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

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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 . It’s 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 0 ( r cos θ ) 2 ( r sin θ ) r 2 = lim r 0 r 3 cos 2 θ sin θ r 2 = lim r 0 r cos 2 θ sin θ = 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 ( π/ 3 , 1 , 2) = ( - 2 3 , - 2 π 3 / 3 , - 1), and hence the equation of the tangent plane is - 2 3( x - π/ 3) - 2 π 3 3 ( y - 1) - ( z - 2) = 0 . 2.4#14. f x = 2 xe x 2 + y 2 , f y = 2 ye x 2 + y 2 , so f xy = 2 e x 2 + y 2 + 4 x 2 e x 2 + y 2 , f yy = 2 e x 2 + y 2 + 4 y 2 e x 2 + y 2 , f xy = f yx = 4 xye x 2 + y 2 . 2.5#2. (a) f ( s, t ) = sin(( s + t )( s 2 + t 2 )), so f s = (3 s 2 + t 2 + 2 st ) cos(( s + t )( s 2 + t 2 )) , f t = ( s 2 + 3 t 2 + 2 st ) cos(( s + t )( s 2 + t 2 )) .

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