Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.7 - Derivatives and Rates of Change - 2.7 Exercises: 15

Answer

The velocity of the particle at time $t=a$ is $v(a)=\frac{-2}{a^3}$ (m/s) The velocity of the particle at time $t=1$ is $v_1=-2$ (m/s) The velocity of the particle at time $t=2$ is $v_2=\frac{-1}{4}$ (m/s) The velocity of the particle at time $t=3$ is $v_3=\frac{-2}{27}$ (m/s)

Work Step by Step

Equation of motion $$s=f(t)=\frac{1}{t^2}$$ * Find the equation for the velocity of the particle at time $t=a$: According to definition, the velocity of the particle at time $t=a$ is $$v(a)=\lim\limits_{h\to0}\frac{f(a+h)-f(a)}{h}$$ $$v(a)=\lim\limits_{h\to0}\frac{\frac{1}{(a+h)^2}-\frac{1}{a^2}}{h}$$ $$v(a)=\lim\limits_{h\to0}\frac{\frac{a^2-(a+h)^2}{a^2(a+h)^2}}{h}$$ $$v(a)=\lim\limits_{h\to0}\frac{a^2-a^2-h^2-2ah}{a^2h(a+h)^2}$$ $$v(a)=\lim\limits_{h\to0}\frac{-h^2-2ah}{a^2h(a+h)^2}$$ $$v(a)=\lim\limits_{h\to0}\frac{-h-2a}{a^2(a+h)^2}$$ $$v(a)=\frac{-0-2a}{a^2(a+0)^2}=\frac{-2a}{a^4}=\frac{-2}{a^3}\hspace{0.1cm}(m/s)$$ *At $t=1$, the velocity of the particle is: $$v_1=\frac{-2}{1^3}=-2\hspace{0.1cm}(m/s)$$ *At $t=2$, the velocity of the particle is: $$v_2=\frac{-2}{2^3}=\frac{-1}{4}\hspace{0.1cm}(m/s)$$ *At $t=3$, the velocity of the particle is: $$v_3=\frac{-2}{3^3}=\frac{-2}{27}\hspace{0.1cm}(m/s)$$
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