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midterm_w09_soln

midterm_w09_soln - Math 237 Midterm Solutions Winter 2009 1...

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Unformatted text preview: Math 237 - Midterm Solutions Winter 2009 1. Short Answer Problems [1] a) Let f : R 2 → R . State the precise definition of lim ( x,y ) → ( a,b ) f ( x, y ) = L . Solution: lim ( x,y ) → ( a,b ) f ( x, y ) = L means for every ǫ > 0 there exists a δ > 0 such that | f ( x, y ) − L | < ǫ whenever < bardbl ( x, y ) − ( a, b ) bardbl < δ. [2] b) Let f : R 2 → R . If f x and f y are both continuous at ( a, b ), then what are two things you can say about f at ( a, b )? Solution: f is differentiable and continuous at ( a, b ). [1] c) Let f ( x, y ) = x 2 + xy + y 3 . What is the greatest rate of change of f at ( − 1 , 1)? Solution: ∇ f ( x, y ) = (2 x + y, x + 3 y 2 ) so the greatest rate of change at ( − 1 , 1) is bardbl∇ f ( − 1 , 1) bardbl = bardbl ( − 1 , 2) bardbl = √ 5. [2] d) If f : R 2 → R has continuous second partial derivatives and Hf ( x, y ) = parenleftbigg 0 0 0 0 parenrightbigg for all points ( x, y ) in a neighborhood of ( a, b ) then what can you conclude about the accuracy of L ( a,b ) ( x, y )? Justify your answer. Solution: Taylor’s theorem says that there exists a point c on the line segment from ( a, b ) to ( x, y ) such that R 1 , ( a,b ) ( x, y ) = 1 2 [ f xx ( c )( x − a ) 2 + 2 f xy ( c )( x − a )( y − b ) + f yy ( c )( y − b ) 2 . By the assumption we have all the second partials at c are 0 hence, the linear approximation is exact. Math 237 - Midterm Solutions Winter 2009 2. Let f ( x, y ) = radicalbig y 2 − x 2 . [2] a) What is the domain and range of f ? Solution: For the domain we need y 2 − x 2 ≥ 0 so y 2 ≥ x 2 . the range is z ≥ 0. [3] b) Sketch the level curves and cross sections of z = f ( x, y ). Solution: Level Curves: k = radicalbig y 2 − x 2 y 2 − x 2 = k 2 , k ≥ Cross sections: z = radicalbig y 2 − c 2 z 2 − y 2 = − c 2 , z ≥ z = √ d 2 − x 2 z 2 + x 2 = d 2 , z ≥ Math 237 - Midterm Solutions Winter 2009 [4] 3. Prove that if f : R 2 → R is differentiable at ( a, b ) then f is continuous at ( a, b ). Solution: If f is differentiable at ( a, b ) then we know that both partial derivatives exist at ( a, b ) and lim ( x,y ) → ( a,b ) | R 1 , ( a,b ) ( x,y ) | √ ( x- a ) 2 +( y- b ) 2 = 0 . where R 1 , ( a,b ) ( x, y ) = f ( x, y ) − f ( a, b ) − f x ( a, b )( x − a ) − f y ( a, b )( y − b )....
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midterm_w09_soln - Math 237 Midterm Solutions Winter 2009 1...

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