practice2as

# practice2as - SOLUTIONS TO 18.02 PRACTICE MIDTERM #2A BJORN...

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SOLUTIONS TO 18.02 PRACTICE MIDTERM #2A BJORN POONEN 1) A particle is moving in the plane, and w = w ( x, y ) is a function of the position of the particle. At a particular instant, x = 2, y = 3, dw dt = 6, and dx dt = 8. Given that w (computed with respect to x and y ) at the point (2 , 3) equals h 5 , 7 i , ﬁnd dy dt at that same instant. Solution: The chain rule says dw dt = ∂w ∂x dx dt + ∂w ∂y dy dt . At the given instant, this becomes 6 = 5 · 8 + 7 dy dt , so dy dt = - 34 / 7. 2) Let P be the point on the curve x 2 - y 3 = 3 . 1 such that P is closest to (2 , 1). Use a gradient to estimate the coordinates of P . Solution: At (2 , 1), the function f ( x, y ) := x 2 - y 3 takes the value 3. To reach the nearest point where f ( x, y ) = 3 . 1, we move in the direction of f = h 2 x, - 3 y 2 i = h 4 , - 3 i at ( x, y ) = (2 , 1). Then |∇ f | = 5 is the rate of change of f in the direction u = h 4 / 5 , - 3 / 5 i , so to get an increase of 0 . 1, we need to go only a distance 0

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## This note was uploaded on 09/14/2011 for the course MATH 18.02 taught by Professor Auroux during the Fall '08 term at MIT.

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practice2as - SOLUTIONS TO 18.02 PRACTICE MIDTERM #2A BJORN...

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