Connected Rates: Filling a Cylindrical Tank

This connected-rates question links the rate of volume to the rate of height. For a cylinder with fixed radius, the volume is a simple multiple of the height, so $\frac{dV}{dh}$ is constant and the chain rule is especially clean.

Water is poured into a cylindrical tank of radius $5$ cm at a constant rate of $20\ \text{cm}^3\,\text{s}^{-1}$. Find the rate at which the water level is rising. [5]

The working

Step 1, write $V$ in terms of $h$ (radius fixed at $5$): $V = \pi r^2 h = \pi (5)^2 h = 25\pi h \quad \text{(M1)}$

Step 2, differentiate to link $V$ and $h$: $\frac{dV}{dh} = 25\pi \quad \text{(A1)}$

Step 3, build the chain for the rate you want: $\frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt} = \frac{1}{25\pi} \times 20 \quad \text{(M1)}$

Step 4, evaluate: $\frac{dh}{dt} = \frac{20}{25\pi} = \frac{4}{5\pi} \approx 0.255\ \text{cm s}^{-1} \quad \text{(A1, A1)}$

Where the marks are won and lost

• Express $V$ as a function of $h$ with the radius as a constant: $V = 25\pi h$. Treating $r$ as a variable (as for a cone) is wrong here, the tank’s radius is fixed.
• The chain uses $\frac{1}{dV/dh}$, so $\frac{1}{25\pi}$, not $25\pi$. Inverting correctly is the usual slip.
• Units and a decimal ($0.255\ \text{cm s}^{-1}$, 3 s.f.) or the exact $\frac{4}{5\pi}$ are both acceptable; a bare number risks the final mark.

Common mistakes

• Using the cone volume $\frac13\pi r^2 h$ or letting $r$ vary.
• Multiplying by $25\pi$ instead of dividing.
• Dropping the units.

Full method: Rates of Change notes. Topic home: Calculus pillar.