Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.9 Related Rates - Exercises - Page 159: 11


$x=4.737\,m$ $\frac{dx}{dt}=0.405\,m/s$

Work Step by Step

$h(2)=h(0)+ dt\times\frac{dh}{dt}$ $=4\,m+(2\,s\times-1.2\,m/s)=1.6\,m$ As $x^{2}+h^{2}=5^{2}$, $x(2)=\sqrt {(5\,m)^{2}-(1.6\,m)^{2}}=4.737\,m$ Differentiating both sides of the equation $x^{2}+h^{2}=5^{2}$ with respect to $t$, we have $2x\frac{dx}{dt}+2h\frac{dh}{dt}=0$ $\implies \frac{dx}{dt}=-\frac{h}{x}\frac{dh}{dt}$ $\frac{dx}{dt}|_{t=2\,s}=-\frac{1.6\,m}{4.737\,m}\times(-1.2\,m/s)=0.405\,m/s$
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