This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: wolz (cmw2833) HW 10 odell (54615) 1 This printout should have 13 questions. Multiplechoice 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 , , 7 ) 2. limit = ( 6 , 5 , ) 3. limit = ( , 5 , 7 ) 4. limit = ( 3 , , ) correct 5. limit = ( 3 , 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 , while lim t g ( t ) = lim t e 5 t = 0 , and lim t h ( t ) = lim t 7 ln t t = 0 , using LHospitals Rule. Consequently, lim t r ( t ) = ( 3 , , ) . 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 xyplane is a circle of radius 3 having center at (4 , 2 , 0). Thus, as as circle in the xyplane 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)....
View Full
Document
 Spring '07
 Sadler
 Calculus

Click to edit the document details