A water container is made in the shape of a hollow inverted right circular cone with semi-vertical angle of 30°, as shown in Figure 1 - Edexcel - A-Level Maths Pure - Question 6 - 2018 - Paper 9

We know that: $\frac{dV}{dt} = 200 \text{ cm}^3 \text{s}^{-1}$ From the derived equation for volume: $V = \frac{1}{9} \pi h^3$ Differentiating both sides with respect to time t: $\frac{dV}{dt} = \frac{d}{dt}\left(\frac{1}{9} \pi h^3\right)$ Using the chain rule: $\frac{dV}{dt} = \frac{1}{9} \pi \cdot 3h^2 \frac{dh}{dt}$ Thus: $\frac{dV}{dt} = \frac{\pi h^2}{3} \frac{dh}{dt}$ Setting the volume rate: $200 = \frac{\pi (15)^2}{3} \frac{dh}{dt}$ Now solving for ( \frac{dh}{dt} ): $200 = \frac{\pi (225)}{3} \frac{dh}{dt}$ $200 = 75\pi \frac{dh}{dt}$ $\frac{dh}{dt} = \frac{200}{75\pi} = \frac{8}{3\pi} \text{ cm/s}$ Therefore, the rate of change of the depth of the water when $h = 15$ is ( \frac{8}{3\pi} \text{ cm/s} ).