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: 9


0.632 m/s

Work Step by Step

Let x be the distance from the bottom of the ladder to the wall and h be the height of its top. The goal is to find $\frac{dh}{dt}$ at $ t=2s $. We are given that $\frac{dx}{dt}=0.8\, m/s $ and $ x(0)=1.5\,m $ Using Pythagoras' theorem, we have $ x^{2}+h^{2}=(5\,m)^{2}$ Differentiating both sides with respect to $ t $, we have $2x\frac{dx}{dt}+2h\frac{dh}{dt}=0$ $\implies \frac{dh}{dt}=-\frac{x}{h}\frac{dx}{dt}=-\frac{x}{h}\times0.8\,m/s $ $ x(2)=x(0)+dt\frac{dx}{dt}=1.5\,m+ 2\,s(0.8\,m/s)=3.1\,m $ $ h(2)=\sqrt {(5\,m)^{2}-(3.1\,m)^{2}}\approx 3.923\,m $ $\frac{dh}{dt}|_{t=2}=-\frac{3.1\,m}{3.923\,m}\times0.8\,m/s=0.632\,m/s $
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