Math252Sols3F08

Math252Sols3F08 - QUIZ 3 (SECTIONS 16.3-16.9) SOLUTIONS...

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QUIZ 3 (SECTIONS 16.3-16.9) SOLUTIONS MATH 252 – FALL 2008 – KUNIYUKI SCORED OUT OF 125 POINTS ± MULTIPLIED BY 0.84 ± 105% POSSIBLE 1) Let fr , t () = t 2 sin r t ± ² ³ ´ µ . Find f r r , t . (5 points) f r r , t = D r t 2 "#" ± sin r t ± ² ³ ´ µ · ¸ ¹ ¹ º » ¼ ¼ = t 2 ½ D r sin r t ± ² ³ ´ µ · ¸ ¹ º » ¼ = t 2 ½ cos r t ± ² ³ ´ µ · ¸ ¹ º » ¼ ½ D r r t ± ² ³ ´ µ = t 2 ½ cos r t ± ² ³ ´ µ · ¸ ¹ º » ¼ ½ D r 1 t "#" ± ½ r ± ² ³ ³ ´ µ = t 2 ½ cos r t ± ² ³ ´ µ · ¸ ¹ º » ¼ ½ 1 t = t cos r t ± ² ³ ´ µ
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2) Let fx , y , z () = e xyz . Find f y x , y , z and use that to find f yz x , y , z . (8 points) f y x , y , z = D y e xyz ± ² ³ ´ = e xyz ± ² ³ ´ µ D y xyz ± ² ³ ´ = e xyz ± ² ³ ´ µ D y xz "#" ± µ y · ¸ ¹ º » ± ² ¼ ¼ ³ ´ ½ ½ = e xyz µ xz = xze xyz f yz x , y , z = D z f y x , y , z ± ² ³ ´ = D z x "#" ± ze xyz ± ² µ µ ³ ´ = x · D z ze xyz ± ² ³ ´ We will use a Product Rule for Differentiation. = xD z z ± ² ³ ´ = 1 ²³ ´µ ´ · e xyz ± ² ³ ´ + z ± ² ³ ´ · D z e xyz ± ² ³ ´ ¸ ¹ º º º » ¼ ½ ½ ½ = xe xyz + z · D z e xyz ± ² ³ ´ = xyz + z · e xyz · D z xy "#" ± z ¸ ¹ º » ¼ ½ ± ² µ µ ³ ´ = xyz + z · e xyz · xy ± ² ³ ´ = xyz + xyze xyz = xe xyz 1 + xyz , or xe xyz + x 2 yze xyz 3) Assume that f is a function of s and t . Write the limit definition of f s s , t using the notation from class. (4 points) f s s , t = lim h ± 0 fs + h , t ² , t h
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4) Find ± z y if z = fx , y () is a differentiable function described implicitly by the equation tan y 3 z = x 2 ± yz . Use the Calculus III formula given in class. Simplify. (12 points) First, isolate 0 on one side: tan y 3 z ± x 2 + yz Let this be Fx , y , z ±² ³³ ³´ ³³³ = 0 Find z y . When using the formula, treat x , y , and z as independent variables. z y = ² F y x , y , z F z x , y , z = ² D y tan y 3 z ² x 2 "#" ± + yz "#" ± ³ ´ µ µ · ¸ ¸ D z tan y 3 z ² x 2 "#" ² + y "#" ² z ³ ´ µ µ · ¸ ¸ = ² sec 2 y 3 z ³ ´ · ¹ D y y 3 z "#" ± º » ¼ ¼ ½ ¾ ¿ ¿ ³ ´ µ µ · ¸ ¸ + z sec 2 y 3 z ³ ´ · ¹ D z y 3 "#" ² z º » ¼ ½ ¾ ¿ ³ ´ µ µ · ¸ ¸ + y = ² sec 2 y 3 z ³ ´ · ¹ z ¹ D y y 3 ³ ´ · + z sec 2 y 3 z ³ ´ · ¹ y 3 ³ ´ · + y = ² sec 2 y 3 z ³ ´ · ¹ z ¹ 3 y 2 ³ ´ · + z sec 2 y 3 z ³ ´ · ¹ y 3 ³ ´ · + y = ² 3 y 2 z sec 2 y 3 z + z y 3 sec 2 y 3 z + y or ² z 3 y 2 sec 2 y 3 z + 1 ³ ´ · yy 2 sec 2 y 3 z + 1 ³ ´ ·
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5) Let f , g , and h be differentiable functions such that z = fu , v () , u = gr , s , t , and v = hr , s , t . Use the Chain Rule to write an expression for ± z t . (5 points) ± z ± t = ± z ± u ± u ± t + ± z ± v ± v ± t 6) The temperature at any point x , y in the xy
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This note was uploaded on 09/08/2011 for the course MATH 252 taught by Professor Staff during the Spring '11 term at Mesa CC.

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Math252Sols3F08 - QUIZ 3 (SECTIONS 16.3-16.9) SOLUTIONS...

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