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: $y^2 - x =1$ --------
1550776639

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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.