20e-lecture4

20e-lecture4 - Lecture 4 Monday April 6th [email protected]..

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Unformatted text preview: Lecture 4 : Monday April 6th [email protected] Key concepts : Tangent hyperplane, Gradient, Directional derivative, Level curve Know how to find equation of tangent hyperplane, gradient, directional derivatives, direction of fastest increase. 4.1 Differentiability Recall that f : R n R is differentiable at a point a R n if all partial derivatives f j ( a ) exist and lim x a f ( x )- f ( a )- n j =1 f j ( a )( x j- a j ) d ( x,a ) = 0 . It is not enough that all the partial derivatives f j ( a ) exist for f to be differentiable at a . Remember here that x a means ( x 1 ,x 2 ,...,x n ) ( a 1 ,a 2 ,...,a n ). Lets do some examples. In all the examples p = ( x,y ). Example 1. The function f ( x,y ) = xy is differentiable at a = (0 , 0), since f x ( a ) = 0 = f y ( a ) as we checked in a previous example, and lim p a xy p x 2 + y 2 = 0 . Lets use the - definition of limits to see that this limit is zero. Let g ( x,y ) be the function in the limit. For every > 0 we must find a > 0 such that d ( p,a ) < ensures | g ( x,y ) | < . We claim that | xy | x 2 + y 2 . To see this, note that it is equivalent to | x | 2 + | y | 2- | xy | 0. But this is true since ( | x | - | y | ) 2 = | x | 2 + | y | 2- 2 | xy | | x | 2 + | y | 2- | xy | . Now we use this to show that f is differentiable: | g ( x,y ) | < fl fl fl xy p x 2 + y 2 fl fl fl < x 2 + y 2 p x 2 + y 2 < p x 2 + y 2 < d ( p,a ) < . 1 So putting = in the definition of the limit we get lim g ( x,y ) = 0 as required for f to be differentiable at (0 , 0). Theres an easier way to check the limit is zero, using squeezing: note that | xy | p x 2 + y 2 | xy | p y 2 = | x | so the function in the limit is between 0 and | x | . Since both of these have a limit of zero as ( x,y ) (0 , 0), so must xy/ p x 2 + y 2 . Example 2. This example is almost a follow up to the last one, where we now put f ( x,y ) = p | xy | . This function is not differentiable at a = (0 , 0), since f x (0 , 0) = f y (0 , 0) = 0 as we saw last lecture, but lim ( x,y ) a p | xy | p x 2 + y 2 does not exist. To see that the limit fails, along the linedoes not exist....
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This note was uploaded on 02/12/2010 for the course MATH 20E taught by Professor Enright during the Spring '07 term at UCSD.

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20e-lecture4 - Lecture 4 Monday April 6th [email protected]..

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