Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.4 Exercises - Page 139: 112

Answer

(a) $$V=\frac{10,000}{\sqrt{1+t}}$$ (b) Rate of depreciation, 1767.77 (c) Rate of depreciation, 625

Work Step by Step

Step-1: It is given that $V \propto \frac{1}{\sqrt{1+t}}$ $$\therefore V=\frac{k}{\sqrt{1+t}}$$ where $k$ is the constant of proportionality. Step-2: It is also given that the initial value of $V$ is $10,000$. Thus, at $t=0, V=10,000\implies k=10,000$. Thus, $$V=\frac{10,000}{\sqrt{1+t}}$$ Step-3: Rate of depreciation, $$\frac{dV}{dt}=(10,000)\cdot \bigg(-\frac{1}{2}\bigg)\cdot(1+t)^{-3/2}$$ $$\frac{dV}{dt} = -\frac{5000}{(1+t)^{3/2}}$$ Step-4: At $t=1$, $$\frac{dV}{dt}=-\frac{5000}{(1+1)^{3/2}}=-1767.77$$ Step-5: At $t=3$, $$\frac{dV}{dt} = -\frac{5000}{(1+3)^{3/2}}=-625$$
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