Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 10 - Parametric Equations and Polar Coordinates - Review - Exercises - Page 730: 22



Work Step by Step

Given: $x=t^{3}+6t+1$ and $y=2t-t^{2}$ $\frac{dx}{dt}=3t^{2}+6$ $\frac{dy}{dt}=2-2t$ $\frac{dy}{dx}=\frac{{dy}/{dt}}{{dx}/{dt}}$ $=\frac{2-2t}{3t^{2}+6}$ Let $m$ be the slope of the tangent line to the given curve. $m=\frac{dy}{dx}|_{t=-1}=\frac{2-2t}{3t^{2}+6}|_{t=-1}$ $=\frac{2-2\times (-1)}{3(-1)^{2}+6}$ Hence, $m=\frac{4}{9}$
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