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: 16



Work Step by Step

Let the distance between the automobile and the farmhouse be $h$ and the distance traveled by the automobile past the intersection of the highway and the road be $x$. Then, using Pythagoras' theorem, we have $2^{2}+x^{2}= h^{2}$ When $x=6\,km$, $h=\sqrt {2^{2}+6^{2}}= 2\sqrt {10}\,km$ Given: $\frac{dx}{dt}=80\,km/h$ Differentiating $2^{2}+x^{2}=h^{2}$ with respect to $t$, we obtain $2x\frac{dx}{dt}=2h\frac{dh}{dt}$ $\implies \frac{dh}{dt}=\frac{x}{h}\frac{dx}{dt}$ $=\frac{6}{2\sqrt {10}}\times80\,km/h=24\sqrt {10}\,km/h\approx75.89\,km/h$
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