Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.8 - Related Rates - Exercises 3.8 - Page 161: 9



Work Step by Step

Let us differentiating $L$ with respect to t, then $\dfrac{dL}{dt}=\dfrac{1}{2}\dfrac{1}{\sqrt{(x^2+y^2)}}(2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt})$ Since,$L=\sqrt{x^2+y^2}$,$\dfrac{dx}{dt}=-1$and $\dfrac{dy}{dt}=3$ so, $\dfrac{dL}{dt}=\dfrac{1}{\sqrt{(x^2+y^2)}}[(x(-1)+y(3))]$ Plug $x=5$ and $y=12$ Thus, $\dfrac{dL}{dt}=\dfrac{1}{\sqrt{(5^2+12^2)}}(5(-1)+12(3))=\frac{31}{\sqrt{169}}=\dfrac{31}{13}$
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