Calculus 8th Edition

$v(a) = -\frac{2}{a^3}$ $v(1) = -2$ $v(2) = -\frac{1}{4}$ $v(3) = -\frac{2}{27}$
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 $v(3)$ $v(1) = -\frac{2}{1^3}$ => $-\frac{2}{1}$ => $-2$ $v(2) = -\frac{2}{2^3}$ => $-\frac{2}{8}$ => $-\frac{1}{4}$ $v(3) = -\frac{2}{3^3}$ => $-\frac{2}{27}$