Math 234 Practice Test 2 Show your work in all the problems 1 In what directions is the derivative of x2 y 2 x2 y 2 f x y at P 1 1 equal to zero 2 Find an equation for the level surface of the function through the given point P a f x y z z x2 y 2 P 3 1 1 b f x y z Z y x dt 1 t2 Z z 2 dt P 1 1 2 1 t t2 1 3 Compute the limits of the following expressions if they exist If you think they don t consider different paths of approach to show that they do not a x y 1 x y 4 3 x 6 y 1 x y 1 b x2 y 2 x y 0 0 xy 6 0 xy 4 Compute all second order partial derivatives of the function f x y x sin y y sin x xy 5 Find dw if w sin xy x et and y ln t 1 Then evaluate at dt t 0 1 Solutions 1 We first compute the gradient vector of f 4xy 2 4x2 y x2 y 2 2 x2 y 2 2 f Evaluating at 1 1 yields f 1 1 1 1 The directions u in which the directional derivative Du f 1 1 is zero are the unit vectors orthogonal to f 1 1 1 1 i e u u1 u2 has to satisfy f 1 1 u 1 1 u1 u2 u1 u2 0 and u21 u22 1 We get 1 1 u 1 1 and u 1 1 2 2 2 a The level surfaces are given by the equations c z x2 y 2 where c is a constant In order to single out the one which passes through the point P we need to insert the coordinates of P and compute the constant c Hence c 1 32 1 2 9 so that the desired equation is 9 z x2 y 2 b We compute the integrals first Z y x dt sin 1 y sin 1 x 2 1 t and Z z 2 dt 1 1 1 2 sec z sec z sec 4 t t2 1 2 1 The latter follows from the fact that sec 2 is the angle where cos 1 2 which is 4 We insert now x 1 y 1 2 and z 1 c f 1 1 2 1 sin 1 1 2 sin 1 1 sec 1 1 4 0 6 2 4 5 12 The equation of the level surface is then 5 sin 1 y sin 1 x sec 1 z 12 4 or 2 sin 1 y sin 1 x sec 1 z 3 3 a We use the following formula to simplify the given expression q q q x y 1 x y 1 x 2 y 1 2 x y 1 x y 1 Then x y 1 1 x y 1 x y 1 and the limit can simply be computed by inserting x y 4 3 which yields 1 4 b No further simplification as in problem a is possible here We look at the level curves c x2 y 2 or x2 y 2 cxy 0 xy Solving this for y quadratic equation we get 1 y c c2 4 x kx 2 3 Note that all these curves pass through the origin so there is no limit for x y 0 0 In order to confirm we insert y kx and we get x2 y 2 x2 k 2 x2 1 k2 xy kx2 k Taking the limit x 0 still yields 1 k 2 k i e the value depends on the direction in which we approach the origin 4 We have f x y x sin y y sin x xy The first order derivatives are given by fx sin y y cos x y fy x cos y sin x x Then fxx y sin x fyy x sin y fxy fyx cos y cos x 1 5 We compute w w y cos xy x cos xy x y and x t et y t 1 t 1 Then dw w w x t y t dt x y et ln t 1 cos et ln t 1 ln t 1 et cos et ln t 1 t 1 1 et cos et ln t 1 t 1 Evaluating at t 0 yields dw dt t 0 cos 1 4
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