01 · Convert an inflow rate
A cylindrical tank has horizontal area 5 m² and net inflow 0.3 m³/min. Find dh/dt.
Hint
Use 5 dh/dt = 0.3.
Worked solution
dh/dt = 0.06 m/min.
Understand · explore · practise
Differential equations with volume and area
Connect volume, height, radius and area in differential equation models. Use the chain rule, solve draining and filling problems, calibrate constants and enforce physical stopping times.
Before you startSeparable differential equations, differentiation of geometric formulas and the chain rule.
01 / Connect the rate to the geometric quantity
A volume rate is not usually a height rate.
dV/dt = (dV/dh)(dh/dt)
Write volume as a function of height before separating a height equation.
The shape determines the conversion factor. A cylindrical tank has constant horizontal area; a cone has a changing area as the water rises. Keep all quantities in compatible length and time units.
A draining tank has a stopping timeExplore
At t = 0 minutes: h = 4 m, V = 16 m³.
dh/dt = −1 m/min; dV/dt = −4 m³/min.
Emptying time = 8 minutes.
Before emptying: h = (√h₀ − t/4)² with 0 ≤ t ≤ 4√h₀.
The tank is draining. The square-root expression must remain nonnegative.
The drawing uses a fixed height scale. Beyond emptying the physical tank stays empty; extending the squared formula would falsely make it refill.
02 / A constant cross-section gives a simple conversion
For horizontal area A, volume is Ah.
A vertical tank has horizontal area 4 m². Water leaves at 2√h m³/min, where h is its height in metres.Worked example
V = 4h → dV/dt = 4 dh/dt
Differentiate the geometric relation.
dV/dt = −2√h
Outflow makes the signed volume rate negative.
dh/dt = −(1/2)√h
Divide by the area 4, not by the volume.
03 / Integrate the height equation
Separate while the height is positive.
Solve dh/dt = −(1/2)√h with h(0) = 4.Worked example
h^(−1/2) dh = −(1/2) dt
For h > 0, separate.
2√h = −t/2 + C
Integrate the power.
C = 4
Use the initial height.
√h = 2 − t/4
This nonnegative left side restricts the time.
h = (2 − t/4)², 0 ≤ t ≤ 8
The tank empties at eight minutes.
Watch: stop the draining formula at emptying
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 · Half the initial height
When does this tank reach h = 2 m?
Hint
Use √2 = 2 − t/4.
Worked solution
t = 8 − 4√2 minutes, about 2.34 minutes. Half the height does not take half the emptying time.
03 · A later height
Find h and V at t = 4 minutes.
Hint
First find h, then multiply by the horizontal area.
Worked solution
h = 1 m and V = 4 m³.
04 / A squared formula can hide a physical restriction
Do not allow an empty tank to refill without inflow.
√h = √h₀ − t/4 requires t ≤ 4√h₀
Squaring does not remove the restriction inherited from the square root.
For times after emptying, the continued square gives an increasing height and no longer satisfies dh/dt = −√h/2. A physical extension with no input is h = 0. Check this constant state directly in the equation.
04 · Test the false continuation
If h = (2 − t/4)² is used at t = 12, compare its derivative with −√h/2.
Hint
The bracket is −1.
Worked solution
The formula gives h = 1 and h′ = +1/2, while the equation requires −1/2. It is invalid after emptying.
05 · Change the initial height
If h₀ = 9, when does the same model empty?
Hint
Use 4√h₀.
Worked solution
12 minutes.
05 / Find an unknown coefficient from a later observation
Use the integrated square-root relation.
A tank follows h′ = −k√h, with h(0) = 4 and h(3) = 1. Find k and the emptying time.Worked example
2√h = −kt + 4
Integrate and apply the initial value.
2 = −3k + 4 → k = 2/3
Use the second observation.
√h = 2 − t/3
Keep the nonnegative branch.
h = 0 at t = 6 minutes
Restrict the draining solution to 0 ≤ t ≤ 6.
06 · Units of the coefficient
If h is in metres and t in minutes, what units must k have in h′ = −k√h?
Hint
Compare m/min with k times √m.
Worked solution
k has units m^(1/2)/min.
06 / Use similar triangles before differentiating cone volume
The surface radius changes with the water height.
A conical tank has water radius r = h/2 and net inflow π/2 m³/min. Initially h = 1 m.Worked example
V = (1/3)πr²h = πh³/12
Replace r using the geometry.
dV/dt = (πh²/4) dh/dt
Differentiate with respect to time.
h² dh/dt = 2
Use the given inflow.
h³/3 = 2t + C
Separate and integrate.
h³ = 6t + 1 → h = (6t + 1)^(1/3)
Apply h(0) = 1; stop at the tank rim if its capacity is reached.
07 · Reach a specified height
When does this model reach h = 2 m, assuming the tank is tall enough?
Hint
Set h³ = 8.
Worked solution
t = 7/6 minutes.
08 · Why not constant height rate?
Why does constant volume inflow not give constant dh/dt in the cone?
Hint
The horizontal surface area changes.
Worked solution
dh/dt = 2/h² decreases as h grows: more volume is needed for each extra unit of height.
07 / A surface-dependent volume loss can give a linear radius
Use the same geometric radius throughout.
A spherical object loses volume at rate k times its surface area, with k > 0.Worked example
V = 4πr³/3; S = 4πr²
Write the two geometric formulas.
4πr² dr/dt = −k(4πr²)
Use the chain rule and the negative loss rate.
dr/dt = −k for r > 0
Cancel before integrating.
r = r₀ − kt
Apply r(0) = r₀.
0 ≤ t ≤ r₀/k
The model ends when the radius reaches zero.
09 · Vanishing time
If r₀ = 3 cm and k = 0.2 cm/min, when does the model reach zero radius?
Hint
Use r₀/k.
Worked solution
15 minutes.
10 · Volume behaviour
Is the volume decreasing linearly with time in this model?
Hint
Substitute the radius formula into volume.
Worked solution
No. V = (4π/3)(r₀ − kt)³ on the physical interval.
08 / Distinguish constant area growth from perimeter-driven growth
The verbal rate statement determines the equation.
A circular region has area A = πr². Its area grows at c times its circumference, c > 0.Worked example
dA/dt = c(2πr)
Translate the rate statement.
2πr dr/dt = 2πcr
Use the chain rule.
dr/dt = c for r > 0
The radius grows at a constant rate.
r = r₀ + ct
Apply an initially positive radius.
11 · Constant area rate instead
If dA/dt = a is constant and r(0) = r₀, find r(t) for a > 0.
Hint
Integrate the area first.
Worked solution
πr² = at + πr₀², so r = √(r₀² + at/π). This differs from linear radius growth.
12 · Evaluate a radius rate
A circular region has dA/dt = 6π cm²/s. Find dr/dt when r = 3 cm.
Hint
Use 2πr dr/dt = 6π.
Worked solution
dr/dt = 1 cm/s.
09 / Allow a rate to depend explicitly on time
A changing inflow is not a constant forcing term.
A tank has horizontal area 3 m² and net inflow 6t + 3 m³/min for t ≥ 0. Initially h = 2 m.Worked example
3 dh/dt = 6t + 3
Convert the signed volume rate.
dh/dt = 2t + 1
Simplify.
h = t² + t + C
Integrate in time.
h = t² + t + 2
Apply the initial height; use only until the tank overflows.
13 · Volume at two minutes
Find h and V at t = 2, if the tank has sufficient capacity.
Hint
Evaluate h, then use V = 3h.
Worked solution
h = 8 m; V = 24 m³.
10 / Verify the geometry, equation and physical interval
The algebra alone does not establish a useful model.
Differentiate the proposed solution and substitute into the original signed rate equation. Check the initial data, dimensional consistency and the assumptions about tank shape, flow and material. Stop or change the model at emptying, overflow or a change of conditions.
14 · Unit conversion
A tank receives 200 litres per minute. Express this in m³/min.
Hint
1000 litres = 1 m³.
Worked solution
0.2 m³/min.
15 · An incorrect conversion
A solution uses dh/dt = dV/dt for a tank whose horizontal area is 4 m². What is missing?
Hint
Differentiate V = 4h.
Worked solution
The factor 4: dV/dt = 4 dh/dt. The two rates also have different dimensions.
11 / Connect geometry to rates before integrating
Retain physical restrictions after algebraic rearrangement.
- Express volume or area using one varying length.
- Use the chain rule to link the given rate to that length.
- Choose the correct sign for net inflow or loss.
- Use initial and later data to determine constants.
- Check any cancelled zero-radius or zero-height case.
- Stop at emptying, vanishing or overflow.
- Verify units and the original differential equation.
16 · Final chain-rule setup
A sphere gains volume at a constant rate q > 0. Write dr/dt in terms of q and r for r > 0.
Hint
Differentiate V = 4πr³/3.
Worked solution
dr/dt = q/(4πr²). A constant volume rate gives a decreasing radius rate as the sphere grows.
Section 1 of 11 · Connect the rate to the geometric quantity