Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.9 - Related Rates - 3.9 Exercises: 10

Answer

a. $\frac{dx}{dt} = \frac{-\sqrt 5}{4}$ b. $\frac{dy}{dt} = \frac{4\sqrt 5}{5}$

Work Step by Step

a. $\frac{d(4x^2 + 9y^2)}{dt} = \frac{d(36)}{dt}$ $8x \times \frac{dx}{dt} + 18y \times \frac{dy}{dt} = 0$ $8(2) \times \frac{dx}{dt} + 18(\frac{2\sqrt 5}{3}) \times \frac{dy}{dt} = 0$ $16 \times \frac{dx}{dt} + 18(\frac{2\sqrt 5}{3}) \times \frac{1}{3} = 0$ $16 \times \frac{dx}{dt} + 4\sqrt 5 = 0$ Solve for $\frac{dx}{dt}$: $16 \times \frac{dx}{dt} = -4\sqrt 5$ $\frac{dx}{dt} = \frac{-4\sqrt 5}{16}$ $\frac{dx}{dt} = \frac{-\sqrt 5}{4}$ b. $8x \times \frac{dx}{dt} + 18y \times \frac{dy}{dt} = 0$ $8(-2)\frac{dx}{dt} + 18(\frac{2\sqrt 5}{3}) \times \frac{dy}{dt} = 0$ $-16(3) + 12\sqrt 5 \times \frac{dy}{dt} = 0$ $-48 + 12\sqrt 5 \times \frac{dy}{dt} = 0$ Solve for $\frac{dy}{dt}$ $\frac{dy}{dt} = \frac{48}{12\sqrt 5}$ $\frac{dy}{dt} = \frac{4}{\sqrt 5}$ $\frac{dy}{dt} = \frac{4\sqrt 5}{5}$
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