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1 Math 2220 Prelim 1 February 17, 2009 Name: Instructor: Lecture time: TA: Section time: INSTRUCTIONS READ THIS NOW This test has 6 problems on 7 pages worth a total of 100 points. Look over your test package right now. If you find any missing pages or problems please ask a proctor for another test booklet. Write your name, your instructor s name, the time your lecture meets, your TA s name, and the time you section meets right now.. Show your work. To receive full credit, your answers must be neatly written and logically organized. If you need more space, write on the back side of the preceding sheet, but be sure to clearly label your work. This is a 90 minute test. You are not allowed to use a calculator or any other aids. You DO NOT need to SIMPLIFY your answers. OFFICIAL USE ONLY Total:

2 Math 2220 (Spring 2009) Prelim 1 (2/17/2009) 2 Problem 1: 21 points Suppose that the functions f : R 3 R 2 g : R 2 R 1 p : R 1 R 3 are differentiable everywhere on their domains; and that u R 3 and v R 2. For each of the expressions below, indicate either (1 point each) that the expression makes no sense, and (2 points each) a brief explanation why, or (1 point each) that it does make sense, and (2 points each) what kind of thing it denotes: a number, a vector (in what space), a function (what are the dimensions of input and output spaces), a matrix (of what dimensions). (a) Df(u)- This is a 2 3 matrix with real entries. (b) g f- This is a function g f : R 3 R 1. (c) f- This does not make sense because the gradient is only defined for functions to R 1. (d) g(u) v- This does not make sense because g needs to be evaluated at vectors in R 2 and u R 3. (e) ( g )- This is the 1 2 matrix 2 g x ] 2 g 2 (f) u v- This does not make sense because vectors must be in the same space to take their dot product. (g) Dp(x)Df(u)- This does not make sense because Dp(x) is a 1 3 matrix and Df(u) is a 2 3 matrix. You cannot multiply two such matrices in that order.

3 Math 2220 (Spring 2009) Prelim 1 (2/17/2009) 3 Problem 2: (15 points) Let f : R 3 R 1 be a C 1 function everywhere on its domain. Define the function g by (x, y) f(x 2 y, x cos(y), e xy ). Find Dg(2, π) given the information that Df(4π, 2, e 2π ) = 2 1 3] Solution. Define h: R 2 R 3 by (x, y) (x 2 y, x cos y, e xy ). Then g = f h so that we can use the chain rule: Dg(2, π) = Df(h(2, π))dh(2, π) = Df(4π, 1, e 2π )Dh(2, π) 2xy x 2 = 2, 1, 3] cos y x sin y ye xy xe xy 4π 4 = 2, 1, 3] 1 0 πe 2π 2e 2π = 8π πe 2π 8 + 6e ] 2π (2,π)

4 Math 2220 (Spring 2009) Prelim 1 (2/17/2009) 4 Problem 3: (16 points) Suppose that an ant is crawling around in the xy plane (where both coordinates are strictly positive) and that the temperature at point (x, y) is ln(x + y 2 ). (a) If the ant is at (1, 2) find any vector pointing in the direction that the ant must travel to decrease the temperature as rapidly as possible. (b) If the ant is at (1, 2) and traveling at 3 units of distance per second in a northeasterly direction (meaning, in the direction of increasing x and y, at a 45 degree angle to the x-axis), at what rate is the temperature changing? (c) If the ant is once again at point (1, 2), find any vector pointing in a direction that the ant can travel in order to keep the temperature constant. Solution. For ease of notation define f : (x, y) ln(x + y 2 ). (a) Since f points in the direction of maximal increase, the ant must travel in the direction f to decrease the temperature most rapidly: ] 1 2y (1,2) f(1, 2) = x+y 2 x+y 2 = ] (b) We are given that the ant s velocity vector is given by v = 3 2, 2 ]. To compute the rate of change of temperature, we compute the directional derivative of f in the direction of v at the point (1, 2). This is just the dot product 1 f(1, 2) v = 5, 4 ] ] 3 3 2, 5 2 ( 1 = ) 5 2 = 3 degrees per second. 2 (c) So long as the ant travels in a direction orthogonal to f(1, 2), the temperature will remain the same. Thus, we need only product a direction vector which has zero dot product with 1 5, 4 5]. One such direction vector is 4, 1].

5 Math 2220 (Spring 2009) Prelim 1 (2/17/2009) 5 Problem 4: (16 points) Let f be the function given by (x, y) x 2 cos(y). (a) Find the tangent plane to the graph of f at the point (1, π/4, 2/2). (b) Use your answer to part (a) to estimate f(.99,.75) Solution. (a) There are several ways to do this problem, and they all lead to the same equation. Because part (b) asks us to approximate a value of f, here we will think of the tangent plane as the first-order Taylor approximation. We need to find the first partials and evaluate at the point (1, π/4): f x = 2x cos y f y = x 2 sin y f x (1, π/4) = 2 f y (1, π/4) = 2/2 Then, the tangent plane has equation z = f(1, π/4) + f x (1, π/4)(x 1) + f y (1, π/4)(y π/4) = 2/2 + 2(x 1) 2/2(y π/4). (b) Using part (a), we can approximate f(.99,.75) by f(.99,.75) 2/2 + 2(.01) 2/2((3 π)/4)

6 Math 2220 (Spring 2009) Prelim 1 (2/17/2009) 6 Problem 5: (16 points) Show that the system xy 2 + xzu + yv 2 = 3 x 3 yz + 2xv u 2 v 2 = 2 (a) can be solved for u and v as functions of x, y, z near the point (x, y, z, u, v) = (1, 1, 1, 1, 1). (b) Find the value of v Solution. for the solution at (x, y, z) = (1, 1, 1). (a) To check for solvability, we form the equations f(x, y, z, u, v) = xy 2 + xzu + yv 2 3 g(x, y, z, u, v) = x 3 yz + 2xv u 2 v 2 2 and the determinant fu f = det v g u g v ] (1,1,1,1,1) = det = det xz 2vy 2uv 2 2x 2vu 2 ] = ] (1,1,1,1,1) Because 0, we are assured solvability by the Implicit Function Theorem. (b) To calculate v, we differentiate f and g implicitly with respect to y: 2xy + xz u + v2 + 2yv v x 3 z + 2x v 2u u v2 2u 2 v v = 0 = 0. Plugging in (x, y, z, u, v) = (1, 1, 1, 1, 1) and simplifying yields u Thus, we conclude that v (1, 1, 1) = 7/4. (1, 1, 1) + 2 v (1, 1, 1) = 3 2 u = 1

7 Math 2220 (Spring 2009) Prelim 1 (2/17/2009) 7 Problem 6: (16 points) Find (if they exist) local and global extreme points and any saddle points of f(x, y) = 2xy on the closed disk x 2 + y 2 4. Be sure to justify why the points you find are local extrema, global extrema or saddle points. Solution. For ease of notation, let D be the closed disk x 2 + y 2 4. Then D is closed and bounded and D is the circle x 2 + y 2 = 4. Since D is closed and bounded, we know that f must assume attain and absolute maximum and an absolute minimum by Theorem 7. We will start by looking for local and global extrema on the interior of D, that is, on D \ D. Since D \ D is open, Theorem 4 guarantees that all extrema of f will occur at critical points, that is, points where f vanishes. However, f = 2y, 2x] vanishes only at the origin (x, y) = (0, 0). To determine the nature of this critical point, we check the determinant of the Hessian matrix at (0, 0): ] fxx f = det xy = det = 4 f xy Since > 0, and f xx (0, 0) = 2 > 0, we can apply Theorem 6 to conclude that (0, 0) is a local minimum. We won t know if (0, 0) is a global minimum until we analyze the critical points on D. To do this, we need Lagrange multipliers. Let g(x, y) = x 2 + y 2. Then D is the level set g = 4. We want to solve the equation f = λ g. This leads to the system of equations f yy ] (0,0) (0,0) 2y = 2λx (1) 2x = 2λy (2) x 2 + y 2 = 4. (3) Note that equations (1) and (2) show that none of x, y, or λ is zero. Combining equations (1) and (2) gives 2x = 2λ 2 x, or λ = ±1. This gives us that x = ±y and so equation (3) shows that there are four critical points on D: (± 2, ± 2) and (± 2, 2). All we need to do now is evaluate f at these four points and compare to f(0, 0): f(0, 0) = 0 f(± 2, ± 2) = 4 f(± 2, 2) = 4. So, we conclude that f has a global maximum at (± 2, ± 2), a global minimum at (± 2, 2) and a local minimum at (0, 0). STOP. THIS IS THE LAST PAGE.

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