# Exam 2 - Miller Kierste – Exam 2 – Due 5:00 pm – Inst...

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Unformatted text preview: Miller, Kierste – Exam 2 – Due: Oct 31 2007, 5:00 pm – Inst: JEGilbert 1 This print-out should have 18 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 Find a vector function that represents the curve of intersection of the cone z = p x 2 + y 2 and the plane z- y = 13. 1. r ( t ) = t i + ‡ t 2- 169 26 · j + ‡ t 2 + 169 26 · k correct 2. r ( t ) = p 13(2 t + 13) i + ( t- 13) j + t k 3. r ( t ) = ‡ t 2- 169 26 · i + t j + ‡ t 2 + 169 26 · k 4. r ( t ) = ‡ t 2- 169 26 · i + t j + ‡ t 2 + 13 26 · k 5. r ( t ) = p 13(2 t- 13) i + t j + ( t + 13) k Explanation: The vector function r ( t ) = x ( t ) i + y ( t ) j + z ( t ) k intersects the cone z = p x 2 + y 2 and the plane z- y = 13 when z ( t ) = q x ( t ) 2 + y ( t ) 2 , z ( t )- y ( t ) = 13 . Now this last condition already eliminates two of the answer choices, leaving only r ( t ) = t i + ‡ t 2- 169 26 · j + ‡ t 2 + 169 26 · k , r ( t ) = p 13(2 t + 13) i + ( t- 13) j + t k , and r ( t ) = p 13(2 t- 13) i + t j + ( t + 13) k . But only the first of these equations defines a curve lying on the cone z = p x 2 + y 2 . Consequently, only r ( t ) = t i + ‡ t 2- 169 26 · j + ‡ t 2 + 169 26 · k lies on both the cone and the plane. keywords: 002 (part 1 of 1) 10 points Determine f xy when f ( x, y ) = 2 xy ln( xy )- 3 xy . 1. f xy = 2(ln( xy- xy ) + 1 2. f xy = 4(ln( xy ) + xy )- 1 3. f xy = 4(ln( xy )- xy ) + 1 4. f xy = 2ln( xy ) + 1 correct 5. f xy = 4ln( xy )- 1 6. f xy = 2ln( xy )- 1 Explanation: Since ln( xy ) = ln x + ln y , we see that f ( x, y ) = 2 xy (ln x + ln y )- 3 xy . But then, f x = 2 y (ln x + ln y ) + 2 xy x- 3 y = 2 y (ln x + ln y )- y , in which case f xy = 2(ln x + ln y ) + 2 y y- 1 = 2(ln x + ln y ) + 1 , Miller, Kierste – Exam 2 – Due: Oct 31 2007, 5:00 pm – Inst: JEGilbert 2 after differentiating with respect to y . Conse- quently, f xy = 2ln( xy ) + 1 . keywords: partial derivative, mixed partial derivative, log function, 003 (part 1 of 1) 10 points Find the arc length of the curve r ( t ) = (1- 2 t ) i + ln(2 t ) j + (3- t 2 ) k between r (1) and r (3). 1. arc length = 8 + ln3 correct 2. arc length = 3 + ln6 3. arc length = 9 + 2ln3 4. arc length = 8- 2ln3 5. arc length = 6- ln3 6. arc length = 8- ln3 Explanation: The length of a curve r ( t ) between r ( t ) and r ( t 1 ) is given by the integral L = Z t 1 t | r ( t ) | dt . Now when r ( t ) = (1- 2 t ) i + ln(2 t ) j + (3- t 2 ) k we see that r ( t ) =- 2 i + 1 t j + 2 t k ....
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Exam 2 - Miller Kierste – Exam 2 – Due 5:00 pm – Inst...

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