HW 10 - wolz(cmw2833 – HW 10 – odell –(54615 1 This...

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Unformatted text preview: wolz (cmw2833) – HW 10 – odell – (54615) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the domain of the vector function r ( t ) = ( t 3 , √ t − 6 , ln(7 − t ) ) . 1. 6 ≤ t ≤ 7 2. 6 < t < 7 3. 6 < t ≤ 7 4. t < 6 , t > 7 5. 6 ≤ t < 7 correct Explanation: A vector function r ( t ) = ( f ( t ) , g ( t ) , h ( t ) ) is defined when each of f ( t ) , g ( t ) and h ( t ) is defined. Now, for the given function, f ( t ) = t 3 is defined for all t , while g ( t ) = √ t − 6 is defined only when t ≥ 6. On the other hand, h ( t ) = ln(7 − t ) is defined only 7 − t > 0. Consequently, the domain of r ( t ) consists of all t , 6 ≤ t < 7 . keywords: vector function, domain, power function, square root function, log function, 002 10.0 points Find lim t →∞ r ( t ) when r ( t ) = (Big 6 tan − 1 t, e − 5 t , 7 ln t t )Big . 1. limit = ( 6 π, , 7 ) 2. limit = ( 6 π, − 5 , ) 3. limit = ( , − 5 , 7 ) 4. limit = ( 3 π, , ) correct 5. limit = ( 3 π, − 5 , ) 6. limit = ( , , 7 ) Explanation: For a vector function r ( t ) = ( f ( t ) , g ( t ) , h ( t ) ) , the limit lim t →∞ r ( t ) = ( lim t →∞ f ( t ) , lim t →∞ g ( t ) , lim t →∞ h ( t ) ) . But for the given vector function, lim t →∞ f ( t ) = lim t →∞ 6 tan − 1 ( t ) = 3 π , while lim t →∞ g ( t ) = lim t →∞ e − 5 t = 0 , and lim t →∞ h ( t ) = lim t →∞ 7 ln t t = 0 , using L’Hospital’s Rule. Consequently, lim t →∞ r ( t ) = ( 3 π, , ) . 003 10.0 points Find a parameterization of the horizontal circle of radius 3 having center (4 , − 2 , 1). wolz (cmw2833) – HW 10 – odell – (54615) 2 1. r ( t ) = (4+3 sin t ) i +(2 − 3 cos t ) j + k 2. r ( t ) = (4 − 3 cos t ) i − 2 j +(1+3 sin t ) k 3. r ( t ) = (4 − 3 cos t ) i − 2 j − (1+3 sin t ) k 4. r ( t ) = (4 − 3 sin t ) i − (2+3 cos t ) j − k 5. r ( t ) = ((4+3 cos t ) i +2 j − (1 − 3 sin t ) k 6. r ( t ) = (4 + 3 sin t ) i − (2 − 3 cos t ) j + k correct Explanation: If the vector function r ( t ) = x ( t ) i + y ( t ) j + z ( t ) k traces out a horizontal circle having center (4 , − 2 , 1), then z ( t ) = 1 for all t because the circle must lie in the horizontal plane z ( t ) = 1. On the other hand, the projection r ( t ) = x ( t ) i + y ( t ) j of this circle on the xy-plane is a circle of radius 3 having center at (4 , − 2 , 0). Thus, as as circle in the xy-plane it has the equation ( x − 4) 2 + ( y + 2) 2 = 9 . Consequently, ( x ( t ) − 4) 2 + ( y ( t ) + 2) 2 = 9 , and so r ( t ) = (4 + 3 sin t ) i − (2 − 3 cos t ) j + k is one parametrization of the horizontal circle of radius 3 and center (4 , − 2 , , 1)....
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This note was uploaded on 02/25/2011 for the course M 408d taught by Professor Sadler during the Spring '07 term at University of Texas.

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HW 10 - wolz(cmw2833 – HW 10 – odell –(54615 1 This...

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