University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.10 - Related Rates - Exercises - Page 187: 7

Answer

$\frac{dy}{dt} = \frac{-3}{2}$

Work Step by Step

Given that $\frac{dx}{dt} = -2$ and $x^2+y^2=25$ on differentiating with respect to time $\frac{dx^2}{dt} + \frac{d y^2}{dt} = \frac{d25}{dt}$ $2x\frac{dx}{dt} + 2y\frac{dy}{dt} =0 $ when x=3, y=-4 then $ -8\frac{dy}{dt} - 2\times (-2)\times3=0$ $\frac{dy}{dt} = \frac {-3}{2}$ Thus, the answer is $\frac{dy}{dt} = \frac{-3}{2}$
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