hw3sol - Math 241 Solution 3 2.1#44. Since y + 2 here is...

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Unformatted text preview: Math 241 Solution 3 2.1#44. Since y + 2 here is just shifting the graph to the negative y-axis by 2, we can just think of z = 4 x 2 + y 2 . Its a paraboloid having an ellipse, 4 x 2 + y 2 = c , as its section curve at height c . So the original graph is the paraboloid with the same shape shifted to the negative y-axis by 2. 2.2#28. lim ( x,y ) (0 , 0) x 2 y x 2 + y 2 = lim r ( r cos ) 2 ( r sin ) r 2 = lim r r 3 cos 2 sin r 2 = lim r r cos 2 sin = 0 . 2.3#15. f ( x,y ) = ( f x ( x,y ) ,f y ( x,y )) = ( x 2 + y 2 + 1)- ( x- y )2 x ( x 2 + y 2 + 1) 2 ,- ( x 2 + y 2 + 1)- ( x- y )2 y ( x 2 + y 2 + 1) 2 . f (2 ,- 1) = 6- 12 6 2 ,- 6 + 6 6 2- (- 1 / 6 , 0) . 2.3#30. Define f ( x,y,z ) = 4 cos xy- z f ( x,y,z ) = (- 4 y sin xy,- 4 x sin xy,- 1) . So f ( / 3 , 1 , 2) = (- 2 3 ,- 2 3 / 3 ,- 1), and hence the equation of the tangent plane is- 2 3( x- / 3)- 2 3 3 ( y- 1)- ( z- 2) = 0 ....
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This note was uploaded on 09/02/2010 for the course MATH 241 taught by Professor Any during the Spring '08 term at University of Illinois at Urbana–Champaign.

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hw3sol - Math 241 Solution 3 2.1#44. Since y + 2 here is...

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