two functions defined on the interval I. The set of all points (x, y) = (f(t), g(t)) for all t in I is called the plane curve. The equations x = f(t) and y = g(t) for all t in I are called the parametric equations and t is called the parameter. Suppose the parametric equations are defined on the interval a t b. As we plot the points (x, y) as t goes from a to b, the successive points trace out the graph in a certain direction. The direction the graph is traced out as t goes from a to b is called the orientation. Thus, in the case of the circle of radius r discussed above, as t goes from 0 to 2π, the graph gets traced out in a counterclockwise direction so the orientation would be counterclockwise. To indicate the orientation on a graph, we draw arrows on the plane the curve showing the direction that curve is traced out.
2 Graph the curve whose parametric equations are given and show its orientation: Ex. 1 x = 0.5t, y = 0.25t 2 4 t 2 Let's begin by making a table of values and deriving a set of points (x, y) to plot: t x = 0.5t y = 0.25t 2 (x, y) ( 2, 4) ( 1.5, 2.25) ( 1, 1) ( 0.5, 0.25) (0, 0) (0.5, 0.25) (1, 1) Now, plot the points and sketch the graph: 00 To verify the graph, we can get rid of the parameter. In the equation x = 0.5t, we can solve for t to get t = 2x. Now, substitute 2x in for t in the equation y = 0.25t 2 and simplify: y = 0.25(2x) 2 y = 0.25(4x 2 ) y = x 2 Since 4 t 2, then 4 2x 2 or 2 x 1. Thus, y = x 2 for 2 x 1 Notice that it matches our graph. Objective 2: Finding a rectangular equation for a curve defined parametrically. In the last example, we took care to pay attention to the domain of the function after getting rid of the parameter. Many times, the graph of the curve after getting rid of the parameter will contain more points than the parameterized curve so we will need to pay careful attention to which piece of the curve actually belongs on the graph.
3 01 Find the rectangular equation of the curve and sketch the graph: Ex. 2 x = cos(t), y = 4sin(t); t 0 First, solve each of the parametric equations for the trigonometric function: x = cos(t) y = 4sin(t) cos(t) = x sin(t) = y 4 Recall, that cos 2 (t) + sin 2 (t) = 1, x so ( ) 2 + ( y 4 ) 2 = 1 or x2 9 + y2 16 = 1 Thus, the graph is an ellipse with center at (0, 0), major axis running along the yaxis, vertices (0, 4) and (0, 4) and bintercepts (, 0) and (, 0), Since t starts at zero, the graph starts at (, 0) and traces out in a counterclockwise fashion. Suppose t had been restricted to a much smaller interval. Although, the process would be exactly the same as we did above, we may need to eliminate part of the graph depending on the restriction on t. If 0 t π was the restriction, we would only have the upper half of the graph. If π 2 t π 2 was the restriction, we would only have the left side. 0 t π π 2 t π 2
4 02 Ex. x = sin(t), y = 2cos(t); π t 0 First, solve each of the parametric equations for the trigonometric function: x = sin(t) y = 2cos(t) sin(t) = x cos(t) = y 2 Recall, that cos 2 (t) + sin 2 (t) = 1, y so ( 2 ) 2 + ( x ) 2 = 1 or x2 9 + y2 4 = 1 Thus, the graph is an ellipse with center at (0, 0), major axis running along the xaxis, vertices (, 0) and (, 0) and bintercepts (0, 2) and (0, 2), Since t starts at π, the graph starts at (0, 2) and traces out in a clockwise fashion. It ends at (0, 2). Ex. 4 x = sec(t), y = tan(t); π 6 t π Recall, that tan 2 (t) + 1 = sec 2 (t), so (y) = (x) 2 or x 2 y 2 = 1 Thus, the graph is a hyperbola with center at (0, 0), transverse axis running along the xaxis, vertices (1, 0) and ( 1, 0) and oblique asymptotes y = ± x Since t starts at π, the graph 6 starts at (sec( π 6 ), tan( π 6 )) = ( 2, ) (1.15, 0.58) and traces out in a clockwise fashion. It ends at (sec( π ), tan( π )) = (2, ) (2, 1.7).
5 0 Objective Time as a parameter. A common use of parametric equations is when t represents the time. We can describe the path of an object using time as the parameter. The advantage with using parametric equations in this situation is that not only can we determine the path of the object, but also the time that the object reaches a certain point on the path. In calculus, we can derive the parametric equations that determine the path of a projectile through the air. Parametric Equations of a Projectile Motion Let an object be propelled upward at an angle θ with the horizontal from a height h above the horizontal. If the initial speed is v o, then, neglecting for air resistance, the resulting motion is called Projectile Motion and the parametric equations for this motion are: x = (v o cos(θ))t and y = 1 2 gt2 + (v o sin(θ))t + h where t is the time in seconds and g is the acceleration due to gravity (on Earth, g 2 ft/s 2 or 9.8 m/s 2 ). Solve the following: Ex. 5 Standing on top of a hill that was six feet above the horizontal, George Johnson hits a golf ball at 125 ft/sec at an angle of 5 from the horizontal. a) Find the parametric equations that describe the position of the ball as a function of time. b) How long was the ball in the air? c) Find the horizontal distance the ball travelled before hitting the ground. d) When is the ball at its maximum height? What is that maximum height? a) The initial velocity v o = 125 ft/sec. Since the velocity is given in ft/sec, we will use g = 2 ft/sec 2. The angle θ = 5 and h is 6 ft. x = (v o cos(θ))t = 125cos(5 )t t y = 1 2 gt2 + (v o sin(θ))t + h = 1 2 2t2 + (125sin(5 ))t t t + 6 h v o θ
6 04 Thus, our parametric equations are: x t and y 16t t + 6 b) When the ball hits the ground, the height the ball above the ground is 0. To find how long the ball was in the air, we need to first find when the ball hit the ground. Setting y = 0 and solving yields: y 16t t + 6 = 0 (use the quadratic formula) t = b± b2 4ac 2a = ± ± ( )2 4( 16)(6) 2( 16) 4.56 seconds or seconds Since the negative answers does not sense, the ball was in the air for about 4.56 seconds. c) Since the ball was in the air for approximately 4.56 seconds, then the horizontal distance travelled is: x (4.56) ft. d) Notice that the equation for y is a quadratic equation with the coefficient of the squared term being negative. Thus, the maximum value (k) will occur at the vertex (h, k). h = b 2a ( 16) seconds Plugging t = into y, we get the maximum height: y 16(2.2405) (2.2405) ft Thus, the ball reached a maximum height of about ft at t seconds. If you graph the function using equations from part a, we need to remember the xvalues are the horizontal distance travelled and not the time. Ex. 6 At pm, Juan gets in his car and drives towards Atlanta at an average speed of 55 mph. Two hour later, LaTonya leaves from the same location and travels along the same route to Atlanta at 77 mph. How long will it take LaTonya to catch Juan? Use a simulation of two motions to verify your answer. Let t be the time that Juan is on the road. Since LaTonya leaves two hours later, her time on the road is t 2. Since distance = rate time,
7 05 then the distance that Juan travels is x 1 = 55t and the distance that LaTonya travels is x 2 = 77(t 2). At the point when LaTonya catches Juan, each will have travelled the same distance, so set x 1 = x 2 and and solve for the time t: x 1 = x 2 55t = 77(t 2) (distribute) 55t = 77t 154 (subtract 77t from both sides) 22t = 154 (divide both sides by 22) t = 7 hours. LaTonya's time on the road is t 2 = 5. Thus, it takes LaTonya 5 hours to catch Juan. To verify the answer, we will use a simulation of two motions. To do this, we will graph the distance each of them travelled at different times until their positions become aligned. To make it easier to see the alignment on the graph, we will let y 1 = 2 and y 2 = 4. Thus, on our graph, the line that is generated at y = 2 will represent the distance Juan has travelled and the line generated at y = 4 will represent the distance that LaTonya travelled. t = 2, t 2 = 0 t =, t 2 = 1 t = 4, t 2 = 2 t = 5, t 2 = t = 6, t 2 = 4 t = 7, t 2 = 5
8 Notice the distance are aligned at t = 7 hours in the simulation. This verifies our answer. 06 Objective 4: Finding Parametric Equations. Given an equation in rectangular form defined as a function y = f(x), the most basic parametric equations to use is the let x = t and y = f(t) for t in the domain of f. Sometimes, the conditions of the application dictate that such a set of parametric equations will not work. In that case, our choice of parametric equations must allow for the variable x to still be able to match all the values in the domain of f. For instance, if the domain of f is [, 2], we cannot choose x = t 2 since this would only allow for values of x 0. Find two different sets of parametric equations for the following: Ex. 7 y = 2x 2 + Let x = t. Then y = 2t 2 +, so the parametric equations are: x = t, y = 2t 2 +, < t < Another pair that would work would be to let x = t. Then y = 2t 6 +, so the parametric equations are: x = t, y = 2t 6 +, < t < Find parametric equations for the following object in motion: Ex. 8 x y2 = 1 where the parameter t is on seconds and a) the motion around the ellipse is clockwise, begins at the point ( 2, 0), and requires 2 seconds to make one revolution. b) the motion around the ellipse is counterclockwise, begins at the point (0, 1), and requires 4 seconds to make one revolution. a) The equation is an ellipse with center at (0, 0), major along the xaxis, vertices of ( 2, 0) and (2, 0) and bintercepts (0, 1) and (0, 1). Since the ellipse begins at ( 2, 0), then at t = 0, x = 2 and y = 0.
9 07 Thus, x = 2cos(ωt) and y = sin(ωt) or y = sin(ωt). Since the orientation is clockwise, then as the curve leaves the point ( 2, 0), the yvalues will be positive, so y = sin(ωt). Hence, our equations are: x = 2cos(ωt) and y = sin(ωt) Since, as the graph starts to get traced out, the xvalues are negative and the yvalues are positive, so ω has to be greater than 0. Also, it takes 2 seconds to make one revolution, the period is 2 = 2π which implies ω = π. Therefore, our parametric equations are: x = 2cos(πt) and y = sin(πt) 0 t 2 b) The equation is an ellipse with center at (0, 0), major along the xaxis, vertices of ( 2, 0) and (2, 0) and bintercepts (0, 1) and (0, 1). Since the ellipse begins at (0, 1), then at t = 0, x = 0 and y = 1. Thus, x = 2sin(ωt) or 2sin(ωt) and y = cos(ωt). Since the orientation is counterclockwise, then as the curve leaves the point (0, 1), the xvalues will be negative, so x = 2sin(ωt). Hence, our equations are: x = 2sin(ωt) and y = cos(ωt) Since, as the graph starts to get traced out, the xvalues are negative and the yvalues are positive, so ω has to be greater than 0. Also, it takes 4 seconds to make one revolution, the period is 4 = 2π ω which implies ω = π. Therefore, our parametric equations are: 2 x = 2sin( π 2 t) and y = cos( π 2 t) 0 t 4 There are some curves that cannot be represented easily in terms of x and y, but are very easy to represent using parametric equations. One example is a curve called a cycloid. Its path is traced out by a point on the rim of a circle of radius a as the circle rolls along a fixed line without slipping. To derive the parametric equations, let P = (x, y) be a point on the rim of a circle of radius a such that P starts at the origin. We will allow the circle to roll along the positive xaxis: ω
10 08 Y P C t B 2a O X A Let C be the center the circle and t be the angle of how much the circle has rotated. Since the circle has rolled the distance from O to A, d(o, A), then the amount of the circle that has travelled on the ground is the arc length from A to P. Since there is no slippage, then arc length AP = d(o, A) The arc length AP = rθ = at since the radius r = a and t is the angle defined above. This implies that d(o, A) = at. Also, at = d(o, A) = d(o, X) + d(x, A) which implies d(o, X) = at d(x, A) In examining the diagram, d(x, A) = d(p, B) which in the length of the opposite side of t in triangle PCB. Thus PB = asin(t) which implies d(o, X) = at d(x, A) = at asin(t) = a(t sin(t)) Since d(o, X) = x, then x = a(t sin(t)) Now, d(a, C) = d(a, B) + d(b, C) or d(a, B) = d(a, C) d(b, C), but d(a, C) = a and d(b, C) is the adjacent side to t in triangle PCB which means d(b, C) = acos(t). Substituting these relationships in, we get: d(a, B) = a acos(t) = a(1 cos(t)) Since y = d(a, B), then y = a(1 cos(t)) Although we have proved this for t being less than or equal to π 2, these equations can be extended to any real number t. Parametric Equations for a Cycloid x = a(t sin(t)), y = a(1 cos(t)), < t <. The period will be 2πa. Suppose an object is moving without friction from a starting point P to an ending point Q that is lower than P, but not directly below P. If gravity is the only force acting on it, then the quickest way it can get to Q is along the path of an inverted cycloid. This is known as the curve of quickest decent. Another example is suppose that several objects are placed on an inverted at different locations of cycloid at the same time and began sliding at the same moment. If none of the objects were placed on the lowest point of the cycloid, they all will reach the lowest point at the same time.
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