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Hw14-5

# Hw14-5 - Math 1920 Solutions for 14.5 4 As g(x y = x2 2 y2...

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Math 1920 Solutions for 14.5 4. As g ( x, y ) = x 2 2 - y 2 2 we get g ( x, y ) = ∂g ∂x , ∂g ∂y = ( x, - y ) g ( 2 , 1) = 2 , - 1 Now g ( 2 , 1) = 1 2 , hence the level curve is given by the equation x 2 2 - y 2 2 = 1 2 x 2 - y 2 = 1 1

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8. As g ( x, y, z ) = 2 z 3 - 3( x 2 + y 2 ) z + tan - 1 xz we get f ( x, y, z ) = ∂f ∂x , ∂f ∂y , ∂f ∂z = - 6 xz + z x 2 z 2 + 1 , - 6 yz, 6 z 2 - 3( x 2 + y 2 ) + x x 2 z 2 + 1 f (1 , 1 , 1) = - 11 2 , - 6 , 1 2 12. There are two steps: finding the gradient, and then finding the direc- tional derivative. As f ( x, y ) = 2 x 2 + y 2 , we get f ( x, y ) = ∂f ∂x , ∂f ∂y = (4 x, 2 y ) f ( - 1 , 1) = ( - 4 , 2) The direction of ~u is ~u | ~u | = ( 3 5 , - 4 5 ). Thus: D ~u f ( - 1 , 1) = f ( - 1 , 1) · ~u | ~u | = - 12 5 - 8 5 = - 4 20. Recall the direction of highest increase is given by the direction of the gradient. As f ( x, y ) = x 2 y + e xy sin y , we get f ( x, y ) = ∂f ∂x , ∂f ∂y = ( 2 xy + ye xy sin y, x 2 + xe xy sin y + e xy cos y ) f (1 , 0) = (0 , 2) f (1 , 0) |∇ f (1 , 0) | = (0 , 1) Thus the direction of highest increase at P 0 (1 , 0) is (0 , 1) and of highest decrease - (0 , 1) = (0 , - 1). The directional derivatives in those directions are given by: D (0 , 1) f (1 ,
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