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 162: 17

Answer

${\dfrac{ds}{dt}}={\dfrac{x}{\sqrt{(x^2+y^2)}}({\dfrac{dx}{dt}})}$ ${\dfrac{ds}{dt}}={\dfrac{1}{\sqrt{(x^2+y^2)}}(x{\dfrac{dx}{dt}+y{\frac{dy}{dt}}})}$ and ${\dfrac{dx}{dt}}={\dfrac{-y}{x}\dfrac{dy}{dt}}$

Work Step by Step

Given: $s={\sqrt{(x^2+y^2)}}$a) Now, take $y$ as constant and differentiating $s$, we get: Thus, ${\dfrac{ds}{dt}}={\frac{1}{2}(x^2+y^2)^{\frac{-1}{2}}2x{\frac{dx}{dt}}}$ ${\dfrac{ds}{dt}}={\frac{x}{\sqrt{(x^2+y^2)}}({\frac{dx}{dt}})}$ Now, differentiating $s$ for both variables $x,y$, we have ${\dfrac{ds}{dt}}={\frac{1}{2}(x^2+y^2)^{\frac{-1}{2}}(2x{\frac{dx}{dt}}+2y{\frac{dy}{dt}})}={\frac{1}{\sqrt{(x^2+y^2)}}(x{\frac{dx}{dt}+y{\frac{dy}{dt}}})}$ or, $0={\dfrac{1}{\sqrt{(x^2+y^2)}}(x{\frac{dx}{dt}+y{\frac{dy}{dt}}})}$ or, ${\dfrac{dx}{dt}}={\dfrac{-y}{x}\frac{dy}{dt}}$ Hence, ${\dfrac{ds}{dt}}={\dfrac{x}{\sqrt{(x^2+y^2)}}({\dfrac{dx}{dt}})}$ ${\dfrac{ds}{dt}}={\dfrac{1}{\sqrt{(x^2+y^2)}}(x{\dfrac{dx}{dt}+y{\frac{dy}{dt}}})}$ and ${\dfrac{dx}{dt}}={\dfrac{-y}{x}\dfrac{dy}{dt}}$
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