Mathematics 317 Solutions to Ass2(2)

Mathematics 317 Solutions to Ass2(2) - Mathematics 317...

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Mathematics 317 Solutions to Ass. #2 1. (a) The equation 0 ) , ( = y x f defines y as an implicit function of x . From the equation, one has . 0 2 0 2 2 2 = + + + = = + = y f y f y f f y f f y yy xy xx dx f d y x dx df Hence , y x f f y - = . 3 2 2 2 2 2 y x yy y xx y x xy y yy xy xx f f f f f f f f f y f y f f y - - + + = - = As shown in class, the radius of curvature . 2 / 3 2 ) 1 ( y y R + = Since , 1 2 2 2 2 y y x f f f y + = + one obtains . 2 2 2 / 3 2 2 2 ) ( yy x xx y xy y x y x f f f f f f f f f R - - + = (b) For the given ellipse, . 1 ) , ( 2 2 2 2 - + = b y a x y x f Hence , , , 2 2 2 2 2 2 a xx b y y a x x f f f = = = . 0 , 2 2 = = xy b yy f f At ( a ,0): , , 0 , 2 2 2 a xx y a x f f f = = = . 0 , 2 2 = = xy b yy f f Hence the radius of curvature at this point is given by ( 29( 29 . 2 2 2 2 / 3 2 / 4 / 2 ) / 4 ( a b a b a R = = 2. (a) Here a curve is parametrized in terms of the polar angle θ (playing the role of t ). In particular, r u j i j i j i r ) ( sin ) ( cos ) ( sin cos θ θ θ θ θ θ θ f f f r r y x = + = + = + = where . sin cos j i u r θ θ + = Then . cos sin j i u r u θ θ θ θ + - = = d d Note that . k u u r = × θ Thus .
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