237_midterm_f10_soln

# 237_midterm_f10_soln - Math 237 F10 Midterm Solutions 1...

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Unformatted text preview: Math 237 F10 Midterm Solutions 1. Short Answer Problems [2] a) Let f : R 2 → R . State the definition of lim ( x,y ) → ( a,b ) f ( x, y ) = L . Solution: For every ǫ > 0 there exists a δ > 0 such that | f ( x ) − L | < ǫ whenever 0 < bardbl ( x, y ) − ( a, b ) bardbl < δ . [2] b) Let f : R 2 → R . State the definition of the directional derivative of f in the direction of the unit vector ˆ u = ( u 1 , u 2 ) at the point ( a, b ). Solution: D ˆ u f ( a, b ) = d ds bracketleftbig f ( a + su 1 , b + su 2 ) bracketrightbig vextendsingle vextendsingle vextendsingle vextendsingle s =0 . [2] c) Let f : R 2 → R such that f ∈ C 2 . State the definition of the second degree Taylor polynomial P 2 , ( a,b ) ( x, y ) of f at ( a, b ). Solution: P 2 , ( a,b ) ( x, y ) = f ( a, b )+ f x ( a, b )( x − a )+ f y ( a, b )( y − b )+ 1 2 f xx ( a, b )( x − a ) 2 + f xy ( a, b )( x − a )( y − b ) + 1 2 f yy ( a, b )( y − b ) 2 [2] d) Let f : R 3 → R . What is the equation for the tangent plane to the level surface f ( x, y, z ) = 3 at ( a, b, c )? Solution: ∇ f ( a, b, c ) · ( x − a, y − b, z − c ) = 0 [2] e) If f x and f y are both continuous at ( a, b ) what two things can we conclude about f ( x, y ) at ( a, b )? Solution: f is differentiable at ( a, b ) and f is continuous at ( a, b ). 2 [3] 2. Let f ( x, y ) = radicalbig 1 + x 2 + y 2 . Sketch level curves and cross sections of z = f ( x, y ). Solution: The level curves are k = f ( x, y ) = radicalbig 1 + x 2 + y 2 . Simplifying gives k 2 − 1 = x 2 + y 2 . The cross sections x = c are z = f ( c, y ) = radicalbig 1 + c 2 + y 2 , z > 1. Simplifying gives z 2 − y 2 = 1 + c 2 , z ≥ 0. The cross sections y = d are z = f ( x, d ) = √ 1 + x 2 + d 2 , z > 1. Simplifying gives z 2 − x 2 = 1 + d 2 , z ≥ 0. 3 3. For each of the following, evaluate the limit or show that it does not exist. [2] a) lim ( x,y ) → ( − 1 , 2) 2 x 3 y 2 | x | + 4 | y | . Solution: Since f ( x, y ) = 2 x 3 y 2 | x | +4 | y | is continuous at ( − 1 , 2) by the continuity theorems we get lim ( x,y ) → ( − 1 , 2) 2 x 3 y 2 | x | + 4 | y | = 2( − 1) 3 (2) 2 | − 1 | + 4 | 2 | = − 8 9 [3] b) lim...
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## This note was uploaded on 12/10/2010 for the course MATH 237 taught by Professor Wolczuk during the Spring '08 term at Waterloo.

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237_midterm_f10_soln - Math 237 F10 Midterm Solutions 1...

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