solution-A5

solution-A5 - MATHEMATICS 222 CALCULUS III ASSIGNMENT 5...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATHEMATICS 222 CALCULUS III ASSIGNMENT 5 ANSWERS 1. The function f is defined by 2 xy x 2 + y 2 ) for ( x,y ) 6 = (0 , 0) and f (0 , 0) = 0 (a) f x (0 , 0) = lim h 2 h (0) h 2 + 0 2- ! /h = 0 Similarly f y (0 , 0) = 0 For any other point, simply apply the quotient rule to get f x ( x,y ) = 2 y ( y 2- x 2 ) ( x 2 + y 2 ) 2 and f y ( x,y ) = 2 x ( x 2- y 2 ) ( x 2 + y 2 ) 2 (b) In order to find the second partial derivatives at ( x,y ) 6 = (0 , 0) apply the quotient rule again to obtain f xx = 4 xy ( x 2- 3 y 2 ( x 2 + y 2 ) 3 f yy = 4 xy ( y 2- 3 x 2 ) ( x 2 + y 2 ) 3 f xy = 2(6 x 2 y 2- x 4- y 4 ) ( x 2 + y 2 ) 3 Substituting in to (recall the correction y 2 f yy ) x 2 f xx + 2 xyf xy + y 2 f yy = 4 xy ( x 2 + y 2 ) 3 h x 4- 3 x 2 y 2 + y 4- 3 x 2 y 2 + 6 x 2 y 2- x 4- y 4 i = 0 2. (a) if V = x 3- 3 xy 2 then V xx = 6 x and V yy =- 6 x which add up to zero as required....
View Full Document

This note was uploaded on 05/01/2008 for the course MATH 222 taught by Professor Karlpeterrussell during the Winter '08 term at McGill.

Page1 / 2

solution-A5 - MATHEMATICS 222 CALCULUS III ASSIGNMENT 5...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online