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

Given that water flows into the vase at a constant rate of 80 cm³s⁻¹, we use the relationship:

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

Substituting the known rates leads to:

$80 = (8\pi h + 16\pi) \cdot \frac{dh}{dt}$

Now substituting h = 6 into the differentiated volume formula:

$\frac{dV}{dh} = 8\pi(6) + 16\pi = 48\pi + 16\pi = 64\pi$

Then, substituting into the relationship:

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

To find (rac{dh}{dt}):

$\frac{dh}{dt} = \frac{80}{64\pi} = \frac{5}{4\pi} \approx 0.398 \text{ cm s}^{-1}$