Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - Chapter Review Exercises - Page 638: 14


$$ \frac{d y}{d x}= \frac{-\sin \theta}{2\tan\theta \sec^2\theta} $$ and at $\theta=\pi/4$, we have $$ \frac{d y}{d x}= -\frac{\sqrt 2}{8} . $$

Work Step by Step

Since $x=\tan^2\theta$ and $y=\cos\theta$, then we have $$ \frac{d y}{d x}=\frac{y^{\prime}(t)}{x^{\prime}(t)}=\frac{-\sin \theta}{2\tan\theta \sec^2\theta} $$ and at $\theta=\pi/4$, we have: $$\frac{d y}{d x}=\frac{-\frac{\sqrt{2}}{2}}{2\times1\times(\sqrt{2})^2}$$ Which simplifies to: $$ \frac{d y}{d x}= -\frac{\sqrt 2}{8}. $$
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