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Unformatted text preview: hernandez (ah29758) M408D Quest Homework 6 pascaleff (54550) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Which one of the following is the graph of the polar function r = 1 2 cos(2 ) ? 1. 2. 3. 4. correct 5. 6. Explanation: To determine the polar graph of r = 1 2 cos(2 ) , look first at its Cartesian graph 2 3 2 2 3 2 1 1 r hernandez (ah29758) M408D Quest Homework 6 pascaleff (54550) 2 Then, by tracing in polar coordinates the value of r as varies, looking especially at the values of where r = 0 and the values of r at = 0 , 2 , , 3 2 , 2 , we obtain the polar graph keywords: polar function, polar graph, sin function, cos function, 002 10.0 points Find the yintercept of the tangent line to the graph of r = 3 e  2 at the point P corresponding to = 0. 1. yintercept = 1 3 correct 2. yintercept = 4 3 3. yintercept = 5 3 4. yintercept = 2 3 5. yintercept = 1 Explanation: The usual pointslope formula can be used to find an equation for the tangent line to the graph of a polar curve r = f ( ) at a point P once we know the Cartesian coordinates P ( x , y ) of P and the slope of the tangent line at P . Now, when r = 3 e  2 , then x ( ) = (3 e  2) cos , while y ( ) = (3 e  2) sin . Thus in Cartesian coordinates, the point P corresponding to = 0 is (1 , 0). On the other hand, x ( ) = 3 e cos  sin (3 e  2) , while y ( ) = 3 e sin + cos (3 e  2) , so the slope at P is given by dy dx vextendsingle vextendsingle vextendsingle =0 = y (0) x (0) = 1 3 . Consequently, by the point slope formula, the tangent line at P has equation y = 1 3 parenleftBig x 1 parenrightBig , and so has yintercept = 1 3 . 003 10.0 points Find the slope of the tangent line to the graph of r = cos 2 at = / 3. 1. slope = 7 3 3 correct 2. slope = 1 5 3 3. slope = 5 3 3 4. slope = 1 7 3 hernandez (ah29758) M408D Quest Homework 6 pascaleff (54550) 3 5. slope = 1 6. slope = 1 Explanation: The graph of a polar curve r = f ( ) can expressed by the parametric equations x = f ( ) cos , y = f ( ) sin . In this form the slope of the tangent line to the curve is given by dy dx = y ( ) x ( ) = f ( ) sin + f ( ) cos f ( ) cos  f ( ) sin . When r = cos 2 , therefore, dy dx = 2 sin 2 sin + cos 2 cos  2 sin 2 cos  cos 2 sin . Thus, at = / 3, dy dx = ( 6 1) / 4 ( 2 3 + 3) / 4 . Consequently, at = / 3, slope = dy dx vextendsingle vextendsingle vextendsingle = / 3 = 7 3 3 ....
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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
 TextbookAnswers

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