21-356 Lecture 3

Example 5 let f x y 1 0 if y x2 x 0 otherwise given

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Unformatted text preview: g example shows. Example 5 Let f (x, y ) := ! 1 0 if y = x2 , x %= 0, otherwise. Given a direction v = (v1 , v2 ), the line L through 0 in the direction v intersects the parabola y = x2 only in 0 and in at most one point. Hence, if t is very small, f (0 + tv1 , 0 + tv2 ) = 0. It follows that !f f (0 + tv1 , 0 + tv2 ) " f (0, 0) 0"0 (0, 0) = lim = lim = 0. t!0 t!0 !v t t " # However, f is not continuous in 0, since f x, x2 = 1 & 1 as x & 0, while f (x, 0) = 0 & 0 as x & 0. Exercise 6 Let f (x, y ) := $ x2 y x4 +y 2 if (x, y ) %= (0, 0) , if (x, y ) = (0, 0) . 0 Find all directional derivatives of f at 0 and prove that f is not continuous at 0. 5 Friday, January 14, 2011 The previous examples show that in dimension N ! 2 partial derivatives do not give the same kind of results as in the case N = 1. To solve this problem, we introduce a stronger notion of derivative, namely, the notion of di!erentiability. We recall that a function T : RN " R is linear if T (x + y) = T (x) + T (y)...
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This document was uploaded on 03/31/2014.

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