Mathematics

This is a related rates problem where you need to connect the changing volume of water to the changing depth. The key insight is to use the cone's geometry to express volume in terms of height alone, then differentiate with respect to time. Since the cone is inverted, the radius and height of the water stay proportional to the full container's dimensions—find that ratio first, substitute it into the volume formula, and you'll have a single equation linking $\frac{dV}{dt}$ and $\frac{dh}{dt}$.

This is a related rates problem where you need to connect the changing volume of water to the changing depth. The key insight is to use the cone's geometry to express volume in terms of height alone, then differentiate with respect to time. Since the cone is inverted, the radius and height of the water stay proportional to the full container's dimensions—find that ratio first, substitute it into the volume formula, and you'll have a single equation linking $\frac{dV}{dt}$ and $\frac{dh}{dt}$.