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 text(:)`

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 a computer tool — to draw the plots requested.

1. Plot `ftext[(]t text[)]` 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 text(.)`

3. Plot `ftext[(]t text[)]` 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 text(.)` The distance fallen at time `t_2` is `s_2 = text[(]5.5text[)]^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 text(.)` Then

Checkpoint 3

Now we observe that our speed calculation for the falling object does not depend on the specific instant `t_1 = 5 text(.)` 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 text(.)`

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 ${\displaystyle \frac{\Delta s}{\Delta t}}$.
	${\displaystyle ={\displaystyle \frac{c{t}_2^2-c{t}_1^2}{t_2-t_1}}}$	We substituted from our formula for `s text(.)`
	${\displaystyle =\frac{c(t_2-t_1)(t_2+t_1)}{t_2-t_1}}$	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 text(.)` 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 text(.)`

Checkpoint 4