University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 160: 103

Answer

See below for detailed proof.

Work Step by Step

$T=2\pi\sqrt{\frac{L}{g}}$ and $\frac{dL}{du}=kL$ We need to show that $$\frac{dT}{du}=\frac{kT}{2}$$ So applying the Chain Rule, we have $$\frac{dT}{du}=\frac{dT}{dL}\frac{dL}{du}=\frac{dT}{dL}\times kL$$ Now we are left to calculate $dT/dL$: $$\frac{dT}{dL}=T'=\Big(2\pi\sqrt{\frac{L}{g}}\Big)'=\Big(2\pi\frac{L^{1/2}}{\sqrt g}\Big)'=2\pi\frac{(L^{1/2})'}{\sqrt g}$$ $$=2\pi\frac{\frac{1}{2}L^{-1/2}}{\sqrt g}=\pi\frac{1}{\sqrt L\sqrt g}=\frac{\pi}{\sqrt{gL}}$$ Therefore, $$\frac{dT}{du}=\frac{\pi}{\sqrt{gL}}\times kL=k\frac{\pi\sqrt L}{\sqrt g}=k\Big(\pi\sqrt{\frac{L}{g}}\Big)$$ As $T=2\pi\sqrt{\frac{L}{g}}$, $\pi\sqrt{\frac{L}{g}}$ equals $T/2$. $$\frac{dT}{du}=\frac{kT}{2}$$
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