Average and instantaneous velocity

Before we start talking about instantaneous rate of change, let's talk about average rate of change. A simple example is average velocity. If you drive 180 miles in 3 hours, then your average speed is 60 mph. We get this by dividing the distance traveled by the time: $$ v_{avg} = \frac{\Delta s}{\Delta t},$$ where $\Delta s$ is the distance traveled and $\Delta t$ is the time elapsed. We use the Greek letter $\Delta$ to mean "change in". If we start at position $s(t_0)$ at time $t_0$ and end up at position $s(t_1)$ at time $t_1$, then \begin{eqnarray*} \Delta s & = & s(t_1) - s(t_0), \cr \Delta t & = & t_1 - t_0. \end{eqnarray*} If you plot position against time on a graph, $v_{avg}$ is the slope of a secant line.

To get the instantaneous veloctity at a particular time $t=a$, we average over shorter and shorter time intervals. That is, we compute the average velocity between time $a$ and $a + \Delta t$, and then take a limit as $\Delta t \to 0$. On a graph, this is taking the slope of secant lines between points that are getting closer and closer. In the limit, we get the slope of a tangent line.

Whether we think in terms of velocity or slope, we get a limit: $$\lim_{\Delta t \to 0} \frac{s(a+\Delta t)-s(a)}{\Delta t}.$$ This can also be written as: $$\lim_{x \to a} \frac{s(x)-s(a)}{x-a},$$ where $x=a+\Delta t$. This quantity (if the limit exists) is called the derivative of $s(t)$ at time $t=a$.

In the animations below, \(a=3\), the \(x\) coordinate of point \(P\); and \(a+\Delta t\) is the \(x\)-coordinate of point \(Q\). The red tangent line is the limit of the blue secant lines. Note that $\Delta t$ can be positive (first animation) or negative (second animation). For the derivative to exist, the limits as $\Delta t \to 0^+$ and $\Delta t \to 0^-$ must give the same answer.

(Animations from Barry McQuarrie)