2.2.2 Graphical Calculation of Rates of Change

In Activity 1 we used the graphical viewpoint to determine instantaneous speed (at two particular instants) from the position function for a falling object. We also learned something that will enable us to use our graphing tool more effectively. We saw that we could get very good approximations to the rates of change at the instants \(t = 2\) and \(t = 5\) by calculating difference quotients of the form

${\displaystyle {\displaystyle \frac{\Delta s}{\Delta t}}}$

(that is, a slope, a rise-over-run) with a small value of \(\Delta t\).

Next we will do that all at once for every value of \(t\). That should give us a good approximation to a function whose value at each \(t\) is the instantaneous rate of change at \(t\). We write

${\displaystyle f\left(t\right)=c{t}^2}$

in order to have a name for the position function for the falling object, i.e.,

${\displaystyle s=f\left(t\right)}$,

where \(s\) is position at time \(t\). We write \(g\)(\(t\)) for the difference quotient with a suitably small \(\Delta t\), say, \(\Delta t = 0.001\):

${\displaystyle {\displaystyle g\left(t\right)=\frac{\Delta s}{\Delta t}=\frac{f(t+0.001)-f\left(t\right)}{0.001}}}$.

Note 2 – Rates of Change

We'll continue to ignore the fact that we actually know a value for the constant \(c\). Instead, we'll study the whole family of functions of the form \(f(t) = kt^2\).

Activity 2   This activity requires that you use a graphing tool. A graphing calculator will work fine, or use the Graph button below.

1. Graph the distance function \(s\) and the approximate speed function \(g\) for \(k = 1\). The graph of \(g\) has a familiar shape; what is it?

2. Graph the distance function \(s\) and the approximate speed function \(g\) for \(k = 1 / 2\) and \(k = -1 / 2\). Again, the graphs of \(g\) have familiar shapes; what are they?

Checkpoint 2

Activity 3

1. For each of the following functions, $$find a formula for the speed function.

• $f\left(t\right)=t^2$

• $f\left(t\right)=\frac{1}{2}{t}^2$

• $f\left(t\right)=-\frac{1}{2}{t}^2$

• $f\left(t\right)=1.8t^2$

2. If \(f(t) = kt^2\) gives the position at time \(t\) for an object moving in a straight line, what is your best guess at a formula for the instantaneous speed of the object at time \(t \)?

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