Learn: a tank fills faster as time passes
A tank starts empty. After t minutes, the amount of water in it is modelled by V(t) = t² litres, for 0 ≤ t ≤ 4. At 1 minute it contains 1 litre; at 2 minutes, 4 litres; at 3 minutes, 9 litres. Assume no leakage and enough tank capacity.
Two questions are different: “How much water has accumulated?” and “How quickly is it filling right now?” Calculus connects them.
Vocabulary and units
- A function assigns an output to each allowed input. Here t is time in minutes and V is volume in litres.
- An average rate divides a change in output by the corresponding change in input.
- A derivative describes instantaneous rate of change. V′(t), read “V prime of t,” is a filling rate measured in litres per minute.
- A definite integral adds up the contribution of a rate over an interval. A rate in litres per minute accumulated over minutes gives litres.
Worked example: average rate and instantaneous rate
From minute 1 to minute 3, the average filling rate is (9 − 1) ÷ (3 − 1) = 4 litres/minute. That one average describes the entire two-minute interval, not every instant.
To examine a short interval beginning at t, let h be a small nonzero change in time. The average rate is [(t + h)² − t²]/h = [2th + h²]/h = 2t + h. As h approaches zero, the rate approaches 2t. Thus V′(t) = 2t litres/minute. At t = 3, the instantaneous rate is 6 litres/minute.
The units are part of the model: when t is measured in minutes, the coefficient in V = t² has units of litres per minute squared. A rate of 6 litres/minute does not mean the tank contains 6 litres.
Explore: area under a rate graph
Reverse the question. Suppose the filling rate is q(t) = 2t litres/minute and the initial volume is zero. On the graph, horizontal distance represents minutes and vertical height represents litres/minute. The area under this nonnegative rate graph represents water added.
At 3 minutes: current rate 6 litres/minute; accumulated water 9 litres.
Worked example: accumulation over three minutes
- The shaded region is a triangle with base 3 minutes and height 6 litres/minute.
- Area = ½ × base × height = ½ × 3 × 6 = 9 litres.
- Integral notation writes the same accumulation as ∫₀³ 2t dt = [t²]₀³ = 9 − 0 = 9 litres.
- In this notation, dt identifies time as the variable of integration; [t²]₀³ means evaluate at 3 and subtract the value at 0.
This geometry works because the rate graph is a straight line here. Integrals also handle curved rate graphs. If the initial volume were 5 litres, add the same 9 litres to get 14 litres after three minutes.
Where these ideas are useful
- Motion: the derivative of position is velocity; the derivative of velocity is acceleration. Integrating velocity gives displacement. When velocity changes sign, total distance instead uses the integral of speed, its absolute value.
- Growth and temperature: derivatives describe how quickly a measured quantity is changing. Predictions depend on how well a model fits the situation.
- Design: derivatives can locate candidates for a largest area or a smallest material use. Constraints and endpoints must also be checked.
A small optimization example
A rectangular practice plot has a 20 m perimeter. If one side is x metres, the other is 10 − x metres, with 0 < x < 10. Its area is A(x) = x(10 − x) = 10x − x² square metres. The derivative A′(x) = 10 − 2x is positive for x < 5 and negative for x > 5, so area increases then decreases. The largest area is 25 m² at x = 5: a 5 m by 5 m square. This conclusion is for the stated fixed-perimeter rectangle model.
Practice: identify the quantity and its unit
1. For V(t) = t², what is the filling rate at t = 2 minutes?
V′(t) = 2t, so the rate is 4 litres/minute. The volume also happens to be 4 here, but its unit is litres.
2. How much water is added from t = 1 to t = 3?
V(3) − V(1) = 9 − 1 = 8 litres. This is an amount, not a rate.
3. A constant rate of 5 litres/minute lasts 4 minutes. Find the volume added.
5 × 4 = 20 litres, the area of a rectangle under the rate graph.
4. Does setting a derivative to zero always prove you found a maximum?
No. It identifies a candidate when the derivative exists. You still need reasoning such as a sign change, comparisons and endpoint checks where applicable.
Review: match the tool to the question
Use rates to discuss how quickly something changes; use accumulation to find the change over an interval. Write units, assumptions and the interval. A useful calculus answer explains what the number means in the original situation.
Optional reference: OpenStax on derivatives as rates of change. Continue with AP Calculus AB or AP Calculus BC.