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practice1as - SOLUTIONS TO 18.02 PRACTICE MIDTERM #1A BJORN...

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Unformatted text preview: SOLUTIONS TO 18.02 PRACTICE MIDTERM #1A BJORN POONEN 1) Find the equation of the tangent plane to the graph of f ( x, y ) := x 2 y- 3 y 2 at the point where x = 2 and y = 1. Solution: We have f x = 2 xy and f y = x 2- 6 y . At (2 , 1), we have f = 1, f x = 4, f y =- 2. So the tangent plane at (2 , 1 , 1) is z- 1 = 4( x- 2) + (- 2)( y- 1) . Alternative forms of the answer include z = 4 x- 2 y- 5 and 4 x- 2 y- z = 5 . 2) Given that x 1 , x 2 , x 3 , y 1 , y 2 , y 3 are real numbers satisfying the conditions x 2 1 + x 2 2 + x 2 3 = 4 y 2 1 + y 2 2 + y 2 3 = 9 , what is the range of possible values of x 1 y 1 + x 2 y 2 + x 3 y 3 ? Solution: Equivalently, given that ~x is a vector of length 2, and ~ y is a vector of length 3, what is the range of possible values of ~x ~ y ? The angle between ~x and ~ y can lie anywhere in the range [0 , ], so cos can lie anywhere in [- 1 , 1], so ~x ~ y = | ~x || ~ y | cos = 6 cos can lie anywhere in the range [- 6 , 6]. 3) Let f ( x, y ) = x 3 + 3 xy- y 3 . (a) (15 pts.) Find all points (if any) at which f ( x, y ) has a local minimum. Solution to (a): We have f x = 3 x 2 + 3 y f y = 3 x- 3 y 2 . These exist everywhere, so at a local minimum, both f x and f y must be 0. This leads to 3 x 2 + 3 y = 0 3 x- 3 y 2 = 0 1 or equivalently y =- x 2 x = y 2 , so x = (- x 2 ) 2 = x 4 , so x = 0 or 1 = x 3 . The first case....
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practice1as - SOLUTIONS TO 18.02 PRACTICE MIDTERM #1A BJORN...

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