prelim2_spring2006solutions

prelim2_spring2006solutions - Math 192 Prelim 2 Solutions...

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Math 192 Prelim 2 Solutions Spring 2006 1. i) f ( x, y, z ) = cos xy + e yz + ln xz → ∇ f = ( - y sin xy + 1 x ) i + ( - x sin xy + ze yz ) j + ( ye yz + 1 z ) k f (1 , 0 , 1 / 2) = i + 1 2 j +2 k, u = PQ/ | PQ | = ( i +2 j +2 k ) / 3. Therefore, D u f | P = f · u = 1 / 3+1 / 3+4 / 3 = 2. ii) f increases most rapidly in the direction of f and the value of the derivative in that direction is |∇ f | = 1 + 1 / 4 + 4 = 21 / 2. 2. Solve ( a 2) 2 = 4 a 2 cos 2 θ 1 / 2 = cos 2 θ θ = π/ 6. Area = 4 π/ 6 0 2 a cos 2 θ a 2 r drdθ = 2 π/ 6 0 (4 a 2 cos 2 θ - 2 a 2 ) = 2 a 2 [2 sin 2 θ - 2 θ ] π/ 6 0 = 2 a 2 ( 3 - π/ 3). 3. Let f = xyz and g = x 2 + 2 y 2 + 3 z 2 . Then f = yzi + xzj + xyk and g = 2 xi + 4 yj + 6 zk . So, f (1 , 1 , 1) = i + j + k and g (1 , 1 , 1) = 2 i + 4 j + 6 k , which are orthogonal to the level surfaces f = 1 and g = 6 respectively. The tangent line is parallel to v = f × ∇ g = 2 i - 4 j + 2 k Tangent line is given by: x = 1 + 2 t, y = 1 - 4 t, z = 1 + 2 t . 4. 2 0 2 - y 0 (12 - y 2 ) dxdy = 2 0 (2 - y )(12 - y 2 ) dy = 20. 5. f x ( x, y ) = 4 x + 3 y, f y ( x, y ) = 3 x - 3 y 2 . So f (2 , - 1) = 3 , f x (2 , - 1) = 5 , f y (2 , - 1) = 3. Therefore, L ( x, y ) = 3 + 5( x - 2) + 3( y + 1) and L (2 . 1 , - 1 . 03) = 3 . 41. 6. 2 0 1 0 1 x 2 12 xze zy 2 dy dx dz = 2 0 1 0 y 0 12 xze zy 2 dx dy dz = 2 0 1 0 6 yze zy 2 dy dz = 2 0 [3 e zy 2 ] 1 0 dz = 3 2 0 ( e z - 1) dz = 3 e 2 - 9. 7. For ( x, y, z ) on the helix, f
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