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: 12



Work Step by Step

Using Pythagoras' theorem, we have $x^{2}+h^{2}=5^{2}$ Differentiating with respect to $t$, we have $2x\frac{dx}{dt}+2h\frac{dh}{dt}=0$ But we are given that the top and bottom of the ladder move at the same speed. That is, $\frac{dh}{dt}=-\frac{dx}{dt}$ which gives $2x\frac{dx}{dt}+2h(-\frac{dx}{dt})=0$ Or $2x\frac{dx}{dt}=2h\frac{dx}{dt}$ $\implies x=h$
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