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BC MATH 153 - Calculus with Parametric Equations

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9.4 Calculus with Parametric Equations Contemporary Calculus 1 9.4 CALCULUS AND PARAMETRIC EQUATIONS The previous section discussed parametric equations, their graphs, and some of their uses for visualizing and analyzing information. This section examines some of the ideas and techniques of calculus as they apply to parametric equations: slope of a tangent line, speed, arc length, and area. Slope, speed, and arc length were considered earlier (in optional parts of sections 2.5 and 5.2), and the presentation here is brief. The material on area is new and is a variation on the Riemann sum development of the integral. This section ends with a presentation of some of the properties of the cycloid. Slope (also see section 2.5) If x(t) and y(t) are differentiable functions of t, then the derivatives dx/dt and dy/dt measure the rates of change of x and y with respect to t: dx/dt and dy/dt tell how fast each variable is changing. The derivative dy/dx measures the slope of the line tangent to the parametric graph (x(t), y(t) ). To calculate dy/dx we need to use the Chain Rule: dyd t = dydx . dxd t . Dividing each side of the Chain Rule by dxdt , we have dydx = dy/dt dx/dt . Slope with Parametric Equations If x(t) and y(t) are differentiable functions of t and dxd t ≠ 0, then the slope of the line tangent to the parametric graph is dydx = dy/dt dx/dt . Example 1: The location of an object is given by the parametric equations x(t) = t3 + 1 feet and y(t) = t2 + t feet at time t seconds. (a) Evaluate x(t) and y(t) at t = –2, –1, 0, 1, and 2, and then graph the path of the object for –2 ≤ t ≤ 2. (b) Evaluate dy/dx for t = –2, –1, 0, 1, and 2. Do your calculated values for dy/dx agree with the shape of your graph in part (a)? Solution: (a) When t = –2, x = (–2)3 + 1 = –7 and y = (–2)2 + (–2) = 2. The other values for x and y are given in Table 1. The graph of (x, y) is shown in Fig. 1. t x y dy/dx–2 –7 2 –3/12= –1/4–1 0 0 –1/3 0 1 0 undefined 1 2 2 3/3 = 12 9 6 5/12Table 19.4 Calculus with Parametric Equations Contemporary Calculus 2 (b) dy/dt = 2t + 1 and dx/dt = 3t2 so dydx = 2t + 13t2 . When t = –2, dydx = –312 . The other values for dy/dx are given in Table 1. Practice 1: Find the equation of the line tangent to the graph of the parametric equations in Example 1 when t = 3. An object can "visit" the same location more than once, and a parametric graph can go through the same point more than once. Example 2: Fig. 2 shows the x and y coordinates of an object at time t. (a) Sketch the parametric graph (x(t), y(t)), the position of the object at time t. (b) Give the coordinates of the object when t = 1 and t = 3. (c) Find the slopes of the tangent lines to the parametric graph when t = 1 and t = 3. Solution: (a) By reading the x and y values on the graphs in Fig. 2, we can plot points on the parametric graph. The parametric graph is shown in Fig. 3. (b) When t = 1, x = 2 and y = 2 so the parametric graph goes through the point (2,2). When t = 3, the parametric graph goes through the same point (2,2). (c) When t = 1, dy/dt ≈ –1 and dx/dt ≈ +1 so dydx = dy/dtdx/dt ≈ –1+1 = –1. When t = 3, dy/dt ≈ +1 and dx/dt ≈ +1 so dydx ≈ +1+1 = +1. These values agree with the appearance of the parametric graph in Fig. 3. The object goes through the point (2,2) twice (when t=1 and t=3), but it is traveling in a different direction each time. Practice 2: (a) Estimate the slopes of the lines tangent to the parametric graph when t = 2 and t = 5. (b) When does y '(t) = 0 in Fig. 2? (c) When does the parametric graph in Fig. 3 have a maximum? A minimum? (d) How are the maximum and minimum points on a parametric graph related to the derivatives of x(t) and y(t)?9.4 Calculus with Parametric Equations Contemporary Calculus 3 Speed If we know how fast an object is moving in the x direction ( dx/dt ) and how fast in the y direction ( dy/dt ), it is straightforward to determine the speed of the object, how fast it is moving in the xy –plane. If, during a short interval of time ∆t, the object's position changes ∆x in the x direction and ∆y in the y direction (Fig. 4), then the object has moved (∆x)2 + (∆y)2 in ∆t time. Then average speed = distance moved time change = (∆x)2 + (∆y)2∆t = ( ∆x∆t )2 + ( ∆y∆t )2 . If x(t) and y(t) are differentiable functions of t, and if we take the limit of the average speed as ∆t approaches 0, then speed = ! lim"t#0{average speed} = ! lim"t#0 "x"t$ % & ' ( ) 2+"y"t$ % & ' ( ) = ( dxd t )2 + ( dyd t )2 . Speed with Parametric Equations If an object is located at (x(t), y(t)) at time t, and x(t) and y(t) are differentiable functions of t , then the speed of the object is ( dxd t )2 + ( dyd t )2 . Example 3: At time t seconds an object is located at ( cos(t) feet, sin(t) feet ) in the plane. Sketch the path of the object and show that it is travelling at a constant speed. Solution: The object is moving in a circular path (Fig. 5). dx/dt = –sin(t) feet/second and dy/dt = cos(t) feet/second so at all times the speed of the object is ( dxd t )2 + ( dyd t )2 = ( –sin(t) )2 + ( cos(t) )2 = sin2(t) + cos2(t) = 1 = 1 foot per second.9.4 Calculus with Parametric Equations Contemporary Calculus 4 Practice 3: Is the object in Example 2 traveling faster when t = 1 or when t = 3? When t = 1 or when t = 2? Arc Length (also see section 5.2) In section 5.2 we approximated the total length L of a curve by partitioning the curve into small pieces (Fig. 6), approximating the length of each piece using the distance formula, and then adding the lengths of the pieces together to get L ≈ ∑ ( ∆x )2 + ( ∆y )2 = ∑ ( ∆x∆x )2 + ( ∆y∆x )2 ∆x , a Riemann sum. As ∆x approaches 0, the Riemann sum approaches the definite integral L = ⌡⌠x = ab 1 + ( dydx )2 dx . A similar approach also works for parametric equations, but in this case we factor


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BC MATH 153 - Calculus with Parametric Equations

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