University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 10 - Section 10.2 - Calculus with Parametric Curves - Exercises - Page 572: 6

Answer

Tangent line: $\quad y=-\displaystyle \frac{1}{2}x-\frac{1}{2}$ $\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=-\frac{\pi}{4}}=\frac{1}{4}$

Work Step by Step

The slope of the tangent line at $t=-\displaystyle \frac{\pi}{4}$ is $\left. \displaystyle \frac{dy}{dx} \right|_{t=-\frac{\pi}{4}}$ $\displaystyle \left\{\begin{array}{l} \frac{dx}{dt}=2\sec^{2}t\tan t,\\ \\ \frac{dy}{dt}=\sec^{2}t \end{array}\right.\quad, \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{\sec^{2}t}{2\sec^{2}t\tan t}=\frac{\mathrm{l}}{2\tan t}=\frac{1}{2}\cot t$ $\left. \displaystyle \frac{dy}{dx} \right|_{t=-\frac{\pi}{4}}=\frac{1}{2}\cot(-\frac{\pi}{4})=-\frac{1}{2}$ With $y'=\displaystyle \frac{dy}{dx}$, we find $\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=-\frac{\pi}{4}}$, $\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{dy'/dt}{dx/dt}=\frac{\frac{d}{dt}[\frac{1}{2}\cot t]}{2\sec^{2}t\tan t}=\frac{\frac{1}{2}(-\csc^{2}t)}{2\sec^{2}t\tan t}=-\frac{\cot^{2}t}{4\tan t}=-\frac{1}{4}\cot^{3}t$ $\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=-\frac{\pi}{4}}=-\frac{1}{4}\cot^{3}(-\frac{\pi}{4})=-\frac{1}{4}(-1)^{3}=\frac{1}{4}$ The point $P$ on the curve corresponding to $t=-\displaystyle \frac{\pi}{4}$ is $\left\{\begin{array}{l} x=\sec^{2}(-\frac{\pi}{4})-1=1\\ \\ y=\tan(-\frac{\pi}{4})=-1 \end{array}\right.\quad, P(1,-1)$ Use the point-slope equation for the tangent line. $y+1=-\displaystyle \frac{1}{2}(x-1)$ $y=-\displaystyle \frac{1}{2}x+\frac{1}{2}-1$ $y=-\displaystyle \frac{1}{2}x-\frac{1}{2}$
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