A derivative measures how fast one quantity changes compared with another. When a question gives y in terms of t, dy/dt is the rate of change of y with respect to time. Its value at a chosen t tells you how fast y is changing at that moment.
This lesson follows forming a normal in the module on tangents, normals and rates, and it uses the rules from differentiation techniques.
How do you turn a rate question into calculus?
- Identify the quantity that is changing, for example volume V.
- Identify the variable it depends on, usually time t.
- Differentiate V with respect to t to get dV/dt.
- Substitute the given time into dV/dt.
- Write the units and a short sentence if the question asks you to interpret.
The phrase “at the instant when” or “at time t = 5” points to a derivative value. The phrase “between t = 0 and t = 5” points to an average, which uses a difference instead.
Worked example
Water flows into a tank. After t minutes the volume is V = 40t − t² litres, for 0 ≤ t ≤ 20. Find the rate at which the volume is increasing when t = 5, and compare it with the average rate over the first 5 minutes.
Step 1, differentiate: dV/dt = 40 − 2t.
Step 2, substitute t = 5: dV/dt = 40 − 10 = 30.
The volume is increasing at 30 litres per minute when t = 5.
Step 3, average rate: V(0) = 0 and V(5) = 200 − 25 = 175. Average rate = 175 ÷ 5 = 35 litres per minute.
The average (35) is larger than the rate at t = 5 (30) because the tank fills more slowly as time goes on.
The mistake to watch for
The common error is to divide V by t and call that the instantaneous rate.
Mistaken working: V(5) = 175, so the rate is 175 ÷ 5 = 35 litres per minute.
This is the average rate over 5 minutes, not the rate at t = 5.
The correction is to ask “at one moment or over a period?” before you start. One moment means differentiate and substitute. A period means subtract values and divide by the time taken.
Check yourself
Try these, then open each answer.
1. The area of a spreading stain is A = 3t² + 2t cm² after t seconds. Find dA/dt when t = 4.
Show answer
dA/dt = 6t + 2. At t = 4: 24 + 2 = 26 cm² per second.
2. A population is modelled by P = 1000 + 50t − 2t², with t in years. Find the rate of change of P when t = 10.
Show answer
dP/dt = 50 − 4t. At t = 10: 50 − 40 = 10 per year. The population is still growing, but slowly.
3. For y = t³ − 6t² + 5, find the values of t at which the rate of change of y is zero.
Show answer
dy/dt = 3t² − 12t = 3t(t − 4). Setting this to zero gives t = 0 or t = 4. Check: 3(0)(−4) = 0 and 3(4)(0) = 0. ✓
Where this leads next
Two quantities change together, so continue with linking two changing quantities through a related-rate model. Practise in the mixed practice set. The non-calculator working trainer supports the arithmetic and the quadratic structure explorer helps you read turning points from graphs.
If marks disappear on word problems, a teacher in online one-to-one Additional Mathematics tuition can go through your past mistakes and show you how to read each question.