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midterm_s08_soln - Math 237 Midterm Solutions Spring 2008 1...

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Math 237 - Midterm Solutions Spring 2008 1. Short Answer Problems [1] a) Let f : R 2 R . What is the precise definition of lim ( x,y ) ( a,b ) f ( x, y ) = L . Solution: We write lim ( x,y ) ( a,b ) f ( x, y ) = L if for every ǫ > 0 there exists a δ > 0 such that | f ( x, y ) L | < ǫ, whenever 0 < | ( x, y ) ( a, b ) | < δ. [2] b) Find the equation of the tangent plane to x 2 2 y 2 z 3 = 0 at the point (3 , 2 , 1). Solution: We have f ( x, y, z ) = x 2 2 y 2 z 3 so f = (2 x, 4 y, 3 z 2 ). Therefore the equation of the tangent plane is 0 = f (3 , 2 , 1) · ( x 3 , y 2 , z 1) = (6 , 8 , 3) · ( x 3 , y 2 , z 1) = 6( x 3) 8( y 2) 3( z 1) [2] c) What is the precise definition of the directional derivative D u f ( a, b ). Solution: D u f ( a, b ) = d ds f (( a, b ) + su ) vextendsingle vextendsingle vextendsingle vextendsingle s =0 where u is a unit vector. OR D u f ( a, b ) = lim h 0 f ( a + hu ) f ( a ) h . [2] d) State the theorem relating f ( a, b ) to the level curve f ( x, y ) = k through ( a, b ). Solution: If f has continuous partial derivatives and f ( a, b ) negationslash = (0 , 0) then f ( a, b ) is orthogonal to the level curve f ( x, y ) = k through ( a, b ).
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Math 237 - Midterm Solutions Spring 2008 2. Let f ( x, y ) = ( x y + 1) 2 . [2] a) What is the domain and range of f ? Solution: The domain of f is R 2 . The range of f is z 0. [3] b) Sketch the level curves and cross sections of z = f ( x, y ). Solution: Level Curves: k = ( x y + 1) 2 y = x + 1 ± k , k 0 Cross sections: z = ( y + 1 + c ) 2 z = ( y (1 + c )) 2 z = ( x d + 1) 2
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Math 237 - Midterm Solutions Spring 2008 3. Determine if each of the following limits exist. Evaluate the limits that exist.
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