A vase with a circular cross-section is shown in Figure 2 - Edexcel - A-Level Maths Pure - Question 6 - 2014 - Paper 7

We know that water flows into the vase at a rate of $\frac{dV}{dt} = 80 \, \text{cm}^3\text{s}^{-1}$. Using the relationship between $V$ and $h$, by the chain rule we have:

$\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}$

Now substituting for $\frac{dV}{dh}$:

$80 = 8\pi(h + 2) \cdot \frac{dh}{dt}$

Substituting $h = 6$ into the equation gives:

$80 = 8\pi(6 + 2) \cdot \frac{dh}{dt}$

This simplifies to:

$80 = 8\pi(8) \cdot \frac{dh}{dt}$

Or:

$80 = 64\pi \cdot \frac{dh}{dt}$

Now, we isolate for $\frac{dh}{dt}$:

$\frac{dh}{dt} = \frac{80}{64\pi} = \frac{5}{4\pi}$

To get a numerical approximation, we calculate:

$\frac{dh}{dt} \approx \frac{5}{4 \cdot 3.14} \approx 0.398 \, \text{cm/s}$