# Unit_2 - UNIT 2 PARAMETRIC EQUATIONS POLAR COORDINATES...

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UNIT 2 PARAMETRIC EQUATIONS, POLAR COORDINATES INTRO: We continue the study of parametric equations. The biggest difference between the equations of this Unit and those of the first Unit will be the use of trigonometric functions to obtain, for example, circular and elliptic trajectories. We will also study points of intersection of two trajectories, and collision points, and will begin the study of polar coordinates. 1. Parametric Equations with Trig Functions Changing to Cartesian equations using trig identities We would like to change parametric equations into a Cartesian equation in the case that different trig functions of the parameter are found in the different coordinate equations. If, for example, both x and y are given in terms of sin t , then the Cartesian equation can be found just as in Unit 1 by solving one equation for sin t and substitute in the second to obtain an equation with just x and y . If, however, x were given, say, as a function of sin t , and y were given as a function of cos t , then the most convenient method of solution is to solve one equation for sin t , the second for cos t , and substitute the results into the trig identity sin 2 t + cos 2 t = 1, thereby obtaining an equation with just x and y . Likewise, if x and y were given in terms of tan t and sec t , respectively, or in terms of cot t and csc t , respectively, then solving each equation for its trig function and substituting into the well-known identities for tan 2 t, sec 2 t or cot 2 t, csc 2 t would be the most convenient way to obtain a Cartesian equation Example : ( x = sin t y = 3 cos t t 0 Using sin t = x , cos t = y/ 3, substitute into the identity sin 2 t + cos 2 t = 1 to obtain x 2 + ( y 3 ) 2 = 1 Example : ( x = sin e t y = 3 cos e t t 0 Using the identity sin 2 e t + cos 2 e t = 1, obtain again x 2 + ( y 3 ) 2 = 1 Example : ( x = sin 2 t y = cos 2 t , 0 t 2 π Here one can use the trickier identity cos 2 t = cos 2 t - sin 2 t = 1 - 2 sin 2 t to obtain y = - 2 x + 1 (line segment with 0 x 1 retraced 4 times)

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Example : ( x = sin t y = 2 sin 3 t + 3 0 t π/ 2 This is not a case where a trig identity needs to be employed. As in Unit 1, solve for sin t and substitute into the second equation. y = 2 x 3 + 3 , 0 x 1 Example : ( x = 3 sin 3 t + 1 y = 2 cos 5 t - 2 0 t π Some algebra is required to isolate the trig functions: sin t = x - 1 3 1 / 3 , cos t = y + 2 2 1 / 5 . Thus, x - 1 3 2 / 3 + y + 2 2 2 5 = 1 Graphing parametric equations In Unit 1 we discussed graphing parametric equations by constructing a table with selected values of the parameter t , say, in one column and the coordinate values in the remaining columns. The key to an efficient plot is to plug in useful points within the specified range of t , not just any points. For example, if x and y are in terms of sin t and cos t for 0 t π , then t = 0 , π/ 4 , π/ 2 , 3 π/ 4 , π may be sufficient, but if x or y includes a term such as sin 4 t , then t = 0 , π/ 8 , π/ 4 , . . . , etc., may at the very least be necessary, or possibly smaller intervals, since a change of π/ 4 in t would result in a change of π in 4 t , too large a change to recognize what
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