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: 17



Work Step by Step

Take the derivative of the given equations and isolate the variables. $ \dfrac{dx}{dt}=\dfrac{2t+1}{1+3x^{1/2}}$ and $\dfrac{y}{2\sqrt {t+1}}+(\sqrt{t+1}+(t/\sqrt y))\dfrac{dy}{dt}+2 \sqrt y=0$ At $t=0$, we have $x=0$ Then $ \dfrac{dx}{dt}=1$ At $t=0$, we have $y=4$ Then $ \dfrac{dx}{dt}=-6$ Now, Slope: $\dfrac{dy}{dx}=\dfrac{-6}{1}=-6$
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