Exam2_practice_solutions

Exam2_practice_solutions - MATH 2401 October 19, 2009...

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Unformatted text preview: MATH 2401 October 19, 2009 Sections F4 & F5 PRACTICE EXAM 2 Name: gtID # : Note: There are ve problems in this exam, each worth 20 points. Write out your solutions neatly and explain your work. Calculators are not allowed. Basic formulas: f ( x ) = f x ( x ) i + f y ( x ) j + f z ( x ) k f u ( x ) = f ( x ) u d dt [ f ( r ( t ))] = f ( r ( t )) r ( t ) df = f ( x ) h df f = f ( x + h )- f ( x ) 1 Problem 1 (20 points). Let f ( x,y ) = ( x + y )sin( y- x ) + x 3 . (a) Find the directional derivative of f ( x,y ) at (1 , 1) in the direction toward the point (3 ,- 1) . Solution: We have f x (1 , 1) = sin( y- x )- ( x + y )cos( y- x ) + 3 x 2 (1 , 1) =- 2 + 3 = 1 , and f y (1 , 1) = sin( y- x ) + ( x + y )cos( y- x ) (1 , 1) = 2 . Thus f (1 , 1) = i + 2 j . The vector v = (3 i- j )- ( i + j ) = 2 i- 2 j gives the indicated direction, so the directional derivative is f v (1 , 1) = f (1 , 1) v k v k = ( i + 2 j ) i- j 2 =- 1 2 . (b) What is the rate of change of f ( x,y ) with respect to t along the curve r ( t ) = ( t- 1) 2 i + (1 + t 3 ) j , at the point (1 , 1) ? Solution: Note that the curve is at the point (1 , 1) for time t = 0 [not t = 2 ]. We have r ( t ) = 2( t- 1) i + 3 t 2 j , so the tangent of the curve at (1 , 1) is r (0) =- 2 i . We have already computed the gradient f ( r (0)) = f (1 , 1) = i + 2 j in part (a), so we conclude that the rate of change is d dt f (1 , 1) = f ( r (0)) r (0) =- 2...
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Exam2_practice_solutions - MATH 2401 October 19, 2009...

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