A conical tank with a radius of 8 meters and a height of 12 meters is filled with water at a rate of 10 cubic meters per minute. Find the rate at which the water level is rising when the water is 6 meters deep.
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Let's denote the radius of the water surface in the tank as R and the height of the water as h. We want to find the rate at which the water level is rising, which is dh/dt.
We can first find an equation that relates R and h. Since the tank is conical, we know that the ratio of the radius to the height is constant. This means that R/h is always the same.
Initially, when the tank is empty, the radius is 0 since there is no water, and the height is 0 as well. Therefore, we have the initial condition R(0) = 0 and h(0) = 0.
As the water level rises, we can write the equation R/h = k, where k is a constant. To find this constant, we can use the initial condition: R(0)/h(0) = 0/0 = k.
Now, let's find an equation that relates R and h given that the rate of change of the volume of water is 10 cubic meters per minute.
The volume of a cone can be given by the formula V = (1/3)πr^2h, where r is the radius and h is the height. Since we are given the rate of change of volume, we can differentiate this equation with respect to time t.
dV/dt = (1/3)π(2rh(dr/dt) + r^2(dh/dt))
Since the radius of the water surface is R and the height is h, we can rewrite the equation as:
10 = (1/3)π(2Rh(dR/dt) + R^2(dh/dt))
We can rearrange the equation to solve for (dh/dt):
10 = (2/3)πRh(dR/dt) + (1/3)πR^2(dh/dt)
We are given that the radius R is 8 meters and the height h is 6 meters. Substituting these values into the equation, we have:
10 = (2/3)π(8)(6)(dR/dt) + (1/3)π(8^2)(dh/dt)
Simplifying, we get:
10 = 16π(dR/dt) + 64π(dh/dt)
Now, we know that R/h = k, so we can differentiate this equation with respect to time t:
(dR/dt)/h - R(dh/dt)/h^2 = 0
This equation tells us that (dR/dt)/h = R(dh/dt)/h^2, and we can rearrange it to solve for (dR/dt):
(dR/dt) = (R/h)(dh/dt)
Substituting this into the previous equation, we get:
10 = 16π[(R/h)(dh/dt)] + 64π(dh/dt)
10 = 16π(R/h + 4)(dh/dt)
Now, we can solve for (dh/dt):
10/(16π(R/h + 4)) = (dh/dt)
Substituting R = 8 and h = 6, we have:
10/(16π(8/6 + 4)) = (dh/dt)
Simplifying, we find:
10/(16π(4/3 + 4)) = (dh/dt)
10/(16π(16/3)) = (dh/dt)
10/(16π/3)(1/16) = (dh/dt)
3/π = (dh/dt)
Therefore, the rate at which the water level is rising when the water is 6 meters deep is 3/π cubic meters per minute.
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