Math 127C, Spring 2006 Final Exam Solutions. x 2 ), g(y 1, y 2 ) = ( y 1 y 2, y1 2 + y2) 2. (g f) (0) = g (f(0))f (0).

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1 Math 27C, Spring 26 Final Exam Solutions. Define f : R 2 R 2 and g : R 2 R 2 by f(x, x 2 (sin x 2 x, e x x 2, g(y, y 2 ( y y 2, y 2 + y2 2. Use the chain rule to compute the matrix of (g f (,. By the chain rule, (g f ( g (f(f (. The Jacobian matrices of f and g are ( ( cos [f x2 (x, x 2 ] e x, [g y2 y (y, y 2 ] 2y 2y 2, and f(, (,. Thus, [f (, ] ( (, [g (, ] 2, and [(g f (, ] ( ( 2 (. 2 2

2 2. Define f : R 2 R 2 by f(x, y ( x 2 y + x, 6x + y 2. What does the inverse function theorem imply about the existence of a local inverse of f at: (a (x, y (, ; (b (x, y (,? The partial derivatives of the components of f exist and are continuous in R 2, so f is continuously differentiable in R 2, and the inverse funtion theorem applies at points where the derivative of f is invertible. The Jacobian matrix of f is ( 2xy + x [f 2 (x, y] 6 2y. (a At (, we have [f (, ] ( The determinant of this matrix is zero, so f (, is singular and there is no conclusion from the inverse function theorem. (b At (, we have [f (, ] ( 6 2 The determinant of this matrix is non-zero, so f (, is nonsingular. The inverse function theorem implies that there are open sets U, V R 2 with (, U and (, 5 V such that f : U V is one-to-one and onto. Moreover, the local inverse f : V U is continuously differentiable.. 2

3 3. Let [, ] [, ], and define f : R by f(x, y y sin (πxy. Why is f Riemann integrable over? Evaluate f dxdy, and justify your calculations. The function is continuous on, since it is the product and composition of continuous functions. Therefore it is Riemann integrable on. Since f is continuous, the Riemann integral on is equal to the iterated integrals by Fubini s theorem. Performing the x-integral first followed by the y-integral, and using the fundamental theorem of calculus, we get ( f dxdy y sin(πxy dx dy [ π ] x cos(πxy dy x [ cos(πy] dy π [y π ] y sin(πy π y π. 3

4 4. Let ω fdx + dy, λ gdx + dz be one-forms in R 3, where f(x, y, z and g(x, y, z are smooth functions. (a Calculate ω λ and express your answer in standard form. (b Calculate d(ω λ and express your answer in standard form. (a Using the anti-symmetry of the wedge product, we compute that ω λ (fdx + dy (gdx + dz gdy dx + fdx dz + dy dz gdx dy + fdx dz + dy dz. (b From (a, we compute that d(ω λ d( gdx dy + fdx dz + dy dz g z dz dx dy + f y dy dx dz (f y + g z dx dy dz. 5. Define f : R 2 R by f(x, y { x 2 sin(/y if y if y Determine, with proof, at what points of R 2 the function f is differentiable. By the chain, product, and quotient rules, the partial derivatives of f exist and are continuous in the open set E {(x, y : y }, so f is continuously differentiable in E. f x then lim y f(x, y does not exist, so f is not continuous at (x,, and therefore f is not differentiable. 4

5 We claim that f is differentiable at (, with f (,. To prove this, note that for h (h, k R 2 with k, we have f( + h, + k f(, h2 sin k h 2 h 2, while if k, we have f( + h, + k f(,. t follows that f( + h f( lim h h which is what we had to prove., Thus, f is differentiable at all points (x, y R 2 except those with y and x. 6. Let [, ] [, ] and define f : R by { if x y, f(x, y if x y. Prove that f is Riemann integrable on, and evaluate f dxdy. For N N, consider a partition P N of into N 2 equal closed rectangles obtained by partitioning each side [, ] into N equal intervals. The infimum of f on every rectangle is, so the lower Riemann sum of f associated with the partition P N is L (f; P N. The supremum of f on a rectangle in the partition is equal to except for those rectangles that intersect the diagonal x y, where the supremum of f is equal to. There are 3N 2 such rectangles (N on the diagonal and 2(N adjacent to the diagonal. The area of each rectangle is /N 2, so it follows that the upper Riemann sum of f associated with P is given by U (f; P N (3N 2 5 N 2

6 The right-hand side of this equation tends to as N. t then follows from standard properties of Riemann sums that sup L (f; P N f dxdy f dxdy inf U (f; P N. N N N N Hence f dxdy f dxdy, so f is Riemann integrable on, and f dxdy. 7. Let N {, 2, 3,...} denote the natural numbers, and define by d, d 2 : N N R d (n, m n m, d 2(n, m n m. (a Prove that d, d 2 are metrics on N. (b Determine whether or not N is complete with respect each of the metrics d, d 2. (a Both d, d 2 are symmetric, non-negative, and equal to zero if and only if n m. For d, we have d (n, m n m n p + p m n p + p m d (n, p + d 2 (p, m. The triangle inequality for d 2 is immediate. 6

7 (b The metric space (N, d is not complete. For example, consider the sequence {x k } with x k k. This is a Cauchy sequence with respect to d : Given ɛ >, choose N 2/ɛ. Then j, k > N implies that d (x j, x k j k j + k < 2 N < ɛ. f x N is any integer, choose ɛ /(2x >. Then for all k > 2x we have d(x k, x x k > 2x ɛ, t follows that {x k } does not converge with respect to d to any x N. (N, d 2 is complete. Any Cauchy sequence {x k } is constant from some point on, since d 2 (x j, x k < /2 implies that x j x k. Hence, every Cauchy sequence converges. 7

8 8. Define a two-form in R 3 by ω ( x 2 + y 2 + z 2 dx dy. Let { (u, v R 2 : u π/2, v 2π } and define the two-cell φ : R 3 (a half-sphere by φ(u, v (x, y, z where x sin u cos v, y sin u sin v, z cos u. Evaluate φ ω. From the definition of the integral of a differential form, we have ( ω x 2 + y 2 + z 2 (x, y (u, v dudv. On the surface, we have φ x 2 + y 2 + z 2 sin 2 u cos 2 v + sin 2 u sin 2 v + cos 2 u. The Jacobian is (x, y (u, v x u x v y u y v cos u cos v cos u sin v cos u sin u. sin u sin v sin u cos v t follows that ω cos u sin u dudv. φ 8

9 Evaluating this integral by use of Fubini s theorem, we get φ ω 2π 2π π. π/2 ( 2π π/2 [ 2 sin2 u cos u sin u dv du cos u sin u du ] π/2 9

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