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

The volume of the water in the cone can be represented as:

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

To find the derivative (\frac{dV}{dh}), we can differentiate the volume formula with respect to h:

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

Now, we need to relate this to the rate at which water is poured into the cone. Since water is poured in at a constant rate of 8 cm³/s:

$\frac{dV}{dt} = 8$

Using the chain rule, we can relate (\frac{dV}{dt}) to (\frac{dV}{dh}):

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

Setting them equal gives us:

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

Solving for (\frac{dh}{dt}):

$\frac{dh}{dt} = \frac{8 \cdot 4}{\pi h^2} = \frac{32}{\pi h^2}$

Now, substituting (t = 3) into the equation: When t = 3 seconds, we compute the height h:

For (t = 3), we find:

$h = \sqrt[3]{\frac{288}{\pi}} = 288 \cdot \frac{1}{12} = 24$

This gives:

$\frac{dV}{dh} = \frac{\pi (3^2)}{4} = \frac{9\pi}{4}$

Using the values:

$\frac{dh}{dt} = \frac{8}{\frac{9\pi}{4}} = \frac{32}{9\pi}$

Substituting this back:

$\frac{dV}{dh} = \frac{6}{\sqrt{6\pi}}$