10.1 Curves Defined by Parametric Equations
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1 CHAPTER 10. PARAMETRIC AND POLAR Curves Defined by Parametric Equations Example 1. Let x and y be given parametric equations below: x(t) =t cos(t) y(t) =t sin(t) (a) Calculate x and y for 11 values of t: t =0, /5, 2 /5,...,10 /5. (b) Plot the x and y values just calculated, and connect the points with a smooth curve. (c) Explain how the functions and the curve just found can be viewed as a polar function. Solution. (a) t =0 x =0 y =0 t = /5 x = /5cos( /5) 0.51 y = /5sin( /5) 0.37 t =2 /5 x =2 /5cos(2 /5) 0.39 y =2 /5sin(2 /5) 1.2 t =3 /5 x =3 /5cos(3 /5) 0.58 y =3 /5sin(3 /5) 1.8 t =4 /5 x =4 /5cos(4 /5) 2 y =4 /5sin(4 /5) 1.5 t =5 /5 x =5 /5cos(5 /5) 3.1 y =5 /5sin(5 /5) = 0 t =6 /5 x =6 /5cos(6 /5) 3 y =6 /5sin(6 /5) 2.2 t =7 /5 x =7 /5cos(7 /5) 1.4 y =7 /5sin(7 /5) 4.2 t =8 /5 x =8 /5cos(8 /5) 1.6 y =8 /5sin(8 /5) 4.8 t =9 /5 x =9 /5cos(9 /5) 4.6 y =9 /5sin(9 /5) 3.3 t =10 /5 x =10 /5cos(2 ) 6.3 y =10 /5sin(10 /5) = 0 (b) We show below the graph of just the points, and then the graph with a smooth curve through them graphics/parametric_archimedean_spiral_with_points-eps-converted-to.pdf graphics/parametric_archimedian_spiral_graph-eps-convert (c) To view these equations as a polar function, simply change notation: parametric polar
2 CHAPTER 10. PARAMETRIC AND POLAR 100 x(t) =t cos(t) x = r cos( ) (t =, r = ) y(t) =t sin(t) y = r sin( ) Example 2. Let x and y be defined by the following parametric equations x = t +sin( t) y = t + cos( t) (a) Plot by hand 7 values of x and y corresponding to t = 0, 0.25, 0.5, 0.75, 1, 1.25 and 1.5. (b) Use your graphing calculator to plot x and y for 0 apple t apple 5. Solution. (a) We start in the same way you probably learned to graph cartesian equations: calculating a table of points and plotting them by hand. Here are the values (rounded to the first decimal place). t x y Now we plot these points, by hand, as shown, graphics/parametric_points_t+sin_and_t+cos-eps-converted-to.pdf and attempt to draw a line connecting them, just like dot-to-dot, going in the order that the points were defined.
3 CHAPTER 10. PARAMETRIC AND POLAR 101 graphics/parametric_points_connected_t+sin_and_t+cos-eps-converted-to.pdf (b) We enter the equations in our calculator We plot this for 0 apple t apple 5 graphics/parametric_plot_t+sin_and_t+cos-eps-converted-to.pdf Example 3. A few years ago I made and printed out a decorative award for my kid successfully passing swimming lessons. I want to have their name, the date, the name of the swimming school, etc., but I also want a decorative curlicue (aka curly cue ) around the outside of the document. Use your calculator to see which of the following will work best: ( ) ( ) x =1.5cos 5/7 (t)+0.1cos(81t) x =1.5cos 5/7 (t)+0.1cos(81t) or y =sin 5/7 (t)+0.2sin(21t) y =sin 5/7 (t)+0.1sin(81t) If you can, see if you can understand why the curves look the way the do.
4 CHAPTER 10. PARAMETRIC AND POLAR 102 Solution. Both sets of parametric equations start with the same basic shape, x =1.5cos 5/7 (t), y =sin 5/7 (t). This shape is called a superellipse because it looks like an ellipse, but a little bit flattened out, sort of half way between an ellipse and a rectangle: graphics/basic_super_ellipse-eps-converted-to.pdf The other formulas that are added, either 0.1cos(81t), or 0.2sin(21t)or 0.1 sin(81t), litterally add a small variation to the basic shape. Imagine moving your hand around the shape of the super-ellipse, but wiggling your pencil in small circular motions as you move your hand, that s what the other shapes will look like. The first one looks like this graphics/super_ellipse_with_weird_curlicue-eps-converted-to.pdf And the second looks like this
5 CHAPTER 10. PARAMETRIC AND POLAR 103 graphics/super_ellipse_with_nice_curlicue-eps-converted-to.pdf The second shape is about right for an award certificate, especially if we drop the axes and put some words in the middle: graphics/super_ellipse_with_nice_curlicue_no_axes-eps-converted-to.pdf Congratulations! Liam you ve passed your advanced swimming certification, September 1, 2010
6 CHAPTER 10. PARAMETRIC AND POLAR 104 Example 4. Return to Example 4 in the previous section. We had r = 1 sin( ) and found the tangent line at = /12 to be given by y = 0.414(x 0.72) Graph both the original polar function and the tangent line in the same calculator window. Solution. We make both of these functions parametric. For the polar graph, let t =, andthenusex = r cos( ) andy = r sin( ): x(t) =(1 y(t) =(1 sin(t)) cos(t) sin(t)) sin(t) For the cartesian line, simply let t = x and then re-write y as a function of t: x = t y = 0.414(t 0.72) Now we can enter both of these at the same time in the calculator. The result should like like the following: graphics/cardiod_with_tangent_line-eps-converted-to.pdf
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