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**Unformatted text preview: **Math 1920, Prelim II Solutions November 18, 2010, 7:30 PM to 9:00 PM 1. Consider the function f ( x, y, z ) = e y ln ( xz ), and the point P (1,0,1). (a) (5 pts) Find the direction of greatest increase of f at P . The direction of greatest increase is given by the gradient vector. f = f x i + f y j + f z k = e y 1 x i + e y ln ( xz ) j + e y 1 z k At P (1,0,1), f = i + k (b) (10 pts) Find the linear approximation of the change in f for a step size of 1 10 unit length in direction 1 , 6 , 3 from P . Change 1 , 6 , 3 into unit vector: u = 1 , 6 , 3 / 4 D u f | P = f u = 1 df = 1 10 D u f = 0 . 1 2. (10 pts) Find the equation of the tangent plane to the surface S at the point closest to the plane P , where P and S are given by 2 x + 4 y + z =- 8 and z = x 2 + y 2 , respectively. Note that the plane P and the surface S do not intersect. At point closest to P , the tangent plane is parallel to P . The gradient vector is normal to the tangent plane, therefore the gradient vector is parallel to the normal of P . Normal vector of plane P : n = 2 i + 4 j + k Surface S : f = x 2 + y 2- z = 0. Therefore gradient vector f = 2 x i + 2 y j- k . f = c n , so considering each of the components we find that c =- 1, x = c =- 1 and y = 2 c =- 2. Thus z = 5. The point on S closest to P is then (- 1 ,- 2 , 5). Tangent plane to S at (- 1 ,- 2 , 5): 2 x + 4 y + z =- 5. If Lagrange Multipliers was attempted, the most workable method would be to find the point on S for which the distance between it and a fixed point on P projected onto the normal of P is a minimum. The constraint would then be the equation of plane P . 3. (a) (10 pts) The equation 2 x 2 + y 2 = 3 defines an ellipse, and for a non-zero constant k the equation x 2- 2 y 2 = k defines a hyperbola. Find the value of k such that for each point in the intersection, the tangent to the ellipse and the tangent to...

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