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Assignment 8 - patino(mp25752 Assignment8 luecke(57510 This...

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patino (mp25752) – Assignment8 – luecke – (57510) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Hint: Try to write the length of the tangent vector as a perfect square. Find the arc length of the curve r ( t ) = (3 + 2 t ) i + e t j + (5 + e t ) k between r (0) and r (7). 1. arc length = ( e 7 e 7 ) 2 2. arc length = ( e 7 + e 7 ) 2 3. arc length = e 7 e 7 correct 4. arc length = e 7 + e 7 5. arc length = 2 e 7 6. arc length = 2 e 7 Explanation: The length of a curve r ( t ) between r ( t 0 ) and r ( t 1 ) is given by the integral L = integraldisplay t 1 t 0 | r ( t ) | dt. Now when r ( t ) = (3 + 2 t ) i + e t j + (5 + e t ) k , we see that r ( t ) = 2 i + e t e t . But then | r ( t ) | = (2 + e 2 t + e 2 t ) 1 / 2 = e t + e t . Thus L = integraldisplay 7 0 ( e t + e t ) dt = bracketleftBig e t e t bracketrightBig 7 0 . Consequently, arc length = L = e 7 e 7 . 002 10.0 points Find the arc length of the curve r ( t ) = (3 2 t ) i + ln(4 t ) j + (4 t 2 ) k between r (1) and r (2). 1. arc length = 2 + ln 8 2. arc length = 3 ln 2 3. arc length = 8 ln 2 4. arc length = 3 4 ln 2 5. arc length = 4 + 4 ln 2 6. arc length = 3 + ln 2 correct Explanation: The length of a curve r ( t ) between r ( t 0 ) and r ( t 1 ) is given by the integral L = integraldisplay t 1 t 0 | r ( t ) | dt. Now when r ( t ) = (3 2 t ) i + ln(4 t ) j + (4 t 2 ) k we see that r ( t ) = 2 i + 1 t j + 2 t k . But then | r ( t ) | = parenleftBig 4 + 1 t 2 + 4 t 2 parenrightBig 1 / 2 = 2 t 2 + 1 t . Thus L = integraldisplay 2 1 parenleftBig 2 t + 1 t parenrightBig dt = bracketleftBig t 2 + ln t bracketrightBig 2 1 . Consequently, arc length = L = 3 + ln 2 .
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patino (mp25752) – Assignment8 – luecke – (57510) 2 003 10.0 points Find the arc length of the curve r ( t ) = ( sin 2 t, 6 t, cos 2 t ) between r (0) and r (3). 1. arc length = 3 37 2. arc length = 3 39 3. arc length = 3 38 4. arc length = 6 10 correct 5. arc length = 18 Explanation: The length of the curve r ( t ) between r ( t 0 ) and r ( t 1 ) is given by the integral L = integraldisplay t 1 t 0 | r ( t ) | dt. Now when r ( t ) = ( sin 2 t, 6 t, cos 2 t ) , we see that r ( t ) = ( 2 cos 2 t, 6 , 2 sin 2 t ) . But then by the Pythagorean identity, | r ( t ) | = (4 + 36) 1 / 2 . Thus L = integraldisplay 3 0 2 10 dt = bracketleftBig 2 10 t bracketrightBig 3 0 .
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