Rate of Change

Throw a ball. Mark where it is, wait, and mark where it is again; divide the change in height by the wait. Now shorten the wait. The motion is the subject; the graph records what happened.

a thrown balla traina buoy on a swell

the instant, t₀0.00 s · 1.25 m

play the motiontake the limit hereat every instantthen of the velocityplay the demonstration

$$\frac{s(t_0 + \Delta t) - s(t_0)}{\Delta t}$$

$$v(t_0) = \lim_{\Delta t \to 0} \frac{s(t_0 + \Delta t) - s(t_0)}{\Delta t}$$

$$s(t) = 1.25 + 9.8\,t - 4.9\,t^2 \qquad v(t) = 9.8 - 9.8\,t$$

the wait, Δt

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the quotient, Δs / Δt

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what the limit returned

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Drag the instant along the graph, or use the slider, then take the limit there. The position at this instant is 1.25 m; how fast it is changing is what the limit will return.

Every number here is two positions and the wait between them. The lens is a uniform zoom around the chosen instant, so slopes are preserved: as the wait halves, the curve inside it straightens and the chord settles on the tangent. The arrow appears only where the limit has been taken; the same scale is used for it on the rail and on the velocity graph. The value returned is the module’s local rate, computed from positions on both sides of the instant; the rule at the right is printed only for comparison.

The instrument plays the throw first. The dot rises and falls on the rail at the left, and the graph records its height at each moment: the curve is a record of the motion, not another path for the ball. Then it goes back to the start, pauses at an instant, and takes the limit there.

Taking the limit is the operation this page is about. Two marks: A at the chosen instant $t_0$, and B a wait $\Delta t$ later. The change in height between them, divided by the wait, is the average velocity over that wait; on the graph it is the slope of the chord through the two marks. That number answers a question about an interval, not about an instant. So the lens halves the wait, fourteen times. B slides toward A, the zoom follows it in, the curve inside the lens straightens, and the quotient’s digits stop changing from the left. What the lens returns is the one number the quotients settle on. That number is the velocity at $t_0$, the derivative, and the chord it came from is now the tangent.

$$v(t_0) = \lim_{\Delta t \to 0} \frac{s(t_0 + \Delta t) - s(t_0)}{\Delta t}$$

Read the formula the way the instrument runs it. The fraction is the quotient at some wait. $\lim_{\Delta t \to 0}$ says: shorten the wait and take what the quotient settles on. The result is written $v(t_0)$, a value at the instant, produced by an operation that never puts the wait at zero.

▸ When the two marks meet

At exactly the same time, the changes in position and time are both zero. Their quotient $0/0$ has no value. But the derivative is the limit of the nearby quotients, not that quotient evaluated at zero. The tangent and the instantaneous velocity do not disappear when the marks meet.

The wait can shrink toward $t_0$ from before it or from after it. For a derivative to exist at an interior point, both approaches must settle on the same number. The instrument computes both sides and returns a value only when they agree; the mathematical definition asks for agreement to any precision, using sufficiently small nonzero waits.

The value comes back as an arrow on the dot. Its length is the speed and its direction is the way the dot is moving, and the same arrow, at the same scale, stands on the zero line of the velocity graph at that instant. Choose at every instant and the operation runs at each time in turn, so the dot carries its arrow everywhere; play the motion and watch the arrow shrink on the way up, vanish at the top, flip, and grow on the way down. The velocity graph is that arrow’s length recorded over time, the way the position graph is the dot’s height recorded over time. Nothing on it is looked up; every point is a quotient run to its limit. For the ball it is a straight line, falling by 9.8 m/s every second.

Then of the velocity runs the operation once more, on the velocity graph, and the result is a second arrow drawn from the first one’s tip: how fast the arrow itself is changing. For the ball it points down with the same length throughout the flight, including at the top where the velocity passes through zero — the acceleration is $-9.8\,\mathrm{m/s^2}$ the whole way.

Switch to the train or the buoy. The train’s position is distance along its track; the buoy’s is height above still water, the sine wave of an earlier page made a motion. The motion changes; the operation does not. At the two moments where the train stops accelerating and starts braking, the quotients from before and after settle on different numbers, so the acceleration has no value there and the instrument leaves a gap.

There is a reverse direction to explore. Velocity multiplied by a short wait gives an estimate of a change in position. Add those signed changes across the motion, and shorten the waits. That leads to summation and integration: recovering the motion from its rates of change.