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MATH 137 Multivariable Calculus Solutions to Midterm 2.

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Georgetown University MATH 137 Multivariable Calculus Fall 2018 Solutions to Midterm 2 1. Find the critical points of f(x; y) = -x3 + 12x - 4y2 and classify each point as a local maximum, local minimum, or saddl ...

Georgetown University MATH 137 Multivariable Calculus Fall 2018 Solutions to Midterm 2 1. Find the critical points of f(x; y) = -x3 + 12x - 4y2 and classify each point as a local maximum, local minimum, or saddle point. To solve for the critical points, we set fx = -3x2 + 12 = 0 fy = -8y = 0 The first equation gives x2 = 4, whence x = ±2. The second equation gives y = 0. Thus there are two critical points, (2; 0) and (-2; 0). We now use the Hessian H = fxxfyy - (fxy)2 to determine the nature of each critical point. We have ) - 02 = 48x. We have Therefore f(2; 0) is a local maximum. Also, H(-2; 0) = 48(-2) = -96 < 0 So (-2; 0; f(-2; 0)) is a saddle point. 2. Use the Lagrange multiplier method to find the maximum and minimum values of subject to the constraint We have We may deduce from the first four equations of this system that x = y = z = t. Hence x2 + y2 + z2 + t2 = 4x2 = 1, which implies x2 = 1 4, or x = ±12. Therefore there are two critical points to consider, namely 1 2; 1 2; 12; 12 and -1 2; -1 2; -1 2; -1 2 . We have Therefore, on the hypersphere x2 + y2 + z2 + t2 = 1, the absolute maximum of f is 2 and the absolute minimum of f is -2. 3. Consider the surface defined by the equation x3 + x2y2 + z2 = 0. Find an equation for the plane tangent to the surface at the point P(-2; 1; 2). This surface is a level surface of the function f(x; y; z) = x3 +x2y2 +z2. Consequently an equation of the tangent plane at P(-2; 1; 2) is Therefore the tangent plane is given by (a) Find the directional derivative of f at P (1; 1) in the direction of Q(2; 3). (b) Determine the direction of maximum increase of f at P (1; 1). This direction is specified by the gradient vector rf(1; 1) = h2; 6i. 5. Show that the level curves f(x; y) = x2 - y2 = k1 and g(x; y) = xy = k2, where k1 and k2 are constants, intersect orthogonally at any point of intersection P (a; b). Let P (a; b) be any point of intersection of two level curves. We have Therefore the level curves intersect orthogonally. 6. Let f(x; y) satisfy rf(x; y) = chx; yi for some constant c and all (x; y) 2 R2. Show that f is constant on any circle of radius a > 0 centered at the origin.

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