A 24ft high conical water tank has its vertex on the ground and radius of the base is 10 ft. If water flows into the tank at a rate of 20 ft3/min, how fast is the depth of water increasing when the depth of the water is 20 ft?

To find the rate at which the depth of the water is increasing, we need to use related rates. In this case, we are given the rate at which water is flowing into the tank (20 ft³/min) and we need to find the rate at which the depth of the water is increasing.

Let's begin by setting up our variables and equations:

Let h be the depth of the water in the conical tank (in feet), and V be the volume of the water in the tank (in cubic feet).
Given:
- The radius of the base of the conical tank is 10 ft.
- The vertex of the cone is on the ground, which means the height of the cone is equal to the total height of the tank.
- The total height of the tank is 24 ft.

We can use the formula for the volume of a cone to relate the variables:
V = (1/3)πr²h

Now, we want to find dh/dt, the rate at which the depth of the water is increasing, when h = 20 ft.

To do this, we need to take the derivative of the volume equation with respect to time (t):

dV/dt = (1/3)π(2rh(dr/dt) + r²(dh/dt))

Since we are given that the water is flowing into the tank at a rate of 20 ft³/min (dV/dt = 20), and we need to find dh/dt when h = 20, we can substitute these values into the equation:

20 = (1/3)π(2(10)(20/3) + 100(dh/dt))

Simplifying the equation:

20 = (20/3)π + 100(dh/dt)
100(dh/dt) = 20 - (20/3)π
dh/dt = (20 - (20/3)π) / 100

Now we can substitute π ≈ 3.14 into the equation and calculate dh/dt:

dh/dt = (20 - (20/3)(3.14)) / 100
dh/dt = (20 - (20/3)(3.14)) / 100
dh/dt ≈ 0.0206 ft/min

Therefore, the depth of the water is increasing at a rate of approximately 0.0206 ft/min when the depth of the water is 20 ft.

The vertex is on the ground, so the tank is in a funnel position.

Let the water height be h, then the radius of the surface of water is r(h)=10h/24=5h/12
The volume at a height of h is
V(h)=(π/3)r(h)² h
=(π/3)(5h/12)² h
=(25π/432)h³
Differentiate with respect to time, t

dV(h)/dt
=(25π/432)*3h²dh/dt
=(25π/144)h² dh/dt

Since dV(h)/dt is known (=20 ft³/min), you can solve for dh/dt.

Note that the unit of dh/dt is in ft/min.