A fast bowler at Eden Gardens lets the ball go, and the big screen at once shows 143 km/h. A radar gun behind the bowler measured that speed. The ball slows a little in the air, so it reaches the batter at a lower speed. The number on the screen belongs to one instant, the moment of release. Yet speed is a distance divided by a time, and an instant has no time in it.
How fast does a quantity change?
Physics is full of quantities that change. A position changes with time, and so do a charge, a momentum and a temperature. To say how fast a quantity \(y\) changes when another quantity \(x\) changes, compare the two changes. Let \(x\) change from \(x_1\) to \(x_2\) while \(y\) changes from \(y_1\) to \(y_2\). The ratio of the two changes is the average rate of change of \(y\) with respect to \(x\) over that interval.
\[ \dfrac{\Delta y}{\Delta x} = \dfrac{y_2 - y_1}{x_2 - x_1} \]
The Greek letter \(\Delta\), read "delta", means "the change in". Its unit is the unit of \(y\) divided by the unit of \(x\).
Many quantities of physics are rates of exactly this kind. The average velocity is \(\Delta x/\Delta t\), a change of position per unit of time. The average current in a wire is \(\Delta q/\Delta t\), the charge that crosses a section per second. The average force on a body is \(\Delta p/\Delta t\), a change of momentum per second. Only the letters change; the idea stays the same.
The average depends on the interval
Drop a stone from rest from a high bridge, and take \(g = 10\,\text{m/s}^2\). After a time \(t\) the stone has fallen a distance \(y = 5t^2\) metres. From \(t = 1\,\text{s}\) to \(t = 2\,\text{s}\) its average speed is \((20 - 5)/(2 - 1) = 15\,\text{m/s}\). From \(t = 1\,\text{s}\) to \(t = 1.1\,\text{s}\) it is \((6.05 - 5)/0.1 = 10.5\,\text{m/s}\). From \(t = 1\,\text{s}\) to \(t = 1.01\,\text{s}\) it is \((5.1005 - 5)/0.01 = 10.05\,\text{m/s}\).
The three averages differ because the stone speeds up during each interval. A longer interval contains more of the later, faster motion. As the interval shrinks, the averages settle on one number, \(10\,\text{m/s}\). That number no longer depends on any interval. It belongs to the instant \(t = 1\,\text{s}\) alone.
What a limit is
We cannot simply put a zero interval into the average. In no time the stone falls no distance, and \(0/0\) has no value at all. So we take another route. Keep the interval \(h\) different from zero, simplify the average, and then ask what number it approaches as \(h\) is made smaller and smaller. That number is called the limit.
For the stone, the average speed from \(t = 1\,\text{s}\) to \(t = 1 + h\) is
\[ \bar v = \dfrac{5(1 + h)^2 - 5}{h} = \dfrac{5 + 10h + 5h^2 - 5}{h} = 10 + 5h \]
The last step cancels \(h\) from the top and the bottom, which is allowed because \(h\) is not zero. Now let \(h\) approach zero. The term \(5h\) approaches zero, so the average approaches \(10\,\text{m/s}\). We write \(\lim_{h \to 0}(10 + 5h) = 10\). It is read "the limit of \(10 + 5h\) as \(h\) tends to zero is 10".
A negative \(h\) gives an interval that ends at \(t = 1\,\text{s}\) instead of starting there. The same formula \(10 + 5h\) then gives averages just below 10. Fig 1.1 showed both sides: the averages close in on the same value from above and from below. A limit exists only when the two sides agree like this.
Zoom in, and the curve becomes a line
There is a second way to see the limit. Look at the graph of \(y = 5t^2\) more and more closely near \(t = 1\,\text{s}\). A smooth curve, seen closely enough, looks straight. The straight line that it comes to look like is the tangent to the curve at that point. The slope of that line is the rate at that instant.
In Fig 1.2 the window reaches a time \(h\) on each side of P. Over that time the tangent rises by \(10h\) metres, and the curve leaves the tangent by \(5h^2\) metres. Make \(h\) four times smaller, and the rise becomes four times smaller. The gap becomes sixteen times smaller. So the gap, compared with the rise, is \(5h^2/10h = h/2\), and that tends to zero. Close enough to P, the curve and its tangent cannot be told apart.
The derivative and its notation
The rate of change of \(y\) at a single value of \(x\) is called the derivative of \(y\) with respect to \(x\). It is written \(dy/dx\), read "d y by d x". The notation is due to Leibniz. The letter d is not a number that multiplies \(y\). It means "a change of, taken in the limit". So \(dy/dx\) still shows its origin: a change in \(y\) divided by a change in \(x\), both shrunk towards zero.
Let a quantity \(y\) depend on a quantity \(x\). Take the average rate of change \(\Delta y/\Delta x\) over an interval that starts at a given value of \(x\). The derivative of \(y\) with respect to \(x\) there is the limit of that average as the length of the interval tends to zero. It is written \(dy/dx\), or \(f'(x)\) when \(y = f(x)\).
There was nothing special about \(t = 1\,\text{s}\). The derivation below repeats the argument at any time \(t\). The result is a new function of time, the speed of the stone at every instant.
\[ \dfrac{\Delta y}{\Delta t} = \dfrac{5(t + h)^2 - 5t^2}{h} \]
Why. the average rate from \(t\) to \(t + h\), for the fall \(y = 5t^2\).
\[ \dfrac{\Delta y}{\Delta t} = \dfrac{5t^2 + 10th + 5h^2 - 5t^2}{h} \]
Why. expand the square, \((t + h)^2 = t^2 + 2th + h^2\), and multiply by 5.
\[ \dfrac{\Delta y}{\Delta t} = 10t + 5h \]
Why. the two \(5t^2\) terms cancel, and \(h\) cancels from the top and the bottom because \(h \neq 0\).
\[ \dfrac{dy}{dt} = \lim_{h \to 0}(10t + 5h) = 10t \]
Read it. as \(h\) tends to zero the term \(5h\) vanishes. This is \(gt\) with \(g = 10\,\text{m/s}^2\), and at \(t = 1\,\text{s}\) it gives \(10\,\text{m/s}\).
When \(y\) is written as a function \(f(x)\), its derivative is also written \(f'(x)\), read "f prime of x". When the variable is time, physicists often put a dot over the quantity instead: \(\dot x\) means \(dx/dt\). All three notations name the same limit. So the velocity is \(v = dx/dt\), the current is \(i = dq/dt\), and the net force is \(F = dp/dt\).
What does the symbol \(dy/dx\) mean?
The stone falls \(y = 5t^2\) metres in \(t\) seconds. Find its average speed from \(t = 2\,\text{s}\) to \(t = 2.1\,\text{s}\), in m/s.
A body moves along a line with \(x = 2t^2\) metres. The average velocity over the interval from \(t = 3\,\text{s}\) to \(t = 3 + h\) simplifies to \(12 + 2h\). What is the velocity at \(t = 3\,\text{s}\), in m/s?
Why do we not put \(h = 0\) straight into the average \(\big(y(t + h) - y(t)\big)/h\)?
The charge that has crossed a section of a wire by the time \(t\) is \(q(t)\). Which expression is the current at the instant \(t = 3\,\text{s}\)?
When there is no derivative
The limit needs a graph with no break and no sharp corner. A graph with no break is called continuous: it can be drawn without lifting the pen. A ball that hits a wall and bounces back is the standard warning. If the bounce is treated as instantaneous, the position graph has a sharp corner at the bounce. The averages just before the corner and just after it approach two different values. So there is no single tangent and no derivative at that instant.
The limit is a statement about a smooth formula. Measured data never shrink to a zero interval. A radar gun or a timer always divides a small change by a small but finite time. That is good enough only when the rate hardly changes over the measuring interval. Over a bounce, a collision or a switch closing, the rate changes sharply, and an average over even a millisecond can mislead.
Back to physics
The same limit turns a momentum into a force. The chapter Momentum and Its Conservation plots a momentum that grows with time and reads the force as the slope of its tangent. The example below finds that slope with the limit alone.
This is Fig 1.1 of Momentum and Its Conservation, lesson 1, done with the tool. The momentum of a body grows as \(p = 2 + 1.5t^2\), with \(p\) in kg m/s and \(t\) in seconds. The net force is the rate of change of momentum, \(F = dp/dt\). Find the net force at \(t = 2\,\text{s}\) from the limit.
From \(t = 2\,\text{s}\) to \(t = 2 + h\), the momentum changes by \(p(2 + h) - p(2) = 2 + 1.5(4 + 4h + h^2) - 8 = 6h + 1.5h^2\).
The average force over that interval is \(\Delta p/\Delta t = (6h + 1.5h^2)/h = 6 + 1.5h\), after cancelling \(h \neq 0\).
As \(h\) tends to zero the average tends to 6. So the net force at \(t = 2\,\text{s}\) is \(6\,\text{N}\), the slope of the tangent drawn in that figure.
The momentum of a body along a line is \(p = 4t^2 + 3t\), in kg m/s with \(t\) in seconds. Over the interval from \(t = 2\,\text{s}\) to \(2 + h\), the average force works out to \(19 + 4h\). Find the net force at \(t = 2\,\text{s}\), in newtons.
Back to the stadium
The radar gun at Eden Gardens measures the ball over a few thousandths of a second. In so short a time the speed of the ball changes by a tiny fraction. So the average over that interval equals the limit to every digit on the screen. The 143 km/h is the speed at the instant of release, in exactly the sense of this lesson.
The derivative as a limit is the definition behind a whole family of physics quantities. Velocity is \(dx/dt\) and acceleration is \(dv/dt\) in Motion in One Dimension. The current at an instant is \(i = dq/dt\) in Electric Current and Resistance. Newton's second law in his own form, \(F = dp/dt\), opens Momentum and Its Conservation. Power is \(P = dW/dt\) in Work, Kinetic Energy and Power, and Faraday's law turns \(d\Phi/dt\) into an emf. Each one begins as an average over an interval and becomes a value at an instant through the same limit.
- Cancelling the d's in \(dy/dx\) to get \(y/x\).The letter d means "a change, taken in the limit". It is not a number and never cancels. For \(y = 5t^2\), \(y/t = 5t\) but \(dy/dt = 10t\).
- Putting \(h = 0\) before simplifying the average.That gives \(0/0\). Simplify first while \(h \neq 0\), then let \(h\) tend to zero.
- Dividing the value at time \(t\) by \(t\) to get the rate at \(t\).That is an average from \(t = 0\). The rate at an instant needs a vanishing interval at that instant.