# Exam 1 - Miller, Kierste Exam 1 Due: Oct 2 2007, 11:00 pm...

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Miller, Kierste – Exam 1 – Due: Oct 2 2007, 11:00 pm – Inst: JEGilbert 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. The due time is Central time. 001 (part 1 oF 1) 10 points Which one oF the Following polar Functions has graph 1. r = sin 3 θ correct 2. r = sin 4 θ 3. r = cos 4 θ 4. r = cos 2 θ 5. r = sin 2 θ 6. r = cos 3 θ Explanation: Consequently, the graph is that oF r = sin 3 θ . keywords: polar graph, six-leaved rose, eight- leaved rose, Four-leaved rose 002 (part 1 oF 1) 10 points Which oF the Following expressions are well- defned For all vectors a , b , c , and d ? I a · ( b × c ) , II | a ( b × c ) , III ( a × b ) · ( c × d ) . 1. I only 2. II and III only 3. III only 4. I and II only 5. II only 6. none oF them 7. I and III only correct 8. all oF them Explanation: The cross product is defned only For two vectors, and its value is a vector; on the other hand, the dot product is defned only For two vectors, and its value is a scalar. ±or the three given expressions, thereFore, we see that I is well-defned because it is the dot prod- uct oF two vectors. II is not well-defned because the frst term in the cross product is a scalar because it is the length oF a vector, not a vector. III is well-defned because both terms in the dot product are cross products, hence vectors. keywords: vectors, dot product, cross prod- uct, T/±, length, 003 (part 1 oF 1) 10 points Which oF the Following sets oF inequalitites describes the region consisting oF all points outside a sphere oF radius 3 centered at the origin and inside a sphere oF radius 6 centered at the origin.

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Miller, Kierste – Exam 1 – Due: Oct 2 2007, 11:00 pm – Inst: JEGilbert 2 1. 9 < x 2 + y 2 + z 2 < 36 correct 2. 3 x 2 + y 2 + z 2 6 3. 3 < x 2 + y 2 + z 2 < 6 4. 3 x 2 + y 2 + z 2 < 6 5. 9 x 2 + y 2 + z 2 36 6. 9 < x 2 + y 2 + z 2 36 Explanation: The region consists of all points whose dis- tance to the origin is greater that 3, but less than 6, i.e , all points satisfying the inequali- ties 3 < p x 2 + y 2 + z 2 < 6 , which can also be written as 9 < x 2 + y 2 + z 2 < 36 . keywords: 004 (part 1 of 1) 10 points Find d 2 y dx 2 for the curve given parametrically by x ( t ) = 3 + t 2 , y ( t ) = t 2 + 2 t 3 . 1. d 2 y dx 2 = 3 t 2. d 2 y dx 2 = 2 t 3 3. d 2 y dx 2 = 5 6 t 4. d 2 y dx 2 = 3 t 2 5. d 2 y dx 2 = 5 t 6 6. d 2 y dx 2 = 3 2 t correct Explanation: Di±erentiating with respect to t we see that x 0 ( t ) = 2 t, y 0 ( t ) = 2 t + 6 t 2 . Thus dy dx = y 0 ( t ) x 0 ( t ) = 2 t + 6 t 2 2 t = 1 + 3 t. On the other hand, by the Chain Rule, d dt dy dx · = dx dt n d dx dy dx ·o = dx dt · d 2 y dx 2 . Consequently,
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## This note was uploaded on 05/08/2008 for the course M 408 M taught by Professor Gilbert during the Fall '07 term at University of Texas at Austin.

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Exam 1 - Miller, Kierste Exam 1 Due: Oct 2 2007, 11:00 pm...

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