Tuesday February 19 Solutions Taylor series the 2nd derivative test and changing coordinates 1 Consider f x y 2 cos x y 2 e x y a Show that 0 0 is a critical point for f SOLUTION f x 0 0 2 sin x ye x y 0 0 0 and f y 0 0 2y xe x y 0 0 0 b Calculate each of f xx f x y f y y at 0 0 and use this to write out the 2nd order Taylor approximation for f at 0 0 SOLUTION The second order Taylor approximation of a function f x y at 0 0 is given by T2 x y f 0 0 f x 0 0 x f y 0 0 y f xx 0 0 2 x 2 f y y 0 0 2 y 2 f x y 0 0 x y For this problem we have f xx 2 cos x y 2 e x y f y y 2 x 2 e x y and f x y e x y x ye x y So f xx 0 0 2 f y y 0 0 and f x y 0 0 1 Also f 0 0 3 So the second order Taylor approximation for f at 0 0 is g x y 3 x 2 y 2 x y 2 Let g x y be the approximation you obtained for f x y near 0 0 in 1 b It s not clear from the formula whether g and hence f has a min max or a saddle at 0 0 Test along several lines until you are convinced you ve determined which type it is In the next problem you ll confirm your answer in two ways SOLUTION Let s test a general line y mx which goes through 0 0 as x 0 Then g x mx 3 x 2 m 2 x 2 mx 2 3 1 m m 2 x 2 The polynomial 1 m m 2 is always positive it opens upward and has its global minimum at m 1 2 where 1 m m 2 0 So g x mx is always a downward opening parabola This suggests that 0 0 is a relative maximum 3 Consider alternate coordinates u v on R2 given by x y u v u v a Sketch the u and v axes relative to the usual x and y axes and draw the points whose u v coordinates are 1 2 1 1 1 1 SOLUTION If we express u and v in terms of x and y we get u 1 2 x y and v 1 2 y x So the u axis is given in x and y coordinates by all multiples of the vector 1 1 and the v axis is given by all multiples of the vector 1 1 The two axes and the points are shown below v H1 1L u H 1 2L H1 1L b Express g as a function of u and v and expand and simplify the resulting expression SOLUTION 3 x 2 y 2 x y 3 u v 2 u v 2 u v u v 3 u 2 2uv v 2 u 2 2uv v 2 u 2 v 2 3 u 2 3v 2 c Explain why your answer in 3 b confirms your answer in 2 SOLUTION This is an elliptic paraboloid in uv coordinates opening downward with maximum at 0 0 3 so it confirms that 0 0 is a local maximum 0 0 goes to 0 0 under the transformation so this reasoning makes sense d Sketch a few level sets for g What do the level sets of f look like near 0 0 SOLUTION The level sets are sketched for g 2 7 2 8 2 9 on the left and for f 2 7 2 8 2 9 on the right The level sets for g are ellipses that approximate the level sets of f close to 0 0 The ellipses shrink as they get closer to g x y 3 which consists of the single solution x y 0 0 e It turns out that there is always a similar change of coordinates so that the Taylor series of a function f which has a critical point at 0 0 looks like f u v f 0 0 au 2 bv 2 In fact this is why the 2nd derivative test works Double check your answer in 2 by applying the 2nd derivative test directly to f SOLUTION The Hessian f xx f y y f x y 2 is 2 2 12 3 0 at 0 0 and f xx 0 0 2 0 So f has a relative maximum at 0 0 as suspected 4 Consider the function f x y 3xe y x 3 e 3y a Check that f has only one critical point which is a local maximum SOLUTION f x 3e y 3x 2 and f y 3xe y 3e 3y f y 0 only if x e 2y and f x 0 only if e y x 2 Solving these simultaneously we see that x must satisfy x 2 2 e y 2 x so x 0 1 or 1 But x e 2y 0 so the only critical point is x 1 y 0 Calculating we see that f xx 1 0 f y y 1 0 6 and f x y 1 0 3 So the Hessian f xx f y y f x y 2 36 9 27 0 at 1 0 Since f xx 1 0 0 the second derivative test tells us that f 1 0 1 is a local maximum b Does f have an absolute maxima Why or why not SOLUTION f does not have an absolute maximum For instance if we take the trace curve y 0 we get f x 0 3x x 3 1 which is unbounded as x Absolute maxima and minima are only guaranteed over a closed and bounded set in the domain The plane R2 is closed but not bounded so there is no guarantee that a continuous function will achieve an absolute maximum or minimum over R2
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