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05aut-m2sols

# 05aut-m2sols - MATH 51 MIDTERM II Name Numeric Student ID...

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MATH 51 MIDTERM II November 17, 2005 Name: Numeric Student ID: Instructor’s Name: I agree to abide by the terms of the honor code: Signature: Instructions: Print your name, student ID number and instructor’s name in the space provided. During the test you may not use notes, books or calculators. Read each question carefully and show all your work ; full credit cannot be obtained without sufficient justification for your answer unless explicitly stated otherwise. Underline your final answer to each question. There are 9 questions. You have 90 minutes to do all the problems. Question Score Maximum 1 10 2 6 3 8 4 6 5 10 6 10 7 15 8 10 9 10 Total 85

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Question 1 of 9, Page 2 of 14 Solutions 1. Consider the function f ( x, y ) = x 4 y 3 and the point P = (1 , 1 , 1) on its graph. (a) Write down the equation of the tangent plane at the graph of the function at the point P . Solution: The tangent plane has equation, in general, ∂f ∂x , ∂f ∂y , - 1 P · ( x - P ) = 0 In this case, ∂f ∂x = 4 x 3 y 3 which is 4 at (1,1). ∂f ∂y = 3 x 4 y 2 which is 3 at (1,1). So the normal vector to the plane is 4 , 3 , - 1 and the tangent plane has equation: 4( x - 1) + 3( y - 1) - ( z - 1) = 0 (b) Using your answer from (a), write down an expression for the change, Δ z , in z = f ( x, y ) depending on Δ x and Δ y , the change in x and y , respectively, near the point P = (1 , 1 , 1). Is the function f ( x, y ) more sensitive to a change in x or to a change in y ? Explain. Solution: Near the point P = (1 , 1 , 1) the tangent plane is a good approximation to the function. Thus Δ z = 4Δ x + 3Δ y hence the function is more sensitive to changes in x . That is, small changes in x result in larger differences in the resulting z than small changes in y . (c) Using your answer to (b), find the approximate value of f (1 . 01 , 1 . 02). Solution: Using the tangent plane approximation at (1,1), we find f (1 . 01 , 1 . 02) = f (1 , 1) + . 01(4) + . 02(3) = 1 + . 04 + . 06 = 1 . 1 since we know using any linear approximation in two variables, this is just the value of the tangent plane at this point.
Question 2 of 9, Page 3 of 14 Solutions 2. (a) The steady state temperature function T ( x, y ) for a thin flat plate satisfies the equation T xx + T yy = 0 . Does the function T ( x, y ) = ln( x 2 + y 2 ) satisfy the equation above? Show your work. Solution: T xx denotes the second partial derivative with respect to x . T x = ∂T ∂x = 2 x x 2 + y 2 using the chain rule. Then using the quotient rule, T xx = 2( x 2 + y 2 ) - 2 x (2 x ) ( x 2 + y 2 ) 2 = 2 y 2 - 2 x 2 ( x 2 + y 2 ) 2 The function is symmetric in x and y so the derivatives in y will be the same with the roles reversed. Substituting, we see that the function does indeed satisfy the above partial differential equation.

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Question 2 of 9, Page 4 of 14 Solutions (b) Given f ( x, y, z ) = z sin x + cos y ln( z + y ) z ,
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