2.4: From average to instantaneous rate of change

Learning Objectives

1. Demonstrate that a data set with more frequent measurements corresponds to smaller time intervals \(\Delta t\) between data points.
2. Describe the connection between average rate of change over a very small time interval and instantaneous rate of change at a single point.

This section could also be titled "Shrinking the time-steps between measurements." So far, the average rates of change were computed over finite intervals. Our ultimate goal is to refine this idea and define a rate of change at each point, i.e. an instantaneous rate of change. But to do so, we first consider how a data set can be refined by making more frequent measurements to improve the notion of a rate of change close to a given point. We discuss two examples below.

Refined temperature data

We can refine the original data of temperature \(T(t)\) for cooling milk from Figure \(2.2\) by taking more closely spaced time points. Table \(2.4\) provides a sample of the refined data.

The leftmost plot in Figure \(2.9\) shows the original data set with measurements every \(\Delta t=2 \mathrm{~min}\). The second and third plots have successively more refined measurements with shorter intervals between time points ( \(\Delta t=1 \mathrm{~min}\) and \(\Delta t=0.5 \mathrm{~min})\). After 10 minutes, fewer points were collected in each case.

Example 2.7 (Refined average rate of change)

Use the data in Table \(2.4\) to compute the average rate of change of the temperature over the time intervals \(2 \leq t \leq 2+h\) where \(h=\Delta t=2,1,0.5 \mathrm{~min}\), respectively. Which calculation most accurately describes the behavior "close to" \(t=2 \mathrm{~min}\) ?

Table 2.4: Partial data for temperature in \({ }^{\circ} \mathrm{F}\) for the three graphs shown in Figure \(2.9\). The pairs of columns indicate that the data has been collected at more and more frequent intervals \(h=\Delta t\).

Figure 2.9: Three graphs of temperature versus time for cooling milk.

Solution

Computing the ratio \(\Delta T / \Delta t\), we obtain, for \(\Delta t=2,1,0.5\) the following average rates of change (in \({ }^{\circ} \mathrm{F}\) per min):

\[\begin{aligned} & \Delta t=2: \quad \frac{\Delta T}{\Delta t}=\frac{T(2+2)-T(2)}{4-2}=\frac{(164.6-176)}{(4-2)}=-5.7, \\ & \Delta t=1: \quad \frac{\Delta T}{\Delta t}=\frac{T(2+1)-T(2)}{3-2}=\frac{(169.5-176)}{(3-2)}=-6.5, \\ & \Delta t=0.5: \quad \frac{\Delta T}{\Delta t}=\frac{T(2+0.5)-T(2)}{2.5-2}=\frac{(172.9-176)}{(2.5-2)}=-6.2 . \end{aligned} \nonumber \]

The last of these has been calculated over the smallest time interval, and most closely represents the rate of change of temperature close to the time \(t=2\) min. Exercise 2(b) leads to a similar comparison of this sort close to \(t=0\), and results in a similar set of finer values for the average rate of change "near" the initial data point.

Refined data for the height of a falling object

We examine increasingly refined data for the height of a falling object in Figure 2.10.

Figure 2.10(a) shows three stroboscopic images, each giving successive vertical positions of an object falling from a height of \(20 \mathrm{~m}\) over a \(2 \mathrm{~s}\) time period. The location of the ball is given first at intervals of \(\Delta t=0.5\) seconds, then at intervals of \(\Delta t=0.1\) and finally \(\Delta t=0.05 \mathrm{~s}\). In Figure 2.10(b), we graph the height \(Y=Y_{0}-c t^{2}\) against time \(t\). (The distance fallen is still described by the function \(y(t)=c t^{2}\), as in Example 2.3.)

Figure 2.10: Data for the height \(Y\) of a falling object starting with \(t=0\) at the top, and following the object until \(t=2\) at its lowest point. The data set is refined \((\Delta t=0.5,0.2,0.1)\) to get more and more accurate tracking of the object. (a) Stroboscopic images. (b) Graphs of \(Y\) versus \(t\).

Mastered Material Check

10. At what time is the temperature approximately \(120^{\circ} \mathrm{F}\) ?

By collecting data at finer time points, we can determine the "velocity" of the object with greater accuracy. Indeed, taking smaller and smaller time steps leads us to define instantaneous velocity.

Instantaneous velocity

We know that the velocity over an interval can be calculated by finding the slope of a secant line connecting the endpoints of that interval. The slopes of the secant lines in Figure 2.10(b) are steeper at the end of the time interval than at the beginning - lending justification to what we intuitively know: a falling object’s velocity increases as time passes.

With this in mind, to define an instantaneous velocity at some time \(t_{0}\), we compute average velocities over decreasing time intervals \(t_{0} \leq t \leq t_{0}+h\), allowing \(h\) to get smaller.

Note: we use the notation \(h \rightarrow 0\) to denote the shrinking the time interval.

For example, we make the strobe flash faster so that \(\Delta t=\left(t_{0}+h\right)-t_{0}=\) \(h \rightarrow 0\). At each stage, we calculate an average velocity, \(\bar{v}\) for the interval \(t_{0} \leq t \leq t_{0}+h\). As we continue to refine the measurements in this way, we arrive at a value for the velocity that we denote the instantaneous velocity. This number represents "the velocity of the ball at the very instant \(t=t_{0}\) ".

Definition 2.6 (Instantaneous velocity)

The instantaneous velocity at time \(t_{0}\), denoted \(v\left(t_{0}\right)\) is defined as

\[v\left(t_{0}\right)=\lim _{h \rightarrow 0} \bar{v} \nonumber \]

where \(\bar{v}\) is the average velocity over the time interval \(t_{0} \leq t \leq t_{0}+h\). In other words,

\[v\left(t_{0}\right)=\lim _{h \rightarrow 0} \frac{y\left(t_{0}+h\right)-y\left(t_{0}\right)}{h} . \nonumber \]

Mastered Material Check

11. From what height was the object dropped in Figure \(2.10\) ?
12. If you wanted 50 equally spaced data points over a \(2 \mathrm{~s}\) interval, what would \(\Delta t\) be?

A brief summary of average and instantaneous velocity in the example of a falling ball.

We shall be more explicit about the meaning of the notation \(\lim _{h \rightarrow 0}\) in the next chapter.

Example 2.8 (Computing an instantaneous velocity)

Use Gallileo’s formula for the distance fallen, Equation (2.1), \(y(t)=c t^{2}\), to compute the instantaneous velocity of a falling object at time \(t_{0}\).

Solution

We have already found the average velocity of the falling object over a time interval \(t_{0} \leq t \leq t_{0}+h\) in Example 2.6, obtaining Equation (2.3),

\[\bar{v}=c\left(2 t_{0}+h\right) . \nonumber \]

Then, by Definition 2.6,

\[v\left(t_{0}\right)=\lim _{h \rightarrow 0} \bar{v}=\lim _{h \rightarrow 0} c\left(2 t_{0}+h\right)=2 c t_{0} . \nonumber \]

Here we have used the fact that the expression \(c\left(2 t_{0}+h\right)\) approaches \(2 c t_{0}\) as \(h\) shrinks to zero. This result holds for any time \(t_{0}\). More generally, we could write that at time \(t\), the instantaneous velocity is \(v(t)=2 c t\). For example, for \(c=4.9 \mathrm{~m} / \mathrm{s}^{2}\), the velocity of an object at time \(t=1 \mathrm{~s}\) after it is released is \(v(1)=9.8 \mathrm{~m} / \mathrm{s}\).