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practicefinalas - SOLUTIONS TO 18.02 PRACTICE FINAL EXAM A...

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SOLUTIONS TO 18.02 PRACTICE FINAL EXAM A BJORN POONEN 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) If a and b are vectors in R 3 such that a × b = 0 , then either a = 0 or b = 0 . Solution. FALSE. Take a = b = i , for example. ± (b) If the first row of a 3 × 3 matrix is the same as the second row, then the determinant of the matrix is 0. Solution. TRUE. Adding - 1 times the first row to the second row results in a matrix with a row of all zeros, which has determinant 0 as can be seen from the Laplace expansion along that row. ± (c) If T is a unit tangent vector at P to a level curve of a differentiable function f ( x,y ), then the directional derivative of f at P in the direction T is 0. Solution. TRUE. If r ( t ) is a parametrization of the level curve at unit speed, then f ( r ( t )) is constant, so its derivative is zero, but on the other hand, the derivative is f · r 0 ( t ) = D T f ( P ). ± (d) Let P = P ( x,y ) and Q = Q ( x,y ) be continuously differentiable functions on the region R in R 2 where 1 < x 2 + y 2 < 4. If P dx + Qdy is an exact differential (i.e., the differential of some function on R ), then Q x = P y holds everywhere on R . Solution. TRUE. If P dx + Qdy = df , then P = f x and Q = f y , so Q x = f yx = f xy = P y . ± (e) Let f ( x,y ) be a function with continuous second partial derivatives on all of R 2 . Let A = f xx (0 , 0), B = f xy (0 , 0), C = f yy (0 , 0). If f ( x,y ) has a local minimum at (0 , 0), then AC - B 2 0. Solution. TRUE. Since f ( x,y ) has a local minimum at (0 , 0), the point (0 , 0) is a critical point. If AC - B 2 < 0, then it would be a saddle point, and hence not a local minimum. So AC - B 2 0. ±
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2) Find a vector w in R 2 whose scalar component in the direction of h 2 / 2 , 2 / 2 i is 2 and whose scalar component in the direction of h- 2 / 2 , 2 / 2 i is - 4. Solution. Let u = h 2 / 2 , 2 / 2 i and v = h- 2 / 2 , 2 / 2 i . Since u and v are orthogonal unit vectors, w = 2 u - 4 v = h 3 2 , - 2 i . ± 3) For what real numbers t does there exist a solution ( x,y,z ) to ( t - 6) x + 2 z = 0 - 9 x + ( t - 3) y + 5 z = 0 - 12 x + ( t + 4) z = 0 other than (0 , 0 , 0)? Solution.
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This note was uploaded on 09/14/2011 for the course MATH 18.02 taught by Professor Auroux during the Fall '08 term at MIT.

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practicefinalas - SOLUTIONS TO 18.02 PRACTICE FINAL EXAM A...

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