1873_solutions

The principal normal n t of at the number t is

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Unformatted text preview: x, 0 D 1 f 0, 0  lim x0 -4 -2 x x f 0, 0 0 0 and, similarly, D 2 f 0, 0  0. Thus, if r x, y  f x, y f 0, 0 x, y Df 0, 0 0, 0 for each point x, y then we have r x, y x, y  x2 x2y2  y2 3/2 and, since the latter expression approaches 0 as x, y 0, 0 . 0, 0 , the function f is differentiable at b. We define x2y2 2 x 2 y 2 f x, y  0 if x, y 0, 0 . if x, y  0, 0 1 0.2 x 0.1 0 1 y 0 -0.5 -1 -1 This function is not continuous at 0, 0 . c. We define x3y3 2 x 2 y 2 f x, y  0 414 if x, y 0, 0 . if x, y  0, 0 5 0 -5 -2 y0 2 4 -4 x 2 4 This function is continuous at 0, 0 . Now we have f x, 0 D 1 f 0, 0  lim x0 x f 0, 0 0 0 and, similarly, D 2 f 0, 0  0. Thus, if r x, y  f x, y f 0, 0 x, y Df 0, 0 0, 0 for each point x, y then we have r x, y x, y  x3y3 x2  y2 5/2 and, since the latter expression approaches 0 as x, y 0, 0 . 0, 0 , the function f is differentiable at d. We define 1 x 2 y 2 exp f x, y  if if 0 x, y 0, 0 x, y  0, 0 . 0.6 0.5 0.4 0.3 0.2 0.1 0 1 0.5 0 x -11 This function is continuous at 0, 0 . Now we have f x, 0 D 1 f 0, 0  lim x0 0.5 x 0 y f 0, 0 0 0 and, similarly, D 2 f 0, 0  0. Thus, if r x, y  f x, y f 0, 0 Df 0, 0 x, y 0, 0 for each point x, y then we have r x, y x, y  exp x2  y2 and, since the latter expression approaches 0 as x, y 415 1 x 2 y 2 0, 0 , the function f is differentiable at 0, 0 . e. We define 1 x 2 y 2 x 2 y 2 sin f x, y  if if 0 x, y 0, 0 x, y  0, 0 . 12 10 8 6 4 2 0 -5 0 y 5 2 4 This function is continuous at 0, 0 . Now we have f x, 0 D 1 f 0, 0  lim x0 -4 x x f 0, 0 0 0 and, similarly, D 2 f 0, 0  0. Thus, if r x, y  f x, y f 0, 0 x, y Df 0, 0 0, 0 for each point x, y then we have r x, y x, y  x 2 y 2 sin 1 x 2 y 2 x2  y2 and, since the latter expression approaches 0 as x, y 0, 0 . 2. Prove that if A is an m f x  A. We see easily that n matrix and if f x  Ax for every x 0, 0 , the function f is differentiable at R n then for every such point x we have D j f i x  a ij at each point x and for all i and j. 3. Suppose that U is a neighborhood of a point a in R n and that f : U R m . Suppose that A is a k m matrix and that g x  Af x for every x U. Prove that if f is differentiable at the point a then so is g and we have g a  Af a . (Your proof should be very short.) This assertion follows at once from the chain rule. 4. State true or false: If the partial derivatives of a function f are all zero at every point in an open connected set U then f must be constant in U. This statement is true. If the partial derivatives are all zero then they are all continuous. Thus the function f is differentiable with a zero derivative at every point of U. 5. Given an example of a nonconstant function f on an open set U We take n  1 and define fx  R n such that f x  O for every x U. 1 if x  0 0 if x  0. 6. State true of false: if U is a neighborhood of a point a in R n and if the directional derivative of a function f 416 exists at a in every direction then f is differentiable at a. This statement is false. The function does not even have to be continuous at a. 7. Prove that if f and g are both differentiable at a point a in R n then so is the function f  g and we have fg a  f a g a . This assertion is very easy. 417...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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