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: 12

Answer

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

Work Step by Step

$\displaystyle \frac{dx}{dt}=-\sin t,\qquad \displaystyle \frac{dy}{dt}=\cos t$ $\displaystyle \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{\cos t}{-\sin t}=-\cot t$ The slope of the tangent line at $t=\pi/2$ is $\left. \displaystyle \frac{dy}{dx} \right|_{t=\pi/2}=0$ With $y'=\displaystyle \frac{dy}{dx}$, we find $\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=\pi/2}$, $\displaystyle \frac{dy'}{dt}=\frac{d}{dt}[-\cot t]=\csc^{2}t$ $\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{dy'/dt}{dx/dt}=\frac{\csc^{2}t}{-\sin t}=-\csc^{3}t$ $\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=\pi/2}=-1$ The point $P$ on the curve corresponding to $t=\pi/2$ $x=\displaystyle \cos\frac{\pi}{2}=0,\quad,y=1+\sin\frac{\pi}{2}=2,$ is $P(0,2)$ Since the tangent line has slope 0, it is horizontal (and passes through P): $y=2$
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