Math 550, Exam 1. 2/10/11.

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1 Math 55, Exam. //. Read problems carefully. Show all work. No notes, calculator, or text. The exam is approximately 5 percent of the total grade. There are points total. Partial credit may be given. Write your full name in the upper right corner of page. Number the pages in the upper right corner. o problem on page, problem on page, etc. Circle or otherwise clearly identify your final answer. Name:. (5 points: Suppose that f(x, y, z x y y z, and that P (,,. (a (7 points: Find the direction of fastest decrease of f at P. Solution: We compute the gradient: f(x, y, z 6xy i + (x y z j y z k; At P, we have (,, 6 i 9 j 4 k. Hence, the direction of fastest decrease is 6 i + 9 j + 4 k. (b (8 points: Find the rate of change of f at the point P in the direction of v i + j + k. v v ( i + j + k. We Solution: We normalize v to obtain the unit vector u have u f(,, f(,, u (6 i 9 j 4 ( k i + j + k 7.. (5 points: Use the curl to explain why F e xy (+xy i+(x e xy +sin(yz j+sin(yz k cannot be a gradient vector field f for any scalar function f(x, y, z on R. Solution: Suppose that F is gradient field. Then we must have curl F. We compute i j k ( curl F / x / / (sin(yz e xy ( + xy x e xy + sin(yz sin(yz i (x e xy + sin(yz ( ( (sin(yz j (exy ( + xy + x (x e xy + sin(yz k (exy ( + xy x i (z cos(yz y cos(yz j( + k(x ye xy + xe xy e xy x ( + xyxe xy (z y cos(yz i. This is contradicts the hypothesis that F is a gradient field.

2 . (5 points: Let F be a vector field given by F(x, y, z (x + y + z i + (y + j + ln(xyz k. Find vector fields G and H with curl G and div H such that F G + H. Solution: We note the following: Suppose that G(x, y, z G (x i + G (y j + G (z k. Then we have curl G i j k / x / / G (x G (y G (z. Suppose that F(x, y, z F (y, z i + F (x, z j + F (x, y k. Then we have div F F x + F + F. Using these facts, we see that G(x, y, z x i + y j + (ln z k and F(x, y, z (y + z i + j + (ln x + ln y k satisfy the conclusion of the problem. 4. ( points: Suppose that F(x, y, z F (x, y, z i + F (x, y, z j + F (x, y, z k is a vector field on R with all partial derivatives continuous. Show that ( F. Solution: We have i j k (( / x / / F F F F F ( F x F since mixed partial derivatives agree: i ( F x F ( F x F ( F j + x F k + ( F x F F x F x F x + F + F x F F x F x, F x F x, F F.

3 5. (5 points: Suppose that [, ] [, ] is the unit square in the u-v plane, and that is the parallelogram with vertices (,, (-,, (,, and (, in the x-y plane. (a (7 points: Find a one - one and onto linear transformation T :. How do you know that T is one - one and onto? (You can be very brief. (b (8 points: What is the Jacobian of T? What is the area of? Solution: One such linear map T sends: (, (,, (, (,, (, (,, (, (,. ( a b We must find A such that T (u, v A(u, v. We have c d A(, (a, c (, ; A(, (b, d (,. ( It follows that A. We verify that A(, (,. The map T is one-to-one and onto since det(a. The Jacobian of T is also ; hence, has area. To see this, observe that A( dx dy du dv A(. 6. ( points: Use a suitable change of variables to evaluate (x + ye x y dx dy, where is the parallelogram bounded by x y, x y, x + y, and x + y. Solution: Let u x y, and let v x+y. The map T given by T (x, y (x y, x+y (u, v maps in the x-y ( plane to the parellelogram [, ] [, ] in the u-v plane. Moreover, we let A and observe, for all (x, y, that T (x, y A(x, y. Hence, the Jacobian of T is J(T det(a. For all (u, v, we have T (u, v A (u, v. Therefore, the Jacobian of T is J(T det(a det(a. We now have (x + ye x y dx dy ve u du dv ve u du dv ( ( ( v dv e u du ] ( v ] e u (e.

4 Extra problems. Attempt only if you have completed the problems above.. By using the transformation x + y u and y uv, show that x e y/(x+y dy dx e. Solution: We first note that T (u, v (u uv, uv and T (x, y ( x + y, y, defined x+y for x, y with < x+y. In particular, T is well-defined at all points (x, y (, in the right triangle with vertices at (,, (,, and (, in the xy-plane. To determine, we study how T maps the boundary of. Here is what we have: (a {(x, y : x, < y }: vertical leg; maps to {(u, v : v, < u }. (b {(x, y : y, < x }: horizontal leg; maps to {(u, v : v, < u }. (c {(x, y : x + y, < x, y }: hypotenuse; maps to {(u, v : u, < v }. (d While T is not well-defined at (,, we observe that the set {(u, v : u, v } in the uv-plane maps by T to the point (,. It follows that T : [, ] [, ] is onto and essentially ; it fails to be on the the line segment {(u, v : u, v }, which maps to the single point (,. We compute v u v u u( v + uv u u. Hence, we compute e y/(x+y dy dx ( e v u du dv. ( e v dv u du e.. Consider the region in the first quadrant bounded by the hyperbolas x y, x y 9, xy, and xy 4. Compute the integral (x +y dx dy by making a suitable change of variables. Solution: We make the change of variable It follows that x y We observe that hence, we have u x y 9, v xy 4. y x (x + y. (x + y (x y + 4(xy u + 4v ; x + y u + 4v. 4

5 The chain rule implies that (u,v (x,y (x + y u + 4v. We compute (x + y dx dy 4 9 u + 4v u + 4v du dv 4 9 du dv 8 8. Note: The solution to this problem used the chain rule for Jacobians, which we now state and prove. Fact: Suppose that T (u, v (x(u, v, y(u, v and that S(r, s (u(r, s, v(r, s. Then we have (r, s (r, s. Proof. We compute, using the ordinary chain rule for partial derivatives: x x (r, s x u r s + x v x u + x v x x u r v r u s v s u + v u + v u v r s u r v r u s v s u v (r, s. u r v r u s v s As a consequence, we have the following important fact. Fact: Suppose that (x,y. Then we have (u,v, and hence, we have. Proof. Let S T in the previous fact. Then we have T (S(r, s (r, s (x, y. It follows that. 5