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Unformatted text preview: hernandez (ah29758) M408D Quest Homework 9 pascaleff (54550) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Find lim t + r ( t ) when r ( t ) = ( 4 cos t, 2 e t , 10 t ln t ) . 1. limit = ( , 2 , ) 2. limit = ( 4 , , 10 ) 3. limit = ( 4 , 2 , 10 ) 4. limit = ( , , ) 5. limit = ( 4 , , 10 ) 6. limit = ( 4 , 2 , ) correct Explanation: For a vector function r ( t ) = ( f ( t ) , g ( t ) , h ( t ) ) , the limit lim t + r ( t ) = (Big lim t + f ( t ) , lim t + g ( t ) , lim t + h ( t ) )Big . But for the given vector function, lim t + f ( t ) = lim t + 4 cos t = 4 , while lim t + g ( t ) = lim t + 2 e t = 2 , and lim t + h ( t ) = lim t + t ln t = 0 , using LHospitals Rule. Consequently, lim t + r ( t ) = ( 4 , 2 , ) . keywords: vector function, limit, trig function log function, exponential function 002 10.0 points A space curve is shown in black on the surface x y z Which one of the following vector functions has this space curve as its graph? 1. r ( t ) = ( t cos t, t, t sin t ) , t > , 2. r ( t ) = ( sin 4 t, cos t, sin t ) 3. r ( t ) = ( t cos t, t sin t, t ) 4. r ( t ) = ( t cos t, t, t sin t ) 5. r ( t ) = ( t cos t, t sin t, t ) , t > , 6. r ( t ) = ( sin 2 t, cos t, sin t ) 7. r ( t ) = ( cos t, sin t, sin 4 t ) correct 8. r ( t ) = ( cos t, sin t, sin 2 t ) Explanation: hernandez (ah29758) M408D Quest Homework 9 pascaleff (54550) 2 The surface is a cylinder such that hori- zontal cross-sections intersect the surface in a circle. Now of the given vector functions only r ( t ) = ( cos t, sin t, sin 2 t ) r ( t ) = ( cos t, sin t, sin 4 t ) lie on such a cylinder; in particular, the sur- face is the graph of x 2 + y 2 = 1 , and the space curve passes through the point r (0) = (1 , , 0) on the x-axis as well as the point r parenleftBig 2 parenrightBig = (0 , 1 , 0) on the y-axis. But the z-coordinate of the space curve takes both positive and nega- tive values between these two points, while sin2 t > 0 for 0 < t < / 2. Consequently, the space curve is the graph of r ( t ) = ( cos t, sin t, sin 4 t ) . keywords: surface, space curve, vector func- tion, 3D graph, circular cylinder, 003 10.0 points Find a vector function whose graph is the curve of intersection of the circular cylinder x 2 + z 2 = 4 and the hyperbolic paraboloid y = x 2 z 2 . 1. r ( t ) = 2 cos t i + 4 sin 2 t j + 2 sin t k 2. r ( t ) = 2 sin t i 4 sin 2 t j + 2 cos t k 3. r ( t ) = 1 2 cos t i 1 4 sin 2 t j + 1 2 sin t k 4. r ( t ) = 1 2 cos t i + 1 2 cos 2 t j + 1 4 sin t k 5. r ( t ) = 2 sin t i + 4 cos 2 t j + 2 cos t k 6. r ( t ) = 2 cos t i + 4 cos 2 t j + 2 sin t k correct 7. r ( t ) = 2 sin 2 t i + 2 sin 2 t j + 2 sin 2 t k 8. r ( t ) = 2 cos 2 t i + 2 cos t j + 2 sin 2 t k Explanation: The graph of the vector function r ( t ) = x ( t ) i + y ( t ) j + z ( t ) k...
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This note was uploaded on 11/15/2011 for the course M 408 D taught by Professor Textbookanswers during the Spring '07 term at University of Texas at Austin.
- Spring '07