A scooter starts from a signal and picks up speed. The needle of the speedometer moves smoothly from \(0\) to \(40\,\text{km/h}\). At every moment the needle shows one number. The number belongs to that moment alone. It is not the average speed since the signal, because that average would be much lower.
The average is not enough
Consider a particle that moves along a line with position \(x = t^2\). Here \(x\) is in metres and \(t\) in seconds. The table below gives the position at a few times.
\[ \begin{array}{c|c|c|c|c|c|c} t\ (\text{s}) & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline x\ (\text{m}) & 0 & 1 & 4 & 9 & 16 & 25 \end{array} \]
The position grows faster and faster. The particle is speeding up.
Use the table. What is the average velocity of the particle from \(t = 0\) to \(t = 2\,\text{s}\), and from \(t = 2\,\text{s}\) to \(t = 4\,\text{s}\)?
The two averages are different, and both are correct. Each one belongs to a whole interval of two seconds. Now ask a sharper question. What is the velocity at the single instant \(t = 2\,\text{s}\)? Neither answer gives it. The first interval ends at \(2\,\text{s}\) and the second starts there, yet their averages differ by \(4\,\text{m/s}\).
Shrinking the interval
An instant has no duration, so we cannot divide by it. We do the next best thing. We take a short interval that starts at \(t = 2\,\text{s}\). Begin with \(0.1\,\text{s}\), from \(t = 2\,\text{s}\) to \(t = 2.1\,\text{s}\). Work out the average velocity over this interval yourself.
For \(x = t^2\), find the average velocity from \(t = 2\,\text{s}\) to \(t = 2.1\,\text{s}\), in m/s.
The answer, \(4.1\,\text{m/s}\), lies between \(2\) and \(6\,\text{m/s}\). It is much closer to the velocity at \(2\,\text{s}\), because the interval is short. Now make the interval ten times shorter, from \(t = 2\,\text{s}\) to \(t = 2.01\,\text{s}\).
For \(x = t^2\), find the average velocity from \(t = 2\,\text{s}\) to \(t = 2.01\,\text{s}\), in m/s.
The averages were \(4.1\,\text{m/s}\) and then \(4.01\,\text{m/s}\). They seem to approach \(4\,\text{m/s}\). To be sure, let the interval be any short time \(h\). With \(x = t^2\), the average velocity from \(t = 2\) to \(t = 2 + h\) is
\[ \bar{v} = \dfrac{(2 + h)^2 - 2^2}{h} = \dfrac{4 + 4h + h^2 - 4}{h} = \dfrac{h\,(4 + h)}{h} = 4 + h \]
Put \(h = 0.1\) and the average is \(4.1\,\text{m/s}\), as you found. Put \(h = 0.01\) and it is \(4.01\,\text{m/s}\). Put \(h = 0.001\) and it is \(4.001\,\text{m/s}\). As \(h\) shrinks, the average approaches \(4\,\text{m/s}\). We say that the limit of the average velocity, as \(h\) tends to zero, is \(4\,\text{m/s}\). This limit is the instantaneous velocity at \(t = 2\,\text{s}\).
We never put \(h = 0\) directly, because the formula would become \(0/0\). We cancelled an \(h\) from the numerator and the denominator, and cancelling is allowed only while \(h\) is not zero. After cancelling, we let \(h\) approach zero. Both steps are needed. This is what the word limit means.
The same limit, drawn
The algebra above can be seen on the position-time graph of \(x = t^2\). Mark the point P at \(t = 2\,\text{s}\). For a chosen \(h\), mark the point Q at \(t = 2 + h\). The straight line through P and Q is called a chord. The slope of the chord, rise over run, equals the average velocity over the interval, \(4 + h\).
As \(h\) becomes smaller, Q slides towards P. The chord turns. In the limit the chord becomes the tangent to the curve at P. The slope of the tangent is the instantaneous velocity. This is the geometric meaning of the derivative.
The instantaneous velocity of a body at time \(t\) is the limit of the average velocity over an interval starting at \(t\), as the length of the interval tends to zero. It is written as the derivative of position with respect to time, \(v = dx/dt\). On a position-time graph, the instantaneous velocity is the slope of the tangent at that instant.
One tangent at every instant
There was nothing special about \(t = 2\,\text{s}\). The same argument works at any time \(t\). For \(x = t^2\), the average velocity over the interval from \(t\) to \(t + h\) is \(2t + h\). The limit as \(h\) tends to zero is \(2t\). So the velocity is a function of time, \(v = 2t\). The tangent slides along the curve, and its slope changes from point to point.
Find the velocity as a function of time, and its value at \(t = 2\,\text{s}\), for (a) \(x = 5t^3\), (b) \(x = 7t + 3\), (c) \(x = 2t^2 + 4t + 1\). Positions are in metres and times in seconds.
Use the rule that the derivative of \(t^n\) is \(n\,t^{n-1}\), and that the derivative of a constant is zero. (a) \(v = dx/dt = 15t^2\). At \(t = 2\,\text{s}\), \(v = 60\,\text{m/s}\).
(b) \(v = 7\,\text{m/s}\) at every instant. The velocity is constant, and the constant \(3\) in the position does not affect the velocity.
(c) \(v = 4t + 4\). At \(t = 2\,\text{s}\), \(v = 12\,\text{m/s}\).
A particle moves along a line with \(x = 3t^2\) metres. Find the instantaneous velocity at \(t = 2\,\text{s}\), in m/s.
Velocity carries a sign; speed does not
Instantaneous speed is the magnitude of the instantaneous velocity. Keep these key points in mind.
- Instantaneous velocity is a vector. On a line it is a signed number.
- A positive velocity means motion in the positive direction. A negative velocity means motion in the negative direction.
- Instantaneous speed is \(|v|\). Speed is never negative.
- Average speed and the size of the average velocity can differ, as in the last lesson. At one instant they cannot, because the path has no time to bend.
A particle moves with \(x = 2t^2 - 3t\) metres. Find the velocity and the speed at \(t = 0.5\,\text{s}\). At what time does the particle change its direction of motion?
\(v = dx/dt = 4t - 3\).
At \(t = 0.5\,\text{s}\), \(v = 2 - 3 = -1\,\text{m/s}\).
The velocity is negative, so the particle is moving in the negative direction at that instant.
The speed is \(|v| = 1\,\text{m/s}\).
The particle changes direction when the velocity passes through zero, that is, when \(4t - 3 = 0\), at \(t = 0.75\,\text{s}\).
Before this time the particle moves in the negative direction; after it, in the positive direction.
A particle moves along a line with \(x = 2t^2 - 3t\) metres. At \(t = 0.5\,\text{s}\), what are its instantaneous velocity and its instantaneous speed?
When differentiating \(t^n\) with respect to \(t\), what happens to the exponent?
Why do we let \(h\) approach zero instead of simply putting \(h = 0\) in the average-velocity formula?
The average-velocity formula divides a displacement by an interval. If the interval is exactly zero, both the displacement and the interval are zero, and \(0/0\) has no meaning. So we keep \(h\) non-zero, simplify the fraction by cancelling \(h\), and then see what value the result approaches as \(h\) becomes very small. That value is the instantaneous velocity.
Newton worked out the idea of a rate of change at one instant in 1665 and 1666, while he stayed at his family farm to escape the plague in Cambridge. He called the changing quantities fluents and their rates fluxions. He did not publish. Leibniz in Germany reached the same idea independently about ten years later and published first, with the notation \(dx/dt\) that we still use. The two men and their followers quarrelled for decades about who had invented calculus. The physics did not care. The slope of a tangent was the velocity for both of them.
- Dividing the position at time \(t\) by \(t\) to get the velocity at \(t\).That gives the average velocity from \(t = 0\), not the instantaneous velocity. Differentiate \(x(t)\) first, then substitute \(t\).
- Reporting a speed with a negative sign.Speed is the magnitude of velocity. Keep the sign for the velocity and drop it for the speed.
- Raising the power when differentiating.The derivative of \(t^n\) is \(n\,t^{n-1}\). Raising the power belongs to integration.