Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.1 Derivatives and Rates of Change - 2.1 Exercises: 15

Answer

$v(a) = -\frac{2}{a^3}$ $v(1) = -2$ $v(2) = -\frac{1}{4}$ $v(3) = -\frac{2}{27}$

Work Step by Step

First find the derivative of the function $s = \frac{1}{t^2}$ $v(a) = \lim\limits_{h \to 0}\frac{s(a+h)-s(a)}{h}$ => $ \lim\limits_{h \to 0}\frac{\frac{1}{(a+h)^2}-\frac{1}{a^2}}{h}$ => $ \lim\limits_{h \to 0}\frac{\frac{a^2 - (a+h)^2}{a^2(a+h)^2}}{h}$ => $\lim\limits_{h \to 0}\frac{a^2 - (a+h)^2}{ha^2(a+h)^2}$ => $ \lim\limits_{h \to 0}\frac{a^2 - (a^2 +2ah + h^2)}{ha^2(a+h)^2}$ => $ \lim\limits_{h \to 0}\frac{- 2ah - h^2}{ha^2(a+h)^2}$ =>$\lim\limits_{h \to 0}\frac{- 2a - h}{a^2(a+h)^2}$ => $ -\frac{2a}{a^2a^2}$ => $-\frac{2}{a^3}$ $v(a) = -\frac{2}{a^3}$ Now use the derivative of $v(a)$ to find $v(1)$, $v(2)$, and $v(3)$ $v(1) = -\frac{2}{1^3}$ => $-\frac{2}{1}$ => $-2$ $v(2) = -\frac{2}{2^3}$ => $-\frac{2}{8}$ => $-\frac{1}{4}$ $v(3) = -\frac{2}{3^3}$ => $-\frac{2}{27}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.