Well, let's dive into this problem! We know that the tank is shaped like a cone with a radius of 5 feet and a height of 12 feet. Now, we want to find the rate of change of the depth of water in the tank when the depth is 7 feet.
To begin, let's first find the volume of the cone-shaped tank. The formula for the volume of a cone is V = (1/3) * π * r^2 * h, where V represents volume, r is the radius of the base, and h is the height of the cone.
Substituting in the given values, we have V = (1/3) * π * 5^2 * 12.
Now, let's find the rate of change of the volume with respect to time. Given that the tank is being filled at a rate of 3 cubic feet per minute, we can write dV/dt = 3.
Next, we need to find the rate of change of the depth of water, which we'll call dh/dt, when the depth is 7 feet.
Now, the volume of a cone can also be expressed as V = (1/3) * π * r^2 * H, where H is the height of the water in the tank.
Since the depth of the water is 7 feet, the height of the water in the tank is equal to the height of the cone minus the depth of the water, which is 12 - 7 = 5 feet.
So, now we can rewrite the volume as V = (1/3) * π * 5^2 * 5.
Differentiating both sides of the equation with respect to time, we get dV/dt = (1/3) * π * 5^2 * dH/dt.
Since dV/dt is given as 3 cubic feet per minute, we can substitute those values in.
3 = (1/3) * π * 5^2 * dH/dt.
Now, we just need to solve for dH/dt, which represents the rate of change of the depth of water.
dH/dt = (3 * 3) / ((1/3) * π * 5^2).
Calculating this, we find dH/dt ≈ 0.114 feet per minute.
So, the rate of change of the depth of water in the tank when the depth is 7 feet is approximately 0.114 feet per minute.
I hope that made a splash in terms of explanation!