{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

practicefinalas

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

This preview shows pages 1–3. Sign up to view the full content.

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 ( 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 + Q dy 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 + Q dy = 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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2) Find a vector w in R 2 whose scalar component in the direction of 2 / 2 , 2 / 2 is 2 and whose scalar component in the direction of - 2 / 2 , 2 / 2 is - 4. Solution. Let u = 2 / 2 , 2 / 2 and v = - 2 / 2 , 2 / 2 . Since u and v are orthogonal unit vectors, w = 2 u - 4 v = 3 2 , - 2 . 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)?
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online