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: 15



Work Step by Step

Take the derivative of the given equations and isolate the variables. $3x^2\dfrac{dx}{dt}+4t=0 \implies \dfrac{dx}{dt}=\dfrac{-4t}{3x^2}$ and $6y^2\dfrac{dy}{dt}-6t=0 \implies \dfrac{dy}{dt}=\dfrac{t}{y^2}$ Now, $\dfrac{dy}{dx}=\dfrac{-3x^2}{4y^2}$ Plug the given values of the $x$ and $y$. we get $x^3+2(2)^2 =9 \implies x=1$ and $2y^3-3(2)^2 =4 \implies y=2$ Thus, $\dfrac{dy}{dx}=\dfrac{-3(1)^2}{4(2)^2}=\dfrac{-3}{16}$
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