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# h12 - moore(jwm2685 HW12 gilbert(55485 This print-out...

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moore (jwm2685) – 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 6 x 2 y at the point (2 , 3) in the direction v = i + j . 1. f v = 1 2 2. f v = 1 3 correct 3. f v = 1 6 4. f v = 2 3 5. f v = 0 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 6 x 2 y are given by ∂f ∂x = 3 6 x 2 y , and ∂f ∂y = 1 6 x 2 y . Thus f ( x, y ) = ∂f ∂x i + ∂f ∂y j = parenleftBig 3 6 x 2 y parenrightBig i parenleftBig 1 6 x 2 y parenrightBig j , and so f (2 , 3) = 1 2 parenleftBig i 1 3 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 i 1 3 j parenrightBig · ( i + j ) . Consequently, f v = 1 2 parenleftBig 1 1 3 parenrightBig = 1 3 . 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 3 y 2 + 3 x 2 y, 6 xy + x 3 )big correct 3. f = (big 3 y 2 + 3 x 2 y, x 3 6 xy )big 4. f = (big 6 xy + x 3 , 3 y 2 + 3 x 2 y )big 5. f = (big x 3 6 xy, 3 y 2 + 3 x 2 y )big 6. f = (big 6 xy + x 3 , 3 x 2 y 3 y 2 )big Explanation: Since f ( x, y ) = (bigg ∂f ∂x , ∂f ∂y )bigg ,

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moore (jwm2685) – 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 (3 , 2). 1. f | P = 4 i 12 k 2. f | P = 4 i 12 jcorrect 3. f | P = 12 j 4 k 4. f | P = 12 j + 4 k 5. f | P = 4 i + 12 j 6. f | P = 12 i + 4 k Explanation: Since f = f x i + f y j = y 2 i + 2 xy j , and P = (3 , 2), we see that f | P = 4 i 12 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 P , II positiveat Q , III positiveat R . 1. none of them 2. III only 3. II only correct 4. I only 5. I and II only 6. II and III only 7. all of them 8. I and 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
moore (jwm2685) – HW12 – gilbert – (55485) 3 going uphill, downhill, or on the level. That will depend on which way we are walking!

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