The Second Derivative

Differentiating twice for velocity and acceleration: time for a given acceleration, when direction reverses, and when two particles move the same way.

In the exam for Edexcel IGCSE Maths 4MA1, Cambridge IGCSE Maths 0580 · 9 real past-paper questions below

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Differentiate Twice

For a moving particle, s is how far it is from O. Differentiate once for velocity, and again for acceleration.

Formula

v = \frac{d s}{d t} a = \frac{d v}{d t} = \frac{d ^{2} s}{d t ^{2}}

Rule

$\frac{d ^{2} s}{d t ^{2}}$ is the second derivative: the derivative of the derivative.

Key idea

Velocity says how fast s changes. Acceleration says how fast v changes.

Key idea

Differentiate one more time and you get jerk: how fast the acceleration changes. Lift engineers keep jerk small so rides feel smooth; most passengers find about $2 m/s^{3}$ fine. The derivatives after jerk even have fun names: snap, crackle and pop!

Time for a Given Acceleration (Higher)

The usual question: find v, then find when the acceleration is a given number.

Worked example

1
$s = 5 t^{3} - 6 t^{2} + 2 t, a = 21$ — The displacement, and the acceleration you are told to match.
2
$v = \frac{d s}{d t} = 15 t^{2} - 12 t + 2$ — Differentiate once for velocity: 5t³ → 15t², −6t² → −12t, 2t → 2.
3
$a = \frac{d v}{d t} = 30 t - 12$ — Differentiate again for acceleration: 15t² → 30t, −12t → −12, and 2 → 0.
4
$30 t - 12 = 21$ — The question says the acceleration is 21, so set the a expression equal to 21.
5
$30 t = 33$ — Add 12 to both sides.
6
$t = 33 \div 30 = 1.1 s$ — Divide both sides by 30. The time is 1.1 seconds.

Common mistake

Setting v = 21. The question asks for acceleration, so differentiate twice and use a.

Key idea

Show each line of differentiating and the equation you set up. Most of the marks are for the working, so a bare answer scores about 1 mark.

When Direction Reverses (Higher)

A particle turns round when it stops for an instant: $v = 0$ . Solve that first.

Worked example

1
$x = 2 t^{3} - 54 t + 10, t \geq 0$ — The displacement. Time cannot be negative.
2
$v = \frac{d x}{d t} = 6 t^{2} - 54$ — Differentiate once. The particle turns round when it stops for an instant, so v = 0.
3
$6 t^{2} - 54 = 0$ — Set v equal to 0.
4
$6 t^{2} = 54$ — Add 54 to both sides.
5
$t^{2} = 9$ — Divide both sides by 6.
6
$t = 3 or t = - 3$ — A square root has a positive and a negative answer.
7
$t = 3$ — Reject t = −3, because time cannot be negative.
8
$a = \frac{d v}{d t} = 12 t$ — Differentiate again: 6t² → 12t, and −54 → 0.
9
$a = 12 \times 3 = 36 m/s^{2}$ — Put t = 3 into a, not into v. v is 0 there, and you already knew that.

Common mistake

Putting t = 3 into v. It gives 0, which you already know. The acceleration comes from a.

Moving the Same Way (Higher)

The sign of v gives the direction. Two particles move the same way when their velocities have the same sign.

Worked example

1
$s_{P} = t^{3} - 5 t^{2} + 7 t, s_{Q} = t^{2} - 6 t + 2, t > 0$ — The two displacements. A particle moves the same way as another when their velocities have the same sign.
2
$v_{P} = \frac{d s _{P}}{d t} = 3 t^{2} - 10 t + 7$ — Differentiate P: 3t², −10t and 7.
3
$v_{P} = (t - 1) (3 t - 7)$ — Factorise. Two numbers that multiply to 3 × 7 = 21 and add to −10 are −3 and −7.
4
$v_{P} = 0 at t = 1 and t = \frac{7}{3}$ — Each bracket equal to 0 gives a root. P changes direction at these times.
5
$v_{Q} = \frac{d s _{Q}}{d t} = 2 t - 6 = 0 \Rightarrow t = 3$ — Differentiate Q, then find when it is 0. Q changes direction at t = 3.
6
$1 < t < \frac{7}{3} : v_{P} < 0, v_{Q} < 0$ — Between 1 and 7/3, v_P is negative (U-shape below the axis), and v_Q is negative because t < 3. Same way.
7
$\frac{7}{3} < t < 3 : v_{P} > 0, v_{Q} < 0$ — P is moving forwards and Q is moving backwards, so they go opposite ways. Leave this gap out.
8
$t > 3 : v_{P} > 0, v_{Q} > 0$ — After t = 3, both are positive. Same way.
9
$1 < t < \frac{7}{3} or t > 3$ — Use strict signs, because a particle at rest is not moving either way. For 0 < t < 1, v_P > 0 but v_Q < 0, so that gap is out.

Key idea

Find where each velocity is zero: at 1, 7/3 and 3. Then check the sign of both velocities in each gap between them.

Least Speed (Higher)

Some questions ask for the least speed. A velocity graph is lowest where it is flat, and its gradient is the acceleration. So set $a = 0$ .

Worked example

1
$s = t^{3} - 6 t^{2} + 15 t - 4$ — The displacement of the particle.
2
$v = 3 t^{2} - 12 t + 15$ — Differentiate once for velocity.
3
$a = 6 t - 12$ — Differentiate again. This is the gradient of the v graph.
4
$6 t - 12 = 0$ — The bottom of the v curve is flat, so its gradient a is 0. v is a U-shape because 3t² is positive.
5
$6 t = 12$ — Add 12 to both sides.
6
$t = 2$ — Divide both sides by 6.
7
$v = 3 (2)^{2} - 12 (2) + 15$ — Now put t = 2 into v, not into a, to find the least velocity.
8
$v = 12 - 24 + 15 = 3 m/s$ — 3 × 4 = 12 and 12 × 2 = 24. The least speed is 3 m/s.

Common mistake

Setting v = 0. This v never reaches 0, so that gives no answer. The least value is where a = 0.

Key idea

Check: $v = 3 (t - 2)^{2} + 3$ . A square is never negative, so v is never below 3. The least speed is 3 m/s, at t = 2.

Which Method? Read the Question

Find what the question names, then count how many times to differentiate. A particle that stops or turns round has $v = 0$ . The least speed is where $a = 0$ .

You see	You do
Velocity	differentiate s once
Acceleration	differentiate twice
Reverses or at rest	v = 0
Same direction	compare the signs of v
Maximum or minimum	Stationary Points

Key idea

Always give units: m/s for v and m/s² for a.

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