Review Sheet Exam 2 Solutions

# Review Sheet Exam 2 Solutions - Tufts University Math 13...

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Unformatted text preview: Tufts University Math 13 Exam 2 Review Solutions Spring 2007 1. Use the chain rule to find ∂z/∂s when s = π and t = 4 , where z = e xy tan y, x = st, y = s t . Solution . ∂z ∂x = ye xy tan y ∂z ∂y = xe xy tan y + e xy sec 2 y ∂x ∂s = t ∂y ∂s = 1 t . When s = π and t = 4, x = 4 π y = π 4 ∂x ∂s = 4 ∂y ∂s = 1 4 ∂z ∂x = π 4 e π 2 ∂z ∂y = 4 πe π 2 + 2 e π 2 = (4 π + 2) e π 2 so ∂z ∂s = ∂z ∂x ∂x ∂s + ∂z ∂y ∂y ∂s = parenleftBig π 4 e π 2 parenrightBig (4) + (4 π + 2) e π 2 parenleftbigg 1 4 parenrightbigg = ( π + π + 1 2 ) e π 2 = parenleftbigg 4 π + 1 2 parenrightbigg e π 2 . 2. Let T ( x, y, z ) = x 2 + y 3 + z 4 . (a) Find the gradient vector vector ∇ T ( x, y, z ) at an arbitrary point ( x, y, z ) . Solution . vector ∇ T ( x, y, z ) = ( 2 x, 3 y 2 , 4 z 3 ) . (b) Find a unit vector in the direction of which T ( x, y, z ) increases the fastest at the point ( − 2 , 1 , 1) . Solution . We want a unit vector vectoru parallel to vector ∇ T ( − 2 , 1 , 1) = (− 4 , 3 , 4 ) , so vectoru = vector ∇ T ( − 2 , 1 , 1) bardbl vector ∇ T ( − 2 , 1 , 1) bardbl = 1 √ 16 + 9 + 16 (− 4 , 3 , 4 ) = 1 √ 41 (− 4 , 3 , 4 ) = (− 4 √ 41 , 3 √ 41 , 4 √ 41 ) . 1 (c) What is the fastest rate of increase of T ( x, y, z ) at the point ( − 2 , 1 , 1) ? Solution . bardbl vector ∇ T ( − 2 , 1 , 1) bardbl = √ 41 . 3. For the function f ( x, y ) = 3 x − x 3 − 3 xy 2 find all critical points and classify each as local minimum, local maximum, or neither. Critical Points: f x ( x, y ) = 3 − 3 x 2 − 3 y 2 = 0 ⇔ x 2 + y 2 = 1 f y ( x, y ) = − 6 xy = 0 ⇔ x = 0 or y = 0 . This yields four critical points: (0 , ± 1) and ( ± 1 , 0) . Classification: f xx ( x, y ) = − 6 x, f xy ( x, y ) = − 6 y, f yy ( x, y ) = − 6 x so the discriminant is D ( x, y ) = f xx f yy − f 2 xy = ( − 6 x )( − 6 x ) − ( − 6 y ) 2 = 36( x 2 − y 2 ) . At (0 , ± 1) , D (0 , ± 1) = 36(0 − 1) < 0, so saddle points at (0 , ± 1) . At ( ± 1 , 0) , D ( ± 1 , 0) = 36(1 − 0) > 0, so extrema: • f xx (1 , 0) = − 6 x = − 6 < 0, so local maximum at (1 , 0) ; • f xx ( − 1 , 0) = − 6 x = 6 > 0, so local minimum at ( − 1 , 0) . 4. (a) By eliminating a variable, find the point(s) closest to the origin (0 , , 0) on the cone z 2 = ( x − 1) 2 + ( y − 2) 2 . It is easier to minimize the square of the distance from the origin, and since these minima occur at the same place as the minima of the distance itself, we look for minima of f ( x, y, z ) = x 2 + y 2 + z 2 Using the fact that on the cone, z 2 = ( x − 1) 2 + ( y − 2) 2 , we can re-write f ( x, y, z ) = f ( x, y ) = x 2 + y 2 + ( x − 1) 2 + ( y − 2) 2 = 2 x 2 − 2 x + 1 + 2 y 2 − 4 y + 4 ....
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## This note was uploaded on 03/26/2008 for the course MATH 13 taught by Professor Weiss during the Fall '07 term at Tufts.

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Review Sheet Exam 2 Solutions - Tufts University Math 13...

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