A negative derivative means the quantity is decreasing as the variable increases. The sign is part of the answer, so write what it means, the size of the rate and the unit.
This lesson is part of recovering the method in a mixed calculus problem. It uses the rates from optimisation and rates-of-change problems.
A worked model: a draining tank
The water in a tank is V = 100 − 8t + 0.1t² litres after t minutes, for 0 ≤ t ≤ 15. Describe how the volume is changing at t = 5.
Differentiate. dV/dt = −8 + 0.2t.
Substitute. At t = 5: dV/dt = −8 + 1 = −7.
Interpret. The volume is decreasing at 7 litres per minute at t = 5.
At t = 10 the rate is −8 + 2 = −6. The tank is still emptying, but more slowly, because the derivative is getting closer to zero. For the whole domain 0 ≤ t ≤ 15, the derivative stays negative, since 0.2t ≤ 3 < 8.
Sign, size and unit
Use this three-part frame in the answer.
| Part | What to state | In the example |
|---|---|---|
| Sign | Increasing or decreasing | Decreasing |
| Size | The number, positive | 7 |
| Unit | Output unit per input unit | Litres per minute |
If the question asks for the rate of decrease, give 7. If it asks for the rate of change, give −7.
Worked model 2: velocity
A particle moves so that its displacement is s = t² − 6t metres from a point O. Find its velocity at t = 1.
v = ds/dt = 2t − 6. At t = 1: v = −4. The particle is moving in the negative direction at 4 m/s. Its velocity is −4 m/s and its speed is 4 m/s.
The mistake that costs marks
The slip is to treat the negative sign as an error and change it, or to write the sign when the question asked for the size.
| Question wording | Wrong | Right |
|---|---|---|
| Rate at which V is decreasing at t = 5 | −7 litres per minute | 7 litres per minute |
| Rate of change of V at t = 5 | 7 litres per minute | −7 litres per minute |
Read the wording, then decide whether to keep the sign.
Check yourself
The stock in a shop is S = 500 − 30t + 0.4t² items after t days, for 0 ≤ t ≤ 20. Find the rate of change at t = 10, and find the value of t when stock is decreasing at 18 items per day.
Answer
dS/dt = −30 + 0.8t. At t = 10: dS/dt = −30 + 8 = −22, so the stock is decreasing at 22 items per day.
Decreasing at 18 per day means dS/dt = −18, so −30 + 0.8t = −18, giving 0.8t = 12 and t = 15. This is inside the domain.
Check: −30 + 0.8(15) = −18.
What to study next
Continue with the integrated practice set, or see how signs work for motion in connecting integration to displacement and distance.
To have a teacher check the sentence that explains your answer, see online one-to-one Additional Mathematics tuition.