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: 20



Work Step by Step

Here, we have $\dfrac{dx}{dt}=2(x-t)$ and $\dfrac{dy}{dt}=(e^t)(t+1)$ Here, $x-t=1$ when $t=0$ $\implies \dfrac{dx}{dt}=2$ Also, $ \dfrac{dx}{dt}(t=0)=1$ Thus, Slope: $\dfrac{dy}{dx}=\dfrac{(e^{0})(0+1)}{2(1)}=\dfrac{1}{2}$
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