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: 19



Work Step by Step

Take the derivative of the given equations and isolate the variables. $\dfrac{dx}{dt}=3t^2+1$ and $\dfrac{dy}{dt}+6t^2=2\dfrac{dx}{dt}+2t$ Then $ \dfrac{dy}{dt}=2t+2$ Now, Slope: $\dfrac{dy}{dx}=\dfrac{2t+2}{3t^2+1}=\dfrac{2(1)+2}{3(1)^2+1}=1$
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