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Rate of change

A particle moves on a line. The variable x gives its position. A point on the line is chosen as the origin, x = 0, with the other points marked off in meters.

The particle's motion is described by a function, x, as a function of time, t: x(t).That is, for each value of time t, the function gives us a unique value of position x.

Let us consider a small time interval \(\Delta t = t_2 – t_1\). At t1 the particle is at x1 and at t2 it is at x2.

\(\Delta x = x_2 – x_1\) is defined as the displacement of the particle.

Note that displacement means change in position. It can be negative as well as positive. If the final position is to the left of the initial position, \(\Delta x < 0\). Furthermore, the magnitude of \(\Delta x\) is not the distance traveled, since in between the two times the particle could have gone to the edge of the galaxy and come back: we do not know.

The quantity \(\frac{\Delta x}{\Delta t}\) is defined as the average velocity during the interval t1 to t2.

The average velocity is negative if the final position x2 is to the left of the initial position x1.

A particle moves so that its position x is given as a function of time t as x(t) = t2.

What is the average velocity of the particle between t = 2 and t = 2.5 sec?

\[ x(2) = 4.0 \\ x(2.5) = 6.25\]

Therefore

\[\Delta x = 2.25 \\ \Delta t = 0.5 \\ V_{ave} = \frac{\Delta x}{\Delta t} = 4.5 m/s\]

This gives us the change in x divided by the change in time. In other words, it gives us the rate of change of position with time.

Note that it gives us only the average velocity for the 0.5 sec interval considered. The particle need not have maintained a constant velocity during that interval.

Let us reduce the time interval, keeping the initial value t1 = 2. Let t2 = 2.01 sec. Then x2 = 4.0401.

\[\Delta x = 0.0401 \\ \Delta t = 0.01 \\ \frac{\Delta x}{\Delta t} = 4.01 \]

4.01 m/s is the average velocity during the 0.01 sec time interval.

Let us reduce the time interval further. Let t2 = 2.0001 sec. Then x2 = 4.00040001. This gives us

\[\Delta x = 0.00040001 \\ \Delta t = 0.0001 \\ \frac{\Delta x}{\Delta t} = 4.0001 m/s\]

We can see (intuitively) that as \(\Delta t\) gets smaller and smaller (approaches zero), \(\frac{\Delta x}{\Delta t}\) approaches 4.

\[\lim_{\Delta t\to0} \frac{\Delta x}{\Delta t} = 4\]

We have not proved it, but take it as reasonable for now.

The above limit is defined as the instantaneous velocity at time t = 2 sec. It is the instantaneous rate of change of position with time at the instant t = 2.

We have a shorthand for

\[\lim_{\Delta t\to0} \frac{\Delta x}{\Delta t}\]

We write it as

\[\frac{dx}{dt}\]

This is not to be interpreted as dx divided by dt, but as the limit of \(\frac{\Delta x}{\Delta t}\) as \(\Delta t\) goes to zero.

Let us now prove that the limit is in fact, 4.

We wish to find the instantaneous velocity dx/dt at some t. At t, x = t2. A little time \(\Delta t\) later

\[t = t + \Delta t, x = (t + \Delta t)^2 \\ \Delta x = (t + \Delta t)^2 - t^2 = 2t\Delta t + (\Delta t)^2 \\ \frac{\Delta x}{\Delta t} = 2t + \Delta t\]

And, therefore

\[\lim_{\Delta t\to0} \frac{\Delta x}{\Delta t} = 2t\]

We have a formula now for dx/dt, the instantaneous velocity, at any time t:

\[v = \frac{dx}{dt} = 2t\]

At t = 2, v = 4. QED.

Another example. Consider a sphere of radius r. Its volume is

\[V = \frac{4\pi}{3r^3}\]

The volume is a function of the radius. Suppose that we want to find the rate of change of the volume with radius.

You can see that when we change the radius from one unit to two, there is a certain volume change and when we change the radius from two to three, there is a larger volume change. The rate at which the volume changes with the radius is not constant. It increases with the radius.

Perhaps you can see that from looking at a sphere. The larger the radius, the larger is the increase in the volume for unit change in the radius.

So we ask the question, what is the rate of change when the radius is r1, or r2, or some other specific value. That is, we want the instantaneous rate of change at some value of r.

This rate is given by

\[\frac{dV}{dr} = \lim_{\Delta r\to0} \frac{\Delta V}{\Delta r}\]

To find this, we proceed as follows: At r, the volume is

\[V = \frac{4\pi}{3r^3}\]

At \(r + \Delta r\) the volume is

\[V = \frac{4\pi}{3(r + \Delta r)^3}\]

The change in the volume is

\[\Delta V = \frac{4\pi}{3} (3r^2\Delta r + 3r(\Delta r)^2 + (\Delta r)^3) \\ \frac{\Delta V}{\Delta r} = \frac{4\pi}{3} (3r^2 + 3r\Delta r + (\Delta r)^2)\]

Therefore

\[\frac{dV}{dr} = \lim_{\Delta r\to0} \frac{\Delta V}{\Delta r} = \frac{4\pi}{3} 3r^2 = 4\pi r^2\]

Not surprisingly, it is an increasing function of r. We can now substitute any value of r into \(4\pi r^2\) to find the instantaneous rate of change of the volume with radius at that value of r. Note that instantaneous does not refer to an instant of time. It means at a specific value of r.

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