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HW12-solutions

# HW12-solutions - zakaria(mmz255 HW12 gilbert(55485 This...

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zakaria (mmz255) – HW12 – gilbert – (55485) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points Find the directional derivative, f v , of f ( x, y ) = radicalbig 2 x + 3 y at the point (6 , 2) in the direction v = i j . 1. f v = 7 12 2. f v = 1 12 3. f v = 1 4 4. f v = 5 12 5. f v = 1 12 correct Explanation: For an arbitrary vector v , f v = f · parenleftbigg v | v | parenrightbigg , where we have normalized the direction vector so that it has unit length. Now the partial derivatives of f ( x, y ) = radicalbig 2 x + 3 y are given by ∂f ∂x = 1 2 x + 3 y , and ∂f ∂y = 3 2 2 x + 3 y . Thus f ( x, y ) = ∂f ∂x i + ∂f ∂y j = parenleftBig 1 2 x + 3 y parenrightBig i + parenleftBig 3 2 2 x + 3 y parenrightBig j , and so f (6 , 2) = 1 2 parenleftBig 1 3 i + 1 2 j parenrightBig . On the other hand, v = i j = v | v | = 1 2 ( i j ) . But then f · parenleftbigg v | v | parenrightbigg = 1 2 parenleftBig 1 3 i + 1 2 j parenrightBig · ( i j ) . Consequently, f v = 1 2 parenleftBig 1 3 1 2 parenrightBig = 1 12 . keywords: 002 10.0points Find the gradient of f ( x, y ) = 3 xy 2 + x 3 y . 1. f = (big 3 x 2 y 3 y 2 , 6 xy + x 3 )big 2. f = (big 6 xy + x 3 , 3 x 2 y 3 y 2 )big 3. f = (big x 3 6 xy, 3 y 2 + 3 x 2 y )big 4. f = (big 3 y 2 + 3 x 2 y, x 3 6 xy )big 5. f = (big 6 xy + x 3 , 3 y 2 + 3 x 2 y )big 6. f = (big 3 y 2 + 3 x 2 y, 6 xy + x 3 )big correct Explanation: Since f ( x, y ) = (bigg ∂f ∂x , ∂f ∂y )bigg ,

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zakaria (mmz255) – HW12 – gilbert – (55485) 2 we see that f = (big 3 y 2 + 3 x 2 y, 6 xy + x 3 )big . keywords: 003 10.0points Determine the gradient of f ( x, y ) = xy 2 at the point P (1 , 2). 1. f | P = 4 i 4 jcorrect 2. f | P = 4 i 4 k 3. f | P = 4 i + 4 j 4. f | P = 4 j + 4 k 5. f | P = 4 j 4 k 6. f | P = 4 i + 4 k Explanation: Since f = f x i + f y j = y 2 i + 2 xy j , and P = (1 , 2), we see that f | P = 4 i 4 j . 004 10.0points The contour map given below for a function f shows also a path r ( t ) traversed counter- clockwise as indicated. 0 1 2 3 -3 -2 -1 0 Q P R Which of the following properties does the derivative d dt f ( r ( t )) have? I positiveat R , II positiveat Q , III positiveat P . 1. none of them 2. all of them 3. I only 4. II and III only 5. I and III only 6. II only correct 7. I and II only 8. III only Explanation: By the multi-variable Chain Rule, d dt f ( r ( t )) = ( f )( r ( t )) · r ( t ) . Thus the sign of d dt f ( r ( t )) will be the sign of the slope of the surface in the direction of the tangent to the curve r ( t ), and we have to know which way the curve is being traversed to know the direction the tangent points. In other words, if we think of the curve r ( t ) as defining a path on the graph of f , then we need to know the slope of the path as we travel around that path - are we
zakaria (mmz255) – HW12 – gilbert – (55485) 3 going uphill, downhill, or on the level. That will depend on which way we are walking!

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