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. Using the familiar d=rt, where rate
means the average velocity, we see that r=dt. We
will use: vavg=ΔsΔt,
where Δs
is the distance traveled and Δt is the time elapsed. We use
the Greek letter Δ to mean "change in". If we start at
position s(t0) at time t0 and end up at position s(t1) at
time t1, then Δs=s(t1)−s(t0),Δt=t1−t0.
If you
plot position against time on a graph, vavg is the slope of a secant line.
To get the instantaneous velocity
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+Δt, and
then take a limit as Δt→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:
limΔt→0s(a+Δt)−s(a)Δt.
This can also be written as: limt→as(t)−s(a)t−a,
where t=a+Δt. 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+Δx is the x-coordinate of point
Q. The movement is the point Q approaching the point
P, which means 3+Δx is getting closer to 3, which
means Δx is getting closer to 0.
The red tangent line (at P) is the limit of the blue secant
lines through P and Q. Note that Δx can be positive
(first animation) or negative (second animation). For the
derivative to exist, the limits as Δx→0+ and Δx→0− must give the same answer.