UNIT 2PARAMETRIC EQUATIONS, POLAR COORDINATESINTRO: We continue the study of parametric equations. The biggest difference between theequations of this Unit and those of the first Unit will be the use of trigonometric functions toobtain, for example, circular and elliptic trajectories. We will also study points of intersectionof two trajectories, and collision points, and will begin the study of polar coordinates.1. Parametric Equations with Trig FunctionsChanging to Cartesian equations using trig identitiesWe would like to change parametric equations into a Cartesian equation in the case thatdifferent trig functions of the parameter are found in the different coordinate equations. If, forexample, bothxandyare given in terms of sint, then the Cartesian equation can be found justas in Unit 1 by solving one equation for sintand substitute in the second to obtain an equationwith justxandy. If, however,xwere given, say, as a function of sint, andywere given as afunction of cost, then the most convenient method of solution is to solve one equation for sint,the second for cost, and substitute the results into the trig identity sin2t+ cos2t= 1, therebyobtaining an equation with justxandy. Likewise, ifxandywere given in terms of tantandsect, respectively, or in terms of cottand csct, respectively, then solving each equation for itstrig function and substituting into the well-known identities for tan2t,sec2tor cot2t,csc2twould be the most convenient way to obtain a Cartesian equationExample:(x= sinty= 3 costt≥0Using sint=x, cost=y/3, substitute into the identitysin2t+ cos2t= 1 to obtainx2+(y3)2= 1Example:(x= sinety= 3 cosett≥0Using the identitysin2et+ cos2et= 1, obtain againx2+(y3)2= 1Example:(x= sin2ty= cos 2t,0≤t≤2πHere one can use the trickier identity cos 2t= cos2t-sin2t= 1-2 sin2tto obtainy=-2x+ 1(line segment with 0≤x≤1 retraced 4 times)