This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: SOLUTIONS TO THE 18.02 FINAL EXAM BJORN POONEN December 14, 2010, 9:00am12:00 (3 hours) 1) For each of (a)(e) below: If the statement is true, write TRUE. If the statement is false, write FALSE. (Please do not use the abbreviations T and F.) No explanations are required in this problem. (a) (5 pts.) If f ( x, y ) is a continuously differentiable function on R 2 , and 2 f x 2 = 0 at every point of R 2 , then there exist constants a and b such that f ( x, y ) = ax + b for all x and y . Solution: FALSE. The values of a and b could depend on y . The function f ( x, y ) = y is a counterexample. (b) (5 pts.) The annulus R in R 2 defined by 9 x 2 + y 2 16 is simply connected. Solution: FALSE. The circle of radius 3 . 5 centered at the origin gives a closed curve in R that cannot be shrunk to a point within R . (c) (5 pts.) If f ( x, y ) is a function whose second derivatives exist and are continuous everywhere on R 2 , and f (0 , 0) = f x (0 , 0) = f y (0 , 0) = f xx (0 , 0) = f yy (0 , 0) = 0 and f xy (0 , 0) 6 = 0, then f has a saddle point at (0 , 0). Solution: TRUE. This follows from the second derivative test. In the notation of that test, we are given A = 0, B 6 = 0, and C = 0, so AC B 2 < 0, so f has a saddle point at (0 , 0). (d) (5 pts.) If A is a 3 3 matrix, and b is a column vector in R 3 , and the square system A x = b has more than one solution x , then det A = 0. Solution: TRUE. If det A 6 = 0, then there would have been one solution, namely x = A 1 b . (e) (5 pts.) If F is a continuously differentiable 3D vector field on R 3 such that curl F = everywhere, and A and B are two points in R 3 , then the value of R C F d r is the same for every piecewise smooth path C from A to B . Solution: TRUE. The region R 3 is simply connected, so all six conditions for conserva tiveness are equivalent, and these are two of them. 1 2) (a) (5 pts.) For which pairs of real numbers ( a, b ) does the matrix A = a b 0 0 1 1 0 0 have an inverse? Solution: We have det A = 0 + b + 0 0 = b, so A is invertible if and only if b 6 = 0. (There is no condition on a .) (b) (10 pts.) For such pairs ( a, b ), compute A 1 . (Its entries may depend on a and b . Suggestion: Once you have the answer, check it!) Solution: Compute nine 2 2 determinants to get the matrix of minors:  1 b b a Apply the checkerboard signs to get the matrix of cofactors: 1 b b a Take the transpose to get the adjoint matrix: 0 0 b 1 0 a b Divide by det A to get the inverse matrix: 1 1 /b a/b 1 ....
View Full
Document
 Fall '08
 Auroux
 Multivariable Calculus

Click to edit the document details