A conical tank( with vertex down) is 10 feet across the top and 18 feet deep. As the water flows into the tank, the change is the radius of the water at a rate of 2 feet per minute, find the rate of change of the volume of the water when the radius of the water is 2 feet.
Let y be the water level height above the vertex. The volume of water is
V = (pi/3)r^2 y
From the dimensions you have provided, r = (5/18) y
y = (18/5) r
V = (pi/3)(18/5)^2 r^3
Calculate dV/dt = (dV/dr)*(dr/dt) and evaluate it when r = 2 ft.
In your case, dr/dt = 2 ft/min
To find the rate of change of the volume of the water, we can use the formula for the volume of a cone, V = (1/3)πr^2h, where V is the volume, r is the radius of the water, and h is the height or depth of the water.
Given that the tank is conical with the vertex down, we can assume that the water level forms a cone inside the tank.
We are given that the tank has a top diameter of 10 feet, which means the radius of the top of the cone is 5 feet. We also know that as the water flows into the tank, the radius of the water changes at a rate of 2 feet per minute.
Let's denote the radius of the water as r(t), where t is the time in minutes. We need to find the rate of change of the volume of the water, dV/dt, when the radius of the water is 2 feet.
To solve this problem, we need to relate the variables r and h. Since the tank is conical, we can use similar triangles to express h in terms of r.
The larger triangle formed by the tank has a height of 18 feet and a base diameter of 10 feet. The smaller triangle formed by the water has a height of h and a base diameter of 2r. These two triangles are similar, so we can set up the following proportion:
(18)/(10) = h/(2r)
Simplifying this equation, we get:
9/5 = h/r
Now, we can express the height or depth of the water, h, in terms of the radius, r:
h = (9/5)r
Substituting this expression for h in the formula for the volume of a cone, we get:
V = (1/3)πr^2((9/5)r)
V = (3/5)πr^3
Now, we can differentiate both sides of this equation with respect to time t to find the rate of change of the volume:
dV/dt = (3/5)d(πr^3)/dt
Using the chain rule, we can differentiate πr^3 with respect to t:
dV/dt = (3/5)(3πr^2)(dr/dt)
Given that dr/dt = 2 feet per minute (the rate at which the radius changes), and we are interested in finding the rate of change of volume when the radius is 2 feet, we can substitute these values:
dV/dt = (3/5)(3π(2^2))(2)
Simplifying this equation, we get:
dV/dt = (3/5)(3π(4))(2)
dV/dt = (3/5)(12π)(2)
dV/dt = (36/5)π
Therefore, the rate of change of the volume of the water when the radius is 2 feet is (36/5)π cubic feet per minute.