# Related Rates: How fast is the water leaking from a conical-shaped tank?

Water is poured at the rate of 8 cubic feet per minute into a conical-shaped tank, 20 ft deep and 10 ft in diameter at the top. If the tank has a leak in the bottom and the water level is rising at the rate of 1 in./min, when the water is 16 ft deep, how fast is the water leaking?

The change in the volume of the water is the difference in the fill and leak rates (taking the leak rate to be a positive number).
$$\frac{dV}{dt} = \frac{dF}{dt} - \frac{dL}{dt}$$ Thus, the leak rate is the difference in the fill rate and the actual rate of increase in the volume of the water. $$\frac{dL}{dt} = \frac{dF}{dt} - \frac{dV}{dt}$$ The volume of the water when its height is $h$ and its radius is $r$ is $$V = \frac{1}{3}\pi r^2h$$ Since the inverted conical tank has a diameter of $10~\text{ft}$ at the top, its radius at the top is $5~\text{ft}$. By similar triangles, the ratio of the radius of the water to the height of the water in the cone is $$\frac{r}{h} = \frac{5~\text{ft}}{20~\text{ft}} = \frac{1}{4} \implies r = \frac{h}{4}$$ Thus, we can express the volume of the water as a function of $h$ by substituting $h/4$ for $r$, which yields $$V(h) = \frac{1}{3}\pi\left(\frac{h}{4}\right)^2h = \frac{1}{48}\pi h^3$$ Differentiating with respect to time yields $$\frac{dV}{dt} = \frac{1}{16}\pi h^2 \frac{dh}{dt}$$ We are given that $$\frac{dh}{dt} = 1~\frac{\text{in}}{\text{min}} = \frac{1}{12}~\frac{\text{ft}}{\text{min}}$$ when $h = 16~\text{ft}$. Thus, the rate at which the volume is increasing is $$\frac{dV}{dt} = \frac{1}{16}\pi (16~\text{ft})^2\left(\frac{1}{12}~\frac{\text{ft}}{\text{min}}\right) = \frac{4}{3}\pi~\frac{\text{ft}^3}{\text{min}}$$

Since the fill rate is $8~\text{ft}^3/\text{min}$, the leak rate is $$\frac{dL}{dt} = \frac{dF}{dt} - \frac{dV}{dt} = 8~\frac{\text{ft}^3}{\text{min}} - \frac{4}{3}\pi~\frac{\text{ft}^3}{\text{min}} = \left(8 - \frac{4}{3}\pi\right)~\frac{\text{ft}^3}{\text{min}}$$

The volume of a cone.

$v=\frac{1}{3}\pi r^2 h$

You have been given height of the water. Sufficient information to find the radius, the rate that the height is changing, and the ratio that the radius changes as the height changes.

$\frac{dv}{dt} = \frac{2}{3}\pi r h \frac{dr}{dh}\frac{dh}{dt} + \frac{1}{3}\pi r^2 \frac{dh}{dt}$

$\frac{dv}{dt} = 8 \frac{ft^3}{min}$ - rate of the leak