A container in the form of a right circular cone (vertex down) has a radius of 4m and height of 16m. If water is poured into the container at the constant rate of 16m^3/min, how fast is the water level rising when the water is 8m deep.

V=(pi/3)(r^2)h

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 = (π/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 = (π/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 = (π/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 = (π/3)(16m^2)(dh/dt)

Let's simplify further:

16m^3/min = (16π/3)(dh/dt)

Now, let's solve for dh/dt:

dh/dt = (16m^3/min) / ((16π/3)m^2)

Great! Now, let's plug in those numbers and calculate the result:

dh/dt = (16m^3/min) / ((16π/3)m^2)
= (3/π) m/min

So, the water level is rising at a rate of (3/π) 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.

To find the rate at which the water level is rising when the water is 8m deep, we need to determine how the volume of water in the cone is changing with respect to the height.

The volume V of a cone can be calculated using the formula V = (1/3)πr^2h, where r is the radius and h is the height.

Differentiating the volume with respect to time t will give us the rate at which the volume is changing:

dV/dt = (dV/dh) * (dh/dt)

Since the cone is being filled at a constant rate, we know that dh/dt = 16 m^3/min.

Now, let's find dV/dh:
V = (1/3)πr^2h
dV/dh = πr^2

Substituting the given values:
r = 4m
h = 8m

dV/dh = π(4^2) = 16π m^2

Now, let's substitute the values into the equation for the rate of change of volume with respect to time:

dV/dt = (dV/dh) * (dh/dt)
dV/dt = (16π m^2) * (16 m^3/min)
dV/dt = 256π m^3/min

Therefore, the rate at which the water level is rising when the water is 8m deep is 256π m^3/min.

To find how fast the water level is rising, we need to find the rate at which the volume of the water is increasing with respect to time. This can be done by taking the derivative of the volume formula with respect to time (t).

The formula for the volume of a cone is V = (π/3) * r^2 * h, where V is the volume, r is the radius, and h is the height. In this case, r = 4m and h is a function of time, h(t).

We are given that water is being poured into the container at a constant rate of 16m^3/min, so dh/dt = 16m^3/min.

To find dh/dt when the water is 8m deep, we need to substitute the given values into the volume formula and then differentiate it with respect to time.

First, substitute r = 4m and h = 8m into the volume formula:
V = (π/3) * (4^2) * 8
V = (64π/3) * m^3

Now, differentiate both sides of the equation with respect to time:
dV/dt = d((64π/3) * m^3)/dt

The derivative of a constant times a function (in this case, the constant is (64π/3)) with respect to time is simply the derivative of the function times the constant:
dV/dt = (64π/3) * dh/dt

Now, substitute the rate of change of the volume (dV/dt) as 16m^3/min:
16 = (64π/3) * dh/dt

Finally, solve for dh/dt, the rate at which the water level is rising:
dh/dt = (16 * 3) / (64π)
dh/dt = 48 / (64π)
dh/dt ≈ 0.2387 m/min

Therefore, the water level is rising at a rate of approximately 0.2387 meters per minute when the water is 8 meters deep.

easy way:

r = (4/16) h = (1/4) h
surface area = pi r^2
= pi (1/16) h^2
dV = surface area * dh
dV = pi (1/16) h^2 dh
DV/dt = (1/16) pi h^2 dh/dt

hard way:
V = (pi/3) r^2 h given
again r^2 = (1/16) h^2
V = (pi/3)(1/16) h^3
dV/dh = (pi/16) h^2
again dV/dt = (pi/16) h^2 dh/dt