Math252Sols2F07

Math252Sols2F07 - QUIZ 2 (CHAPTER 15, 16.1, 16.2) SOLUTIONS...

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QUIZ 2 (CHAPTER 15, 16.1, 16.2) SOLUTIONS MATH 252 – FALL 2007 – KUNIYUKI SCORED OUT OF 125 POINTS MULTIPLIED BY 0.84 105% POSSIBLE 1) Find the length of the curve parameterized by: x = t 2 , y = 5 2 t 2 , z = 2 t , 0 t 2 . Major Hint (which you may use without proof): According to the Table of Integrals, if a is a positive real constant, a 2 + u 2 du = u 2 a 2 + u 2 + a 2 2 ln u + a 2 + u 2 + C Leave your answer as a simplified exact answer; do not approximate it using a calculator. You do not have to apply log properties at the end. Distance is measured in meters. Show all work! (20 points) L = dx dt 2 + dy dt 2 + dz dt 2 dt 0 2 = 2 t ( ) 2 + 5 t ( ) 2 + 2 ( ) 2 dt 0 2 = 4 t 2 + 5 t 2 + 4 dt 0 2 = 4 + 9 t 2 dt 0 2 = 2 ( ) 2 + 3 t ( ) 2 dt 0 2 Let a = 2 , so that a 2 = 4 . Perform a classic u -substitution: Let u = 3 t . Then, du = 3 dt dt = 1 3 du
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Change the limits of integration: t = 0 u = 3 0 ( ) = 0 t = 2 u = 3 2 ( ) = 6 L = 4 + u 2 1 3 du 0 6 = 1 3 4 + u 2 du 0 6 = 1 3 u 2 4 + u 2 + 4 2 = 2 ln u + 4 + u 2 0 6 by the Table of Integrals hint ( ) = 1 3 6 ( ) 2 4 + 6 ( ) 2 + 2ln 6 ( ) + 4 + 6 ( ) 2 0 ( ) 2 4 + 0 ( ) 2 + 2ln 0 ( ) + 4 + 0 ( ) 2 = 1 3 3 40 + 2ln 6 + 40 0 + 2ln 4 ( ) = 1 3 3 2 10 ( ) + 2ln 6 + 2 10 2ln 2 = 1 3 6 10 + 2ln 6 + 2 10 ( ) 2ln2 meters or 2 10 + ln 3 + 10 ( ) 2/3 meters See Note below. ( ) Note: 1 3 6 10 + 2ln 6 + 2 10 ( ) = 1 3 6 10 + ln 6 + 2 10 ( ) 2 ln 2 ( ) 2 = 1 3 6 10 + ln 6 + 2 10 ( ) 2 2 ( ) 2 = 1 3 6 10 + ln 6 + 2 10 2 2 = 2 10 + 1 3 ln 3 + 10 ( ) 2 = 2 10 + ln 3 + 10 ( ) 7.53685
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2) A curve C is parameterized by the vector-valued function (VVF) given by r t ( ) = e 2 t ,3 t + 1, t 2 . Find a tangent vector to C at the point e 10 ,16, 25 ( ) . (10 points) Find the t -value corresponding to the given point.
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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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Math252Sols2F07 - QUIZ 2 (CHAPTER 15, 16.1, 16.2) SOLUTIONS...

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