MATH 2203 Exam 3 (Version 2) Solutions March 6, 2015 S. F. Ellermeyer Name Instructions. Your work on this exam will be graded according to two criteria: mathematical correctness and clarity of presentation. In other words, you must know what you are doing (mathematically) and you must also express yourself clearly. In particular, write answers to questions using correct notation and using complete sentences where appropriate. Also, you must supply sufficient detail in your solutions (relevant calculations, written explanations of why you are doing these calculations, etc.). It is not sufficient to just write down an answer with no explanation of how you arrived at that answer. As a rule of thumb, the harder that I have to work to interpret what you are trying to say, the less credit you will get. You may use your calculator but you may not use any books or notes. 1. This is a matching question. Match each of the given functions (a f) with their contour plots (labelled 1 6). Grading of this question will be as follows: Number correct 0 1 2 3 4 5 6 Points Awarded 0 1.1 2.2 3.3 6.7 13.3 20 a. The function fx,y x y 2 matches the contour plot 2. b. The function fx,y x y matches the contour plot 6. c. The function fx,y x 2 y 2 matches the contour plot 5. d. The function fx,y x y matches the contour plot 1. e. The function fx,y y x 2 matches the contour plot 4. f. The function fx,y x y matches the contour plot 3.
2. For the function fx,y 1 x y, 2 compute the partial derivatives f x and f y and also compute the second order mixed partial derivatives f xy and f yx. (You should find that f xy f yx.) You must show all details of your computations. Just giving answers will not suffice. Solution: Treating y as a constant and using the quotient rule, we obtain f x x y2 0 11 1 x y 2 2 x y 2 2 Using the quotient rule and treating x as a constant, we obtain f y x y2 0 12y 2y x y 2 2 x y 2 2 Thus and f xy f x y x y2 2 0 1 2x y 2 2y x y 2 4 f yx f y x x y2 2 0 2y 2x y 2 1 x y 2 4 We see that f xy f yx as expected. 3. Suppose that z fy y gx x hs,t. 4y x y 2 3 4y x y 2 3. a. Draw a chain diagram and write a Chain Rule for dz/ds and a Chain Rule for dz/dt. b. Let z y 2 3y y sinx x s 2 t 3 t. Write a general formula for dz/ds and use your formula to evaluate dz/ds s,t1,1. Solution: The chain diagram is
and the Chain Rules are and Using we obtain dz ds dy dz dy dx dx ds dz dt dz dy dy dx dx dt. z y 2 3y y sinx x s 2 t 3 t, dz ds dy dz dy dx dx ds 2y 3cosx2st 3. At the point s,t 1,1, wehave z 0 y sin0 0 x 0, and thus dz 3cos02 6. ds s,t1,1 4. A graph of the function fx,y x y is shown below. (The graph is a plane.)
a. Find the general gradient vector fx,y. b. FindanequationforC 3, the level curve of f that corresponds to z 3. Draw a picture of C 3. c. If you did part b correctly, then you will observe that the point 3,6 lies on the level curve C 3. Find f3,6 and show this vector in the picture that you drew in part b. d. What is the relationship between C 3 and f3,6? Solution: The gradient vector of f at any point x,y is fx,y f x i f y j i j. The level curve C 3 is x y 3 ory x 3, which is a line (pictured below). Also, the gradient vector f3,6 i j is pictured below. It is orthogonal to the level curve C 3.
5. For the function fx,y x y (which is the same function as in problem 4): a. Let u be the unit vector u 5 i 2 5 j and compute D 5 5 u f3,6. b. Find the unit vector, u, such that f is increasing most rapidly in the direction of u at the point 3,6. Inotherwords,findusuch that D u f3,6 is as large as possible. Then compute D u f3,6. c. Find the unit vector, u, such that f is decreasing most rapidly in the direction of u at the point 3,6. Inotherwords,findusuch that D u f3,6 is as small (meaning as negative) as possible. Then compute D u f3,6. d. Find a unit vector, u, such that f is not changing in the direction of u at the point 3,6. Inotherwords,findusuch that D u f3,6 0. Solution: a. For u 5 i 2 5 j, wehave 5 5 D u f3,6 f3,6 u i j 5 5 i 2 5 5 j 5 5 2 5 5 5 5. b. D u f3,6 is as large as possible when u points in the direction of f3,6 i j. Since i j 2, then a unit vector that points in the direction of f3,6 is
2 u 2 i j 2 2 i 2 2 j. Also D u f3,6 f3,6 i j 2. c. D u f3,6 is as small (negative) as possible when u points in the opposite direction of f3,6 i j. The unit vector that points in this direction is u 2 2 i 2 2 j. Also D u f3,6 f3,6 i j 2. d. D u f3,6 0 when u is orthogonal to f3,6. The two directions in which D u f3,6 0 are thus and u 2 2 i 2 2 j u 2 2 i 2 2 j.