2020_FALL_MATH2011 Homework-3.pdf - Man Kit CHAN WeBWork...

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Man Kit CHAN MATH 2011 (Fall 2014-2015) WeBWork Homework-3 due 11/03/2020 at 05:00pm HKT You may need to give 4 or 5 significant digits for some (floating point) numerical answers in order to have them accepted by the computer. 1. (2 points) Consider the function f ( x , y ) = ( 2 xy ( x 2 + y 2 ) 2 , ( x , y ) 6 = ( 0 , 0 ) 0 , ( x , y ) = ( 0 , 0 ) (a) Use a computer to draw a contour diagram for f . Which of the following is the contour diagram? ? figure 1 figure 2 figure 3 figure 4 1. 2. Is f differentiable at ( 0 , 0 ) ? ? yes no it is not possible to tell (e) Find the partial derivative f y at ( 0 , 0 ) by calculating it di- rectly with a limit: f y = lim h 0 1 h ( f ( , ) - f ( , )) = Do the partial derivatives f x and f y exist and are they contin- uous at ( 0 , 0 ) ? (Hint: to test continuity, you may want to use a similar calculation as you used to test the continuity of f) ? yes no it is not possible to tell they exist but are not continuous (Be sure that you can justify all of your answers in this prob- lem.) SOLUTION
(b) Is f differentiable at all points ( x , y ) 6 = ( 0 , 0 ) ?
(c) Calculate the partial derivatives of f for ( x , y ) 6 = ( 0 , 0 ) : f x = f y = Do the partial derivatives f x and f y exist and are they continuous at all points ( x , y ) 6 = ( 0 , 0 ) ? ? The function f is differentiable at all points ( x , y ) 6 = ( 0 , 0 ) , as it is a rational fraction with denominator ( x 2 + y 2 ) 2 = 0 only when ( x , y ) = ( 0 , 0 ) . (c) The partial derivatives of f at points ( x , y ) 6 = ( 0 , 0 ) are given by f x ( x , y ) = 2 y ( x 2 + y 2 ) 2 - 8 x 2 y ( x 2 + y 2 ) (d) A first test for whether f is differentiable at ( 0 , 0 ) is to see if it is continuous there. Calculate each of the following limits x 2 + y 2 ) 4 . (b) The function f is differentiable at all points ( x , y ) 6 = ( 0 , 0 ) , as it is a rational fraction with denominator ( x 2 + y 2 ) 2 = 0 only when ( x , y ) = ( 0 , 0 ) . (c) The partial derivatives of f at points ( x , y ) 6 = ( 0 , 0 ) are given by f x ( x , y ) = 2 y ( x 2 + y 2 ) 2 - 8 x 2 y ( x 2 + y 2 ) ( x 2 + y 2 ) 4 .
to determine if f is continuous at ( 0 , 0 ) : lim h 0 0 f ( h , 0 ) = (d) The function f is not continuous at ( 0 , 0 ) . To see this, we can calculate the indicated limits, which should all have the same value if the function is continuous:
lim h 0 f ( h , h ) = (In each case, enter DNE if the limit does not exist.) lim h 0

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