FINAL REVIEW #3-solutions

# FINAL REVIEW #3-solutions - suleimenov (bs26835) – FINAL...

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Unformatted text preview: suleimenov (bs26835) – FINAL REVIEW #3 – rusin – (55565) 1 This print-out should have 32 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Third and final installment of the review for the final exam. Some questions may duplicate previous HW questions. These reviews have been very long and even so, are not exactly ”comprehensive” – there are many variations of the ideas that are not on these reviews. But I think if you are com- fortable with all the topics on these reviews, you will be in good shape for the exam. Good luck! 001 10.0 points Find a vector function that represents the curve of intersection of the paraboloid z = x 2 + 3 y 2 and the parabolic cylinder y = x 2 . 1. r ( t ) = √ t i + t j + ( 3 t 2 + t 4 ) k 2. r ( t ) = √ t i + t j + ( t 2 + 3 t 4 ) k 3. r ( t ) = t 2 i + t j + ( t + 3 t 2 ) k 4. r ( t ) = t i + t 2 j + ( 3 t 2 + t 4 ) k 5. r ( t ) = t i + t 2 j + ( t 2 + 3 t 4 ) k correct 6. r ( t ) = t i + √ t j + ( t + 3 t 2 ) k 7. r ( t ) = t i + √ t j + ( 3 t + t 2 ) k 8. r ( t ) = t 2 i + t j + ( 3 t + t 2 ) k Explanation: The vector function r ( t ) = x ( t ) i + y ( t ) j + z ( t ) k represents the inersection of the paraboloid z = x 2 + 3 y 2 and the parabolic cylinder y = x 2 when z ( t ) = x ( t ) 2 + y ( t ) 2 , y ( t ) = x ( t ) 2 . This last condition already eliminates 4 of the possible vector functions, leaving only r ( t ) = t i + √ t j + ( t + 3 t 2 ) k , r ( t ) = t i + t 2 j + ( 3 t 2 + t 4 ) k , r ( t ) = √ t i + t j + ( t 2 + 3 t 4 ) k , and r ( t ) = t i + t 2 j + ( t 2 + 3 t 4 ) k . But of these, only r ( t ) = t i + t 2 j + ( t 2 + 3 t 4 ) k lies on the paraboloid z = x 2 + 3 y 2 . keywords: 002 10.0 points Find an equation in vector form for the tangent line to the graph of r ( s ) = ( cos s, 2 e 3 s , sin s + 2 e − 3 s ) at the point (1 , 2 , 2). 1. L ( t ) = ( 1 , 2 − 2 t, 2 + 5 t ) 2. L ( t ) = ( t, 2 − 3 t, 2 − 5 t ) 3. L ( t ) = ( t, 2 − 2 t, 2 + 6 t ) 4. L ( t ) = ( 1 , 2 + 6 t, 2 − 6 t ) 5. L ( t ) = ( t, 2 + 3 t, 2 − 6 t ) 6. L ( t ) = ( 1 , 2 + 6 t, 2 − 5 t ) correct Explanation: suleimenov (bs26835) – FINAL REVIEW #3 – rusin – (55565) 2 Since (1 , 2 , 2) = r (0), the tangent line has vector form L ( t ) = r (0) + t r ′ (0) . But r ′ ( s ) = (− sin s, 6 e 3 s , cos s − 6 e − 3 s ) , so at r (0) the tangent vector is r ′ (0) = ( , 6 , 1 − 6 ) = ( , 6 , − 5 ) . Consequently, L ( t ) = ( 1 , 2 + 6 t, 2 − 5 t ) . keywords: 003 10.0 points Find the point of intersection of the tangent lines to the curve r ( t ) = ( sin πt, 4 sin πt, cos πt ) at the points where t = 0 and t = 0 . 5. 1. (1 , − 4 , 1) 2. (0 , , 0) 3. (1 , 4 , 1) correct 4. (1 , 3 , 0) 5. (0 , 4 , 1) Explanation: The slope of each component of the tangent line to a vector function r ( t ) = ( x ( t ) , y ( t ) , z ( t ) ) at the point t = 0 is x ′ (0) , y ′ (0) , z ′ (0) , respectively. Thus, we have x ′...
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## This note was uploaded on 09/15/2011 for the course M 55565 taught by Professor Rusin during the Spring '11 term at University of Texas at Austin.

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FINAL REVIEW #3-solutions - suleimenov (bs26835) – FINAL...

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