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: 18

Answer

$-4$

Work Step by Step

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