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

The rate at which volume is changing is: $\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}$

Given that $\frac{dV}{dt} = 80\text{ cm}^3\text{s}^{-1}$, we can substitute this and express $\frac{dh}{dt}$ as:

$80 = \left( 8\text{π}h + 16\text{π} \right) \cdot \frac{dh}{dt}.$

Now we can substitute $h = 6$:

$\frac{dh}{dt} = \frac{80}{8\text{π}(6) + 16\text{π}} = \frac{80}{48\text{π} + 16\text{π}} = \frac{80}{64\text{π}} = \frac{5}{4\text{π}}.$

Calculating the numerical value (using $\text{π} \approx 3.14$) gives: $\frac{dh}{dt} \approx \frac{5}{4 \cdot 3.14} \approx 0.398 \text{ cm/s}.$

Thus, when the depth is $h = 6$ cm, the rate of change of the depth of the water is approximately $0.398$ cm/s.