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: 16


$\dfrac{10\sqrt 3}{9}$

Work Step by Step

Take the derivative of the given equations and isolate the variables. $ \dfrac{dx}{dt}=(\dfrac{-1}{2 \sqrt t})\dfrac{1}{2 \sqrt {5-\sqrt t}}$ and $(\dfrac{\sqrt t}{t-1})+(t-1)\dfrac{dy}{dt}=\dfrac{1}{2\sqrt t}$ At $t=4$, we have $ \dfrac{dx}{dt}=\dfrac{-1}{8\sqrt {3}}$ and $ \dfrac{dx}{dt}=\dfrac{-5}{36}$ Now, Slope: $\dfrac{dy}{dx}=\dfrac{\dfrac{-5}{36}}{\dfrac{-1}{8\sqrt {3}}}=\dfrac{10\sqrt 3}{9}$
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