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HW_9_Key

# HW_9_Key - Miller Kierste Homework 9 Due 3:00 am Inst...

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Miller, Kierste – 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 – find 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 , 2 (1 + t ) 2 E . 1. I = h 4 , ln 2 , 2 π i 2. I = h ln 2 , π, 1 i 3. I = h π, ln 2 , 1 i correct 4. I = h π, 2 , ln 2 i 5. I = h 4 , 2 π, 2 ln 2 i 6. I = h 4 ln 2 , 2 , 2 π i Explanation: For 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 , 2 (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 2 (1 + t ) 2 dt = h - 2 1 + t i 1 0 = 1 . Consequently, I = h π, ln 2 , 1 i . keywords: vector function, definite integral, inverse trig integral, log function, substitution 002 (part 1 of 1) 10 points Find the arc length of the curve r ( t ) = h sin 2 t, 6 t, cos 2 t i between r (0) and r (3). 1. arc length = 3 37 2. arc length = 6 10 correct 3. arc length = 18 4. arc length = 3 38 5. arc length = 3 39 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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Miller, Kierste – Homework 9 – Due: Oct 26 2007, 3:00 am – Inst: JEGilbert 2 Now when r ( t ) = h sin 2 t, 6 t, cos 2 t i , we see that r 0 ( t ) = h 2 cos 2 t, 6 , - 2 sin 2 t i . But then by the Pythagorean identity, | r 0 ( t ) | = (4 + 36) 1 / 2 . Thus L = Z 3 0 2 10 dt = h 2 10 t i 3 0 . Consequently, arc length = L = 6 10 . keywords: curve, 3-space, length, trig func- tion 003 (part 1 of 1) 10 points Find the arc length of the curve r ( t ) = (4 + 2 t ) i + e t j + (1 + e - t ) k between r (0) and r (4). 1. arc length = 2 e 4 2. arc length = e 4 + e - 4 3. arc length = e 4 - e - 4 correct 4. arc length = ( e 4 - e - 4 ) 2 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 ) = (4 + 2 t ) i + e t j + (1 + 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 ) = 6 t 2 , 12 t, 6 ln t fi . 1. T ( t ) = ¿ t 2 2 t 2 + 1 , 2 t 2 t 2 + 1 , 1 2 t 2 + 1 2. T ( t ) = ¿ t 2 2 t 2 + 1 , t 2 t 2 + 1 , 1 2 t 2 + 1 3. T ( t ) = ¿ t 2 t 2 + 1 , t t 2 + 1 , 1 t 2 + 1 4. T ( t ) = ¿ 2 t 2 2 t 2 + 1 , 2 t 2 t 2 + 1 , 1 2 t 2 + 1 correct 5. T ( t ) = ¿ 2 t 2 t 2 + 1 , 2 t t 2 + 1 , 1 t 2 + 1 Explanation: keywords: Stewart5e, vector function, curve in 3-space, tangent, unit tangent vector
Miller, Kierste – Homework 9 – Due: Oct 26 2007, 3:00 am – Inst: JEGilbert 3 005 (part 1 of 3) 10 points Determine the curvature, κ ( t ), of the curve defined parametrically in the plane by ( x ( t ) , y ( t )).

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