HW9S - Gauthier, Joseph Homework 9 Due: Oct 26 2007, 3:00...

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Gauthier, Joseph – Homework 9 – Due: Oct 26 2007, 3:00 am – Inst: JEGilbert 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points Determine the vector I = Z 1 0 r ( t ) dt when r ( t ) = D 4 1 + t 2 , 2 t 1 + t 2 , 6 (1 + t ) 2 E . 1. I = h π, ln 2 , 3 i correct 2. I = h 4 , 2 π, 6 ln 2 i 3. I = h 4 ln 2 , 2 , 6 π i 4. I = h ln 2 , π, 3 i 5. I = h 4 , ln 2 , 6 π i 6. I = h π, 2 , 3 ln 2 i Explanation: ±or a vector Function r ( t ) = h f ( t ) , g ( t ) , h ( t ) i , the components oF the vector I = Z 1 0 r ( t ) dt are given by Z 1 0 f ( t ) dt, Z 1 0 g ( t ) dt, Z 1 0 h ( t ) dt, respectively. But when r ( t ) = D 4 1 + t 2 , 2 t 1 + t 2 , 6 (1 + t ) 2 E , we see that Z 1 0 f ( t ) dt = Z 1 0 4 1 + t 2 dt = h 4 tan - 1 t i 1 0 = π , while Z 1 0 g ( t ) dt = Z 1 0 2 t 1 + t 2 dt = h ln(1 + t 2 ) i 1 0 = ln 2 , and Z 1 0 h ( t ) dt = Z 1 0 6 (1 + t ) 2 dt = h - 6 1 + t i 1 0 = 3 . Consequently, I = h π, ln 2 , 3 i . keywords: vector Function, defnite integral, inverse trig integral, log Function, substitution 002 (part 1 oF 1) 10 points ±ind the arc length oF the curve r ( t ) = h 3 sin 2 t, 8 t, 3 cos 2 t i between r (0) and r (3). 1. arc length = 3 82 2. arc length = 24 3. arc length = 3 73 4. arc length = 3 91 5. arc length = 30 correct Explanation: The length oF the curve r ( t ) between r ( t 0 ) and r ( t 1 ) is given by the integral L = Z t 1 t 0 | r 0 ( t ) | dt.
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Gauthier, Joseph – Homework 9 – Due: Oct 26 2007, 3:00 am – Inst: JEGilbert 2 Now when r ( t ) = h 3 sin 2 t, 8 t, 3 cos 2 t i , we see that r 0 ( t ) = h 6 cos 2 t, 8 , - 6 sin 2 t i . But then by the Pythagorean identity, | r 0 ( t ) | = (36 + 64) 1 / 2 . Thus L = Z 3 0 10 dt = h 10 t i 3 0 . Consequently, arc length = L = 30 . keywords: curve, 3-space, length, trig func- tion 003 (part 1 of 1) 10 points Find the arc length of the curve r ( t ) = (2 + 2 t ) i + e t j + (3 + e - t ) k between r (0) and r (4). 1. arc length = ( e 4 - e - 4 ) 2 2. arc length = e 4 + e - 4 3. arc length = 2 e - 4 4. arc length = e 4 - e - 4 correct 5. arc length = ( e 4 + e - 4 ) 2 6. arc length = 2 e 4 Explanation: The length of a curve r ( t ) between r ( t 0 ) and r ( t 1 ) is given by the integral L = Z t 1 t 0 | r 0 ( t ) | dt. Now when r ( t ) = (2 + 2 t ) i + e t j + (3 + e - t ) k , we see that r 0 ( t ) = 2 i + e t - e - t . But then | r 0 ( t ) | = (2 + e 2 t + e - 2 t ) 1 / 2 = e t + e - t . Thus L = Z 4 0 ( e t + e - t ) dt = h e t - e - t i 4 0 . Consequently, arc length = L = e 4 - e - 4 . keywords: curve, 3-space, arc length, exp function 004 (part 1 of 1) 10 points Find the unit vector T ( t ) tangent to the curve r ( t ) = 4 t 2 , 8 t, 4 ln t f . 1.
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This homework help was uploaded on 03/19/2008 for the course M 408M taught by Professor Gilbert during the Fall '07 term at University of Texas at Austin.

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HW9S - Gauthier, Joseph Homework 9 Due: Oct 26 2007, 3:00...

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