Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 10 - Parametric Equations and Polar Coordinates - 10.1 Exercises - Page 665: 18

Answer

Cartesian equation: $y^2 - x =1$ --------

Work Step by Step

(a) $x= tan^2 \theta$ $-x = -tan^2 \theta$ $y = sec \theta$ $y^2 = sec^2 \theta$ Adding both equations: $y^2 - x =sec^2 \theta - tan^2 \theta$ $y^2 - x = \frac{1}{cos^2 \theta} - \frac{sin^2 \theta}{cos^2 \theta} = \frac{1 - sin^2 \theta}{cos^2 \theta} $ $y^2 -x = \frac{cos^2 \theta}{cos^2 \theta}$ $y^2 - x =1$ ----- (b) 1. Plot points determined by values for $\theta$ from $\frac{-\pi}{2}$ to $\frac {\pi} 2$ 2. Join them to produce a curve. 3. Draw an arrow indicating which direction the curve goes from $\theta = \frac{-\pi}{2}$ to $\theta = \frac{\pi}{2}$ ** Notice, from $\frac{-\pi} 2$ to 0, the curve goes from the right to the left, and between $0$ and $\frac{\pi}{2}$, it goes from left to right.
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