Question: A right circular cone with a radius of 5 cm and a height of 12 cm is being filled with water at a constant rate of 3 cm^3/s. At what rate is the depth of the water increasing when the water is 8 cm deep?

Answer: To solve this problem, we can use related rates, which involves finding the rate of change of one quantity in terms of the rate of change of another related quantity. In this case, the volume of water in the cone is changing, and we want to find the rate at which the depth of the water is increasing.

First, let's denote the radius of the water surface in the cone as $r$, the depth of the water as $h$, and the volume of water in the cone as $V$. We can use the formula for the volume of a cone to relate $r$ and $h$: [ V = \frac{1}{3}\pi r^2 h ]

We are given that the radius of the cone is 5 cm and the height is 12 cm, and the water is being filled at a constant rate of 3 cm^3/s. To find the rate at which the depth of the water is increasing, we can take the derivative of the volume with respect to time: [ \frac{dV}{dt} = \frac{1}{3}\pi (2r)(dr/dt)h + \frac{1}{3}\pi r^2 (dh/dt) ]

Now we substitute the given values: [ 3 = \frac{1}{3}\pi (2 \times 5)(dr/dt)8 + \frac{1}{3}\pi 5^2 (dh/dt) ] [ 3 = \frac{10}{3}\pi(dr/dt)8 + \frac{25}{3}\pi(dh/dt) ]

We are given that the depth of the water is 8 cm, so we can find the radius using similar triangles. Using the similar triangles property, we know that the ratio of the radius to the height is constant, so: [ r/h = 5/12 ] [ r = \frac{5}{12}h ]

Now substitute $r = \frac{5}{12} \times 8 = \frac{10}{3}$ into the equation and solve for $dr/dt$: [ 3 = \frac{10}{3}\pi(dr/dt)8 + \frac{25}{3}\pi(dh/dt) ] [ 3 = \frac{10}{3}\pi(dr/dt)8 + \frac{25}{3}\pi(dh/dt) ] [ 3 = \frac{10}{3}\pi(dr/dt)8 + \frac{25}{3}\pi(dh/dt) ] [ 3 = \frac{10}{3}\pi(dr/dt)\times 8 + \frac{25}{3}\pi(dh/dt) ] [ 3 = \frac{80}{3}\pi(dr/dt) + \frac{25}{3}\pi(dh/dt) ] [ dh/dt = \frac{3 - \frac{80}{3}\pi(dr/dt)}{\frac{25}{3}\pi} ] [ dh/dt = \frac{9 - 80 dr/dt}{25\pi} ]

Thus, the rate at which the depth of the water is increasing when the water is 8 cm deep is $\frac{9 - 80 dr/dt}{25\pi}$.