2.2.3 Algebraic Calculation of Rates of Change

Using a high-tech graphing tool, supplemented by guessing, we have conjectured a formula for the instantaneous speed of a falling object whose position at time \(t\) is \(s = ct^2\):

Our next approach is to use a low-tech tool: algebra. That's harder than looking at computer-drawn pictures but also, as we shall see, more satisfying, because there will not be any guessing at formulas.

On the preceding pages we used time intervals centered on \(t = 5\), the time of interest. For purposes of our algebraic calculation, we will use intervals that have \(t_1 = 5\) and \(t_2 =\)some other time. This has the advantage that only one of the two time values changes when we change \(\Delta t\).

Activity 4   Use your graphing tool — a graphing calculator or our Graph tool— to draw the plots requested, and do the follow-up calculations.

1. Plot \(f(t)=4.90 t^2\) from \(t_1 = 5\) to \(t_2 = 5.5\).

2. Verify that the average speed (rate of change of distance) in (a) is \(51.45 \).

   

3. Plot \(f(t)\) from \(t_1 = 5\) to \(t_2 = 5.05\).

4. Calculate the average speed in (c).

We can use algebra to make these calculations for a general \(c\). With \(t_1 = 5\), we consider first the small change in time, \(\Delta t = 0.5\) or \(t_2 = 5.5 \). The distance fallen at time \(t_2\) is \(s_2 = (5.5)^2c = 30.25 c\) meters. Thus the average speed from \(t_1\) to \(t_2\) is

Check to see that this agrees with part (b) of Activity 4 when \(c = 4.9\).

This only approximates the speed at \(t_1\), but we get a closer approximation by choosing a smaller \(\Delta t\), namely \(\Delta t = 0.05\). Then

Checkpoint 3

Now we observe that our speed calculation for the falling object does not depend on the specific instant \(t_1 = 5\). Indeed, the whole process often turns out to be easier if we do it algebraically (for an arbitrary but unspecified \(t_1\)) rather than arithmetically, as we show in the next example.

Example 1

1. Calculate \(\Delta s / \Delta t\) algebraically for \(s = ct^2 \).

Solution

Algebraic Step		Reason
${\displaystyle \frac{\Delta s}{\Delta t}}$	$={\displaystyle \frac{s_2-s_1}{t_2-t_1}}$	This is the definition of ${\displaystyle \frac{\Delta s}{\Delta t}}$ .
	${\displaystyle ={\displaystyle \frac{c{t}_2^{\hspace{0.17em}2}-c{t}_1^{\hspace{0.17em}2}}{t_2-t_1}}}$	We substituted from our formula for \(s\).
	${\displaystyle =\frac{c(t_2-t_1)(t_2+t_1)}{t_2-t_1}}$	We factored the numerator.
	${\displaystyle =c(t_2+t_1)}$	We canceled the factor ${\displaystyle t_2-t_1}$.
	${\displaystyle =c(t_1+\Delta t+t_1)}$	We substituted for ${\displaystyle t_2}$.
	${\displaystyle =c(2t_1+\Delta t)}$	We collected terms.

2. What happens to the result as \(\Delta t\) becomes smaller and smaller?

Solution

Now we can see clearly what happens as \(\Delta t\) becomes smaller and smaller. The \(\Delta t\) term in our computed average speed disappears, and the average speed approaches \(2ct\). That agrees with our result when \(t_1\) is \(5\) (Checkpoint 1), and it should also confirm your conjectured formula in Activity 3, since it tells us the instantaneous speed at every instant \(t_1\).

Checkpoint 4