Return to List of Lessons Lesson 10. Parametric Curves (A) Parametric Curves If a curve fails the Vertical Line Test, it can’t be expressed by a function. In this case you will encounter a problem if you try to find the slope of a tangent to the curve, or the area enclosed by the curve. In calculus this problem can be solved if the curve is expressed by a pair of parametric equations: x = f (t) , y = g(t) , where t is the parameter with the range of a ≤ t ≤ b in general. Each value of the parameter t determines a point (x, y) = ( f (t), g(t)). The initial point is ()f (a), g(a) and the terminal point is ( f (b), g(b)). As t varies, the point ()x, y will trace out a curve that is called parametric curve. In many applications of parametric curves, t stands for time, but does not always necessarily represent time. In this section we will learn how to sketch the parametric curves using Mathematica. Example1 Sketch and identify the curve given by the parametric equations x = t 2 − 2t, y = t +1 for − 2 ≤ t ≤ 4 . Each value of the parametert gives a point on the curve, which can be evaluated by: ^, ,,,, Note that in the first bracket listed two parametric functions, and in the second bracket , , , means t runs from -2 to 4 in steps of 1. We can find the following points on the parametric curve , , , , , , , , , , , , , They correspond to the set of t values 2, 1, 0, 1, 2, 3, 4 in order. So this curve will start from the point (8, −1) and end at the point (8, 5) . We can plot this curve using “ParametricPlot”. ^ , , , , 1 5 4 3 2 1 2 4 6 8 -1 The curve traced out may be a parabola. This can be confirmed by eliminating the parameter t as follows. We obtain t = y −1 from the second equation and substitute into the first equation. It gives x = t 2 − 2t = ()()y −1 2 − 2 y −1 = y 2 − 4y + 3 and so the curve represented by the given parametric equations is a parabola that matches the Cartesian equation x = y 2 − 4y + 3 . The range of t value will affect the graph a parametric curve significantly. Look at the example again by modifying the range of t values to 0 ≤ t ≤ 4 . Example2 Sketch and identify the curve given by the parametric equations x = t 2 − 2t, y = t +1 for 0 ≤ t ≤ 4 . We evaluate the points on the curve by running t from 0 to 4 in steps of 1, , , , , , and these points on curve will be: , , , , , , , , , . The two points 8,1,3,0 from Example 1 are missing because the values of parameter t = −2 and t = −1are removed from the original list. Again we can plot the curve using “ParametricPlot”. ^,,,, 5 4 3 2 2 4 6 8 2 Example3 What curve is represented by the parametric equations x = cos t, y = sin t for 0 ≤ t ≤ 2π . , , , , 1.0 0.5 -1.0 -0.5 0.5 1.0 -0.5 -1.0 The graph represented by the pair of parametric equations x = cos t, y = sin t is obviously a unit circle centered at the origin. Meanwhile it is not hard to realize that by the Pythagorean Theorem x 2 + y 2 = cos 2 t + sin 2 t = 1, its Cartesian equation is truly a unit circle. Since t value runs from 0 to 2π in this case, the circle will go around the origin exact one revolution. If the range of t value is modified, the shape of parametric curve will change. Check out the next example. Example4 What curve is represented by the parametric equations x = cos t, y = sin t for 0 ≤ t ≤ π ? ,,,,,, 1.0 0.5 -1.0 -0.5 0.5 1.0 -0.5 -1.0 3 We can see that the lower half of the unit circle is gone because the range of t value defined over the interval 0 ≤ t ≤ π which represents the upper half of a circle sketched in Quadrants 1 and 2. (B) The Cycloid The cycloid is one of the most interesting parametric curves that we should know. If a circle has radius r and rolls along a straight line (the x-axis), then a point P on the circumference of the circle will trace out the curve of the cycloid, which can be represented by the parametric equations: x = r()θ − sinθ , y = r(1− cosθ ) for − ∞ < θ < ∞ Note that the cycloid can be nicely and neatly represented by the parametric equations, but it is very complicated to be expressed by a Cartesian equation. Example5 Plot the cycloid given by x = 2(θ − sinθ ), y = 2(1− cosθ ) for − 4π ≤ θ ≤ 4π . , , , , 4 3 2 1 -20 -10 0 10 20 In this case it might be a problem when you try to type in the Greek letterθ . Here it is the direction of where to find the letterθ from the Classroom Assistant. To learn more about the cycloid, please visit the site http://mathworld.wolfram.com/Cycloid.html to visualize the formation of the cycloid by tracing a movement of a fixed point on the circumference of a circle. 4 (C) Tangents of a Parametric Curve For a parametric curve given by the equations: x = f (t) , y = g(t) , and a ≤ t ≤ b , the tangent at any dy dy g '(t) t value can be calculated by = dt = if f '(t) ≠ 0 . dx dx f '(t) dt Example6 Suppose a curve C is defined by the parametric equation x = t 2 , y = t 3 − 3t . (1) Plot the curve, (2) find the equation(s) of the tangent line(s) to the curve at the point (3, 0). (3) Plot the tangent line(s) at the point (3, 0). We will work it out step by step. (1) , ,,, 5 2 4 6 8 -5 We observe that there are two curves that pass through the point (3, 0) , and correspondingly there are two tangent lines passing through this point. Since (3, 0) is a Cartesian point, namely x = 3 and y = 0 . We need to find out the corresponding value of the parameter t. Type in , to find the t values: √3, √3 5 Type in , to find the t values: 0, √3, √3. Taking the solutions of t in common from the two sets, we have t = 3 or t = − 3 . dy g '(t) (2) Now let’s calculate the first derivative function by = , which leads to the slopes of the dx f '(t) tangent lines. We will run the following commands to find the slopes of tangents for t = 3 and t = − 3 . _: Shift + Enter _: Shift Enter Type in to get √ Type in to get √ √ Type in to get √ Thus we found that the slopes of the two tangent lines at the point (3, 0) are 3 and − 3 . Then by the Point-Slope Form, we can formulate the equations of the two tangent lines that pass through the point (3, 0) as: y = 3(x − 3) and y = − 3(x − 3). (3) We will plot these two tangent line(s) at the point (3, 0) in the same viewing box of the curve. , , , √, , √,, , , , 6 4 y= 3(x-3) 2 3, 0 -4 -2 2 4 H L -2 y= - 3 (x-3) -4 (D) The Area Under a Parametric Curve If the curve is traced out once by the parametric equations x = f (t) , y = g(t) for a ≤ t ≤ b , then b we can calculate the area under the curve using the following formula: A = g(t) f '(t) dt . It is ∫ a very fast to find the area under a parametric curve by Mathematica. Example7 Find the area under one arch of the cycloid x = r(θ − sinθ ), y = r()1− cosθ for r = 2 . , , , , 4 3 2 1 2 4 6 8 10 12 It is easy to realize that one arch of the cycloid is traced out by running t from 0 to 2π . We can b find the area under one arch by applying the Area Formula A = g(t) f '(t) dt . ∫ a 7 _: _: , , , Therefore when r = 2 , the area under one arch of the cycloid equals 12π . (E) Arc Length of a Parametric Curve Suppose the curve C is described by the parametric equations x = f (t) , y = g(t) for a ≤ t ≤ b . If f 'and g ' are both continuous on [a, b] and C is traversed exactly once as t increases from a to b, then the length of C can calculated by 2 2 b ⎛ dx ⎞ ⎛ dy ⎞ b 2 2 L = ⎜ ⎟ + ⎜ ⎟ dt , or L = ()()f '(t) + g '(t) dt ∫ a ⎝ dt ⎠ ⎝ dt ⎠ ∫ a Example9 Find the length of one arch of the cycloid x = r(θ − sinθ ), y = r()1− cosθ for r = 2 . Like we did in the previous example, first we define the two parametric equations, and then apply b the Arc Length Formula L = ()()f '(t) 2 + g '(t) 2 dt . ∫ a _: _: , , , Therefore when r = 2 , the length of one arch of the cycloid equals 8r = 16 . Return to List of Lessons 8 .