Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 10 - Review - Exercises - Page 690: 21

Answer

$2$

Work Step by Step

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