Water is poured into an empty cone at a constant rate of 8cm³/s After t seconds the depth of the water in the inverted cone is h cm, as shown in the diagram below - AQA - A-Level Maths Pure - Question 8 - 2022 - Paper 3

To find ( \frac{dV}{dh} ), we start from the volume formula:

$V = \frac{\pi h^3}{12}$

Next, we differentiate V with respect to h:

$\frac{dV}{dh} = \frac{d}{dh}\left(\frac{\pi h^3}{12}\right) = \frac{\pi}{12} \cdot 3h^2 = \frac{\pi h^2}{4}$

Using the given rate of water being poured, we know that:

$\frac{dV}{dt} = 8 \text{ cm}^3/s$

Also, we apply the chain rule:

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

Substituting in the values we have:

$8 = \frac{\pi h^2}{4} \cdot \frac{dh}{dt}$

Solving for ( \frac{dh}{dt} ), we rearrange:

$\frac{dh}{dt} = \frac{32}{\pi h^2}$

Now substituting ( t = 3 ) seconds, we first need to find the height h at that time:

Assuming the volume remains constant:

$V = \frac{1}{3} \text{Area} \cdot \text{Height} = \frac{\pi r^2 h}{3}$ (for a cone)

We can use the earlier derived relationship and substitute h that was found to calculate V accurately and substitute it back to determine the rate.