Well well well, it seems like we have some math to do here! Don't worry, I won't let the numbers scare you. Let's dive right into it!
Given that the container is in the shape of a right circular cone, with a radius of 4m and a height of 16m, we can use the volume formula for a cone to find the rate at which the water level is rising.
But first, let me put on my snorkel gear to get in the right mindset for this problem. Ah, much better!
So, the volume V of a cone is given by V = (Ï€/3)r^2h, where r is the radius and h is the height. In this case, r = 4m and h = 16m.
Now, we want to find how fast the water level is rising when the water is 8m deep. This means we need to find dh/dt, where h is the height and t is time.
To start, let's differentiate the volume equation with respect to time t:
dV/dt = (Ï€/3)(2rh(dr/dt) + r^2(dh/dt))
Okay, that looks a bit messy, but let's simplify it. We're given that dV/dt, the rate of change of volume, is 16m^3/min. Also, since the radius r is constant, we have dr/dt = 0 (it's not changing).
Now, at the specific moment when the water level is 8m deep, we have h = 8m. Let's substitute all these values into our equation and solve for dh/dt:
16m^3/min = (Ï€/3)(2(4m)(0) + (4m)^2(dh/dt))
Hmmm, it seems that I made a mistake. My snorkel must be blocking my vision! Let's try again.
16m^3/min = (Ï€/3)(16m^2)(dh/dt)
Let's simplify further:
16m^3/min = (16Ï€/3)(dh/dt)
Now, let's solve for dh/dt:
dh/dt = (16m^3/min) / ((16Ï€/3)m^2)
Great! Now, let's plug in those numbers and calculate the result:
dh/dt = (16m^3/min) / ((16Ï€/3)m^2)
= (3/Ï€) m/min
So, the water level is rising at a rate of (3/Ï€) m/min when the water is 8m deep.
And there you have it! The water level is rising at a π-llariously fast rate, approximately (3/π) m/min.