Water flows at a rate of 2 cu. ft. per minute into a vessel in the form of an inverted right circular cone of altitude 2 ft. and the radius of base 1 ft. At what rate is the surface rising when the vessel is one-eighth filled?
To find the rate at which the surface is rising, we need to find the rate at which the volume of water in the vessel is increasing.
First, let's find the volume of the cone at any point in time. The volume of a cone can be calculated using the formula V = (1/3)πr^2h, where V is the volume, r is the radius of the base, and h is the height.
In this case, the radius (r) of the cone's base is 1 ft and the height (h) is 2 ft. The volume of the cone is V = (1/3)π(1^2)(2) = (2/3)π ft^3.
Next, let's find the rate at which the volume of water in the vessel is increasing. The rate at which water is flowing into the vessel is given as 2 cu. ft. per minute. Therefore, the rate of change of volume is dV/dt = 2 ft^3/min.
Now, let's find the rate at which the height of the water in the vessel is changing (dh/dt) when the vessel is one-eighth filled.
The volume of water in the vessel when it is one-eighth filled is (1/8) * (2/3)π = (1/12)π ft^3.
Since V = (1/3)πr^2h, we can rewrite this equation to solve for h as h = (3V)/(πr^2).
Now, let's differentiate this equation with respect to t to find the rate at which the height is changing when the vessel is one-eighth filled:
dh/dt = (3(1/12)π)/(π(1^2)) * (dV/dt)
= (1/4) * (2 ft^3/min)
= 1/2 ft/min.
Therefore, the surface of the water in the vessel is rising at a rate of 1/2 ft per minute when the vessel is one-eighth filled.