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 with the volume formula:
$V = \frac{\pi h^3}{12}$
Differentiating both sides with respect to h gives:
$\frac{dV}{dh} = \frac{3\pi h^2}{12} = \frac{\pi h^2}{4}$
Now, since water is being poured into the cone at a constant rate of 8 cm³/s, we can use the chain rule:
$\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}$
Setting $\frac{dV}{dt} = 8$, we have:
$8 = \frac{\pi h^2}{4} \cdot \frac{dh}{dt}$
Rearranging gives:
$\frac{dh}{dt} = \frac{8 \cdot 4}{\pi h^2} = \frac{32}{\pi h^2}$
Next, substituting t = 3 into the volume equation to find h:
Using the constant rate of water flow:
$V = 8t = 8 \cdot 3 = 24 \text{ cm}^3$
Now set this equal to the volume formula: $\frac{\pi h^3}{12} = 24$
Solving for h leads to
$h^3 = 24 \cdot 12 / \pi = \frac{288}{\pi}$
Taking the cube root:
$h = \sqrt[3]{\frac{288}{\pi}}$
Now substitute this value into \frac{dh}{dt}:
$\frac{dh}{dt} = \frac{32}{\pi (\sqrt[3]{\frac{288}{\pi}})^2}$
After calculations, you would arrive at:
$\frac{dh}{dt} = \frac{6}{\sqrt{6}\pi}$