Stand in a DTC bus in Delhi without holding the rail. As it pulls away from a stop you are pushed backward, and as it brakes you lurch forward. On a smooth straight road at a steady \(50\,\text{km/h}\) you feel nothing at all. Your body does not feel the speed of the bus. It feels how fast that speed changes.
Differentiating twice
The derivative of a position \(x(t)\) is the velocity \(v(t)\), and that is itself a function of time. So it can be differentiated again. The result is the rate of change of the velocity, the acceleration. Because it comes from differentiating twice, it is called the second derivative of the position. It is written \(d^2x/dt^2\), read "d two x by d t squared".
The notation keeps track of the units. The "d two" on top says that d has acted twice on \(x\). The "\(dt^2\)" below says that we divided by a time twice. So \(d^2x/dt^2\) is in metres per second per second, m/s², the unit of acceleration. Other notations are \(f''(x)\), read "f double prime", and \(\ddot x\) for a second time derivative.
Take the toy car of lesson 2, with \(x = t^3 - 6t^2 + 9t + 1\) metres. Its velocity is \(v = 3t^2 - 12t + 9\). Differentiate again: \(a = 6t - 12\). At \(t = 1\,\text{s}\) the velocity is zero but the acceleration is \(-6\,\text{m/s}^2\). The car is at rest for an instant, and it is about to move in the negative direction.
The chain need not stop at two. The third derivative of position, the rate of change of acceleration, is called the jerk, in m/s³. For \(x = t^3\) it is a steady \(6\,\text{m/s}^3\). Metro engineers keep the jerk small, so that standing passengers are not thrown off balance when the train starts.
A particle moves along a line with \(x = t^3 - 6t^2 + 9t + 1\) metres. Find its acceleration at \(t = 1\,\text{s}\), in m/s².
A car moves at a constant velocity of \(15\,\text{m/s}\). What must be true of its position function \(x(t)\)?
For \(y = x^3\), what is \(d^2y/dx^2\) at \(x = 2\)?
The bend of a graph
The second derivative is the rate at which the slope changes. Picture the tangent as it slides along a graph. If the slope is growing, the tangent turns anticlockwise and the curve bends upward, like the inside of a cup. The graph is then concave up, and it lies above its tangents. If the slope is falling, the tangent turns clockwise and the curve bends downward, like a dome. It is then concave down, and it lies below its tangents.
Read Fig 4.1 against its lower graph. Before \(t = 2\,\text{s}\) the acceleration is negative and the curve bends down. After \(2\,\text{s}\) it is positive and the curve bends up. At \(t = 2\,\text{s}\) the acceleration passes through zero and the bend changes over. Such a point is called a point of inflection. At it the tangent crosses the curve instead of staying on one side.
Where \(d^2y/dx^2 > 0\), the graph of \(y\) against \(x\) is concave up and lies above its tangents. Where \(d^2y/dx^2 < 0\), it is concave down and lies below them. A point where \(d^2y/dx^2\) changes sign is a point of inflection.
A stone thrown straight up at \(20\,\text{m/s}\) has the height \(h = 20t - 5t^2\). Its graph is a dome with its top at \(t = 2\,\text{s}\). Before you press play below, decide what the second derivative is at the very top.
At the top of the stone's flight the tangent to \(h = 20t - 5t^2\) is flat. What is \(d^2h/dt^2\) there?
The slope in the figure above falls from \(+20\) to \(-20\,\text{m/s}\) in four seconds, by the same \(10\,\text{m/s}\) every second. At the top it passes through zero without pausing. So the second derivative is \(-10\,\text{m/s}^2\) at the top, as everywhere else. A zero first derivative says nothing about the second.
Which of these positions, in metres with \(t\) in seconds, have zero acceleration at every instant?
Where does the graph of \(y = x^3 - 6x^2\) have its point of inflection?
The second derivative is not the square of the first: \(d^2y/dx^2\) and \((dy/dx)^2\) are different things. And a zero second derivative at a point does not by itself make a point of inflection. For \(y = x^4\), \(d^2y/dx^2 = 12x^2\) is zero at \(x = 0\), yet the graph is a cup on both sides. The sign must change.
Back to physics
Motion in One Dimension finds the acceleration from two position formulas in its lesson on acceleration. Each is two differentiations with the power rule.
This is the worked example of Motion in One Dimension, lesson 5, done with the tool. Find the acceleration for (a) \(x = t^3 + 2t\) and (b) \(x = 5 - 2t^2\). Positions are in metres and times in seconds.
(a) The first derivative is \(v = dx/dt = 3t^2 + 2\). The second is \(a = dv/dt = 6t\). The acceleration is not constant; it grows with time, and the x-t graph is concave up for \(t > 0\).
(b) The first derivative is \(v = -4t\). The second is \(a = -4\,\text{m/s}^2\), constant and negative. The x-t graph is a dome, and the velocity becomes more and more negative.
A block on a spring moves as \(x = 0.05\sin 20t\), with \(x\) in metres and \(t\) in seconds. Its acceleration is \(d^2x/dt^2\). What is the greatest size of the acceleration, in m/s²?
Back to the bus
A bus that cruises at a steady speed has a straight x-t graph, and its second derivative is zero. You feel nothing. As it pulls away, its x-t graph bends upward and the second derivative is positive. The floor carries your feet forward, and you feel pushed backward. As it brakes, the graph bends downward, the second derivative is negative, and you lurch forward. What you feel in the bus is the sign and the size of \(d^2x/dt^2\).
The second derivative is the acceleration of Motion in One Dimension, and the bend of an x-t graph in Graphical Understanding of Motion. It defines simple harmonic motion, \(d^2x/dt^2 = -\omega^2x\), in Kinematics of SHM and Dynamics of SHM, and the damped motion of the oscillations unit. The wave equation of the waves unit ties a second derivative in time to one in space. The charge on an LC circuit obeys an equation of the same form. In Potential Energy and Conservation of Energy, the sign of the second derivative of the energy tells stable equilibrium from unstable.
- Squaring the first derivative to get the second.Differentiate the first derivative again. For \(y = x^3\), \(d^2y/dx^2 = 6x\), not \(9x^4\).
- Believing that a zero velocity means a zero acceleration.At the top of a throw \(dh/dt = 0\) but \(d^2h/dt^2 = -g\). The slope is still changing.
- Taking a concave-down x-t graph to mean "slowing down".The bend gives the sign of \(a\) only. Speeding up or slowing down depends on whether \(v\) and \(a\) have the same sign.