Question

We are currently using limits in Calculus AP.

Q. Find the function for the velocity where the time-position function is given by s(t) = 2t - t^3 feet. (t is given in minutes.)

Answer 1

So, velocity is the derivative of position. So, if you are using limits, the limit definition of the derivative is \[\lim_{h\to 0}=\frac{s(t+h)-s(t)}{h} \]
So:
\[ \begin{align}&=\lim_{h\to 0}\frac{[2(t+h)-(t+h)^3]-[2t-t^3]}{h}\\&=\lim_{h\to 0} \frac{2t+2h-(t+h)(t+h)^2-2t+t^3}{h}\\&= \lim_{h\to 0}\frac{2t+2h-(t+h)(t^2+2ht+h^2)-2t+t^3}{h}\\&=\lim_{h\to 0}\frac{\cancel{2t}+2h\cancel{-t^3}-2ht^2-th^2-t^2h-2h^2t-h^3\cancel{-2t}\cancel{+t^3}}{h}\\&\lim_{h\to 0}\frac{h(2-2t^2-th-t^2-2ht-h^2)}{h}\\&=\lim_{h\to 0}(2-2t^2-th-t^2-2ht-h^2)\\&=2-2t^2-t^2\\&=2-3t^2
\end{align}\]