The Slope of the Tangent and the Power Rule

The slope of a straight line

A straight line rises by the same amount for every unit it runs to the right. The ratio of the rise to the run is its slope. A line that goes up to the right has a positive slope, and one that goes down has a negative slope. A level line has zero slope. When both axes are drawn to the same scale, the slope is \(\tan\theta\), where \(\theta\) is the angle the line makes with the \(x\) axis.

A curve has no single slope, because its direction changes from point to point. At each point P, however, it has a tangent. Lesson 1 showed that the curve and its tangent cannot be told apart close to P. So the slope of the curve at P means the slope of its tangent at P. That slope is the derivative.

The derivative is itself a function

Let the toy car's position be \(x = t^3 - 6t^2 + 9t + 1\) metres, with \(t\) in seconds. In Fig 2.1 a tangent slides along the x-t graph. The slope of the tangent at each instant is plotted below it. The slopes form a second curve, the velocity-time graph. The derivative of a function is a new function.

Read Fig 2.1 at a few instants. At \(t = 0\) the curve is steep and rising, so the car moves forward fast, at \(9\,\text{m/s}\). At \(t = 1\,\text{s}\) the tangent is flat, so the car is at rest for an instant. Between \(1\) and \(3\,\text{s}\) the curve falls, so the velocity is negative and the car moves backward. It moves backward fastest at \(t = 2\,\text{s}\), where the falling curve is steepest.

Rising, falling or flat

The sign of the derivative gives a quick test. If \(dy/dx > 0\) at a point, then \(y\) increases as \(x\) increases through that point. If \(dy/dx < 0\), \(y\) decreases. If \(dy/dx = 0\), the tangent is flat and \(y\) is momentarily unchanged. So a sign tells us which way a quantity is going without drawing the graph.

The tangent and the normal

The line through P at right angles to the tangent is the normal to the curve at P. If the tangent has slope \(m\), the normal has slope \(-1/m\), because two perpendicular lines have slopes whose product is \(-1\). For \(y = x^2\) at \(x = 2\), the point is \((2, 4)\) and the tangent slope is \(4\). The tangent line is \(y = 4x - 4\). The normal has slope \(-1/4\), and its line is \(y = -x/4 + 9/2\). Physics meets the normal often: the normal force on a curved track, and a ray that strikes a curved mirror along its normal.

One rule for every power

Lesson 1 found by a limit that \(5t^2\) has the derivative \(10t\). Doing a limit every time would be slow. Fortunately every power of \(x\) follows one pattern, and the derivation below proves it for a whole-number power. The power comes down in front as a factor, and the new power is one less.

The proof above needs a whole-number power, but the rule holds for every constant power. So \(\sqrt{x} = x^{1/2}\) has the derivative \(\tfrac12 x^{-1/2} = 1/(2\sqrt{x})\). And \(1/x = x^{-1}\) has the derivative \(-x^{-2} = -1/x^2\). Physics uses this last one constantly, because fields and potentials fall off as powers of the distance.

Three simple rules complete the set. The derivative of a constant is zero, because a constant has a flat graph. A constant factor stays in front, because stretching a graph upward stretches every slope by the same factor. A sum is differentiated term by term, because the rises of the two parts add. For example, \(2x^{12} + 3x^{23}\) has the derivative \(24x^{11} + 69x^{22}\).

Back to physics

Motion in One Dimension differentiates three position functions in its lesson on instantaneous velocity. With the rules of this lesson, each takes one line.

When two changing quantities multiply

A truck is loaded with sand from a chute while it moves. Its momentum is \(p = mv\), and both the mass and the speed change. How fast does the momentum change? Draw the momentum as the area of a rectangle, of width \(m\) and height \(v\). In a short time the width grows by \(\Delta m\) and the height by \(\Delta v\).

The gain of area in Fig 2.2 is \(m\,\Delta v + v\,\Delta m + \Delta m\,\Delta v\). Divide it by the time \(\Delta t\) and let \(\Delta t\) tend to zero. The first two terms become \(m\,dv/dt\) and \(v\,dm/dt\). The corner term is a product of two small changes, so it vanishes faster than the others. What remains is the product rule.

So \(F = d(mv)/dt = m\,dv/dt + v\,dm/dt\). For a body of fixed mass the second term is zero, and the law becomes \(F = ma\). For a belt or a truck that gains sand at a steady speed, the first term is zero, and \(F = v\,dm/dt\). The quotient rule follows from the product rule. For example, \(x^3/(x^2 + 1)\) has the derivative \((x^4 + 3x^2)/(x^2 + 1)^2\).

Back to the toy car

The car of the opening has \(x = t^3 - 6t^2 + 9t + 1\). The power rule gives its velocity at once: \(v = 3t^2 - 12t + 9\). At \(t = 0.5\,\text{s}\) this is \(0.75 - 6 + 9 = 3.75\,\text{m/s}\), forward. The velocity is zero at \(t = 1\,\text{s}\) and \(t = 3\,\text{s}\), where the car stops and turns. No speedometer was needed, only the curve and one line of rules.