Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 11: Parametric Equations and Polar Coordinates - Section 11.2 - Calculus with Parametric Curves - Exercises 11.2 - Page 657: 9

Answer

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

Work Step by Step

The point $P$ on the curve corresponding to $t=-1$ is $\left\{\begin{array}{l} x=2(1)+3=5\\ \\ y=1 \end{array}\right.\quad, P(5,1)$ The slope of the tangent line at $t=-1$ is $\left. \displaystyle \frac{dy}{dx} \right|_{t=-1}$ $\displaystyle \left\{\begin{array}{l} \frac{dx}{dt}=4t,\\ \\ \frac{dy}{dt}=4t^{3} \end{array}\right.\quad, \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{4t^{3}}{4t}=t^{2}$ $\left. \displaystyle \frac{dy}{dx} \right|_{t=-1}=1$ Use the point-slope equation for the tangent line. $y-1=1\cdot(x-5)$ $y=x-4$ With $y'=\displaystyle \frac{dy}{dx}$, we find $\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=-1}$, $\displaystyle \frac{dy'}{dt}=\frac{d}{dt}[t^{2}]=2t$ $\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{dy'/dt}{dx/dt}=\frac{2t}{4t}=\frac{1}{2}$ $\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=-1}=\frac{1}{2}$
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.