The method depends on what is given and what is asked. From a total to a rate, differentiate. From a rate to a total, integrate.
This lesson opens recovering the method in a mixed calculus problem. It needs the rules from differentiation and evaluating definite integrals.
The given-and-asked table
| You are given | You are asked for | Method |
|---|---|---|
| Displacement s(t) | Velocity | Differentiate |
| Velocity v(t) | Acceleration | Differentiate |
| Velocity v(t) | Displacement over an interval | Integrate |
| Flow rate R(t) | Total volume over an interval | Integrate |
| Gradient dy/dx | The curve through a point | Integrate |
| A curve | Tangent, gradient or turning point | Differentiate |
Worked example 1: one function, two questions
A particle moves with velocity v = 3t² − 4t m/s. Find its acceleration at t = 2, and its displacement from t = 0 to t = 2.
Acceleration. It asks for a rate from a velocity, so differentiate: a = dv/dt = 6t − 4. At t = 2, a = 12 − 4 = 8 m/s².
Displacement. It asks for a total from a velocity, so integrate: the integral of (3t² − 4t) from 0 to 2 is [t³ − 2t²] from 0 to 2 = (8 − 8) − 0 = 0 m.
The displacement is zero, which means the particle ends where it started. The distance travelled is not zero, because the particle moved forward then back.
Worked example 2: a flow rate
Water flows into a tank at R = 6 − 0.5t litres per minute for 0 ≤ t ≤ 8. Find the total volume that flows in.
The question asks for a total from a rate, so integrate: the integral of (6 − 0.5t) from 0 to 8 is [6t − 0.25t²] = 48 − 16 = 32 litres.
The rate is always positive here (6 − 4 = 2 at t = 8), so no water flows out.
The mistake that costs marks
The common slip is to differentiate the rate when the question asks for a total.
| Question | Wrong | Right |
|---|---|---|
| Total water in 8 minutes | dR/dt = −0.5 | ∫₀⁸ R dt = 32 litres |
Underline the quantity requested before you start. If it is a total accumulated over time, you are integrating.
Check yourself
A cyclist’s speed is v = 4t − t² m/s for 0 ≤ t ≤ 4. Find the distance travelled in the first 3 seconds, and the acceleration at t = 1.
Answer
Distance: the speed is positive on this interval, since 4t − t² = t(4 − t) > 0 for 0 < t < 4. Integrate: [2t² − t³/3] from 0 to 3 = (18 − 9) − 0 = 9 m.
Acceleration: a = dv/dt = 4 − 2t. At t = 1, a = 2 m/s².
What to study next
Continue with using a tangent condition to recover an unknown coefficient. Practise both tools on the polynomial integration and area explorer.
If you would like a teacher to drill the method choice with you, see online one-to-one Additional Mathematics tuition.