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Unformatted text preview: Gauthier, Joseph – Exam 2 – Due: Oct 31 2007, 5:00 pm – Inst: JEGilbert 1 This printout should have 18 questions. Multiplechoice 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 A line ‘ passes through the point P (4 , 4 , 3) and is parallel to the vector h 1 , 2 , 3 i . At what point Q does ‘ intersect the xy plane? 1. Q (0 , 3 , 6) 2. Q (5 , 6 , 0) 3. Q (0 , 6 , 5) 4. Q (3 , 2 , 0) correct 5. Q (0 , 5 , 2) 6. Q (2 , 3 , 0) Explanation: Since the xyplane is given by z = 0, we have to find an equation for ‘ and then set z = 0. Now a line passing through a point P ( a, b, c ) and having direction vector v is given parametrically by r ( t ) = a + t v , a = h a, b, c i . But for ‘ , a = h 4 , 4 , 3 i , v = h 1 , 2 , 3 i . Thus r ( t ) = h 4 + t, 4 + 2 t, 3 + 3 t i , so z = 0 when t = 1. Consequently, ‘ intersects the xyplane at Q (3 , 2 , 0) . keywords: line, parametric equations, direc tion vector, point on line, intercept, coordi nate plane 002 (part 1 of 1) 10 points Evaluate the integral I = Z 1 a ( t ) · b ( t ) dt when a ( t ) = 2 e t 2 i + 3 j ln(1 + t ) k and b ( t ) = 4 t i + 3 j 2 k . 1. I = 8 e 3 + 2ln2 2. I = 4 e 3 + 4ln2 3. I = 8 e + 3 2ln2 4. I = 4 e + 3 + 4ln2 correct 5. I = 4 e + 3 2ln2 6. I = 8 e 3 4ln2 Explanation: When a ( t ) = 2 e t 2 i + 3 j ln(1 + t ) k and b ( t ) = 4 t i + 3 j 2 k we see that a · b = 8 te t 2 + 9 + 2 t ln(1 + t ) . Thus I = Z 1 { 8 te t 2 + 9 + 2 t ln(1 + t ) } dt . Now, by substitution, Z 1 8 te t 2 dt = h 4 e t 2 i 1 = 4( e 1) , Gauthier, Joseph – Exam 2 – Due: Oct 31 2007, 5:00 pm – Inst: JEGilbert 2 while Z 1 9 dt = h 9 t i 1 = 9 , and Z 1 2ln(1 + t ) dt = 2 h (1 + t )ln(1 + t ) (1 + t ) i 1 = 2(2ln 2 1) , using integration by parts. Consequently, I = 4 e + 3 + 4ln 2 . keywords: vector function, definite integral, exp function, log function, integration by parts, substitution 003 (part 1 of 1) 10 points Which one of the following is the graph of the 3 Dcurve given parametrically by x = cos t , y = sin t , z = sin 4 t ? 1. x z y correct 2. x z y 3. x z y 4. x z y 5. x z y 6. x z y Explanation: Gauthier, Joseph – Exam 2 – Due: Oct 31 2007, 5:00 pm – Inst: JEGilbert 3 Consequently, the graph of the 3 Dcurve x = cos t , y = sin t , z = sin4 t is x z y keywords: 3Dcurve, helix, circular cylinder cylinder, cone , 3Dgraph 004 (part 1 of 1) 10 points Which one of the following equations has graph x z y 1. y 2 x 2 z 2 = 1 2....
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 Fall '07
 Gilbert
 Derivative, Cos, Gauthier

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