M408D Quest Homework 6-solutions

# M408D Quest Homework 6-solutions - hernandez (ah29758)...

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Unformatted text preview: hernandez (ah29758) M408D Quest Homework 6 pascaleff (54550) 1 This print-out should have 15 questions. Multiple-choice 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 y-intercept of the tangent line to the graph of r = 3 e - 2 at the point P corresponding to = 0. 1. y-intercept =- 1 3 correct 2. y-intercept =- 4 3 3. y-intercept =- 5 3 4. y-intercept =- 2 3 5. y-intercept =- 1 Explanation: The usual point-slope 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 y-intercept =- 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.

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M408D Quest Homework 6-solutions - hernandez (ah29758)...

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