If the radius of the water level is increasing at a rate of 3 ft./min when the height level of the water is 4 ft., what is rate at which water is flowing into the tank?

Question

A conical tank stands point down and has a height of 16 ft. and a base radius of 12 ft. Water is flowing into the tank.

A) If the radius of the water level is increasing at a rate of 3 ft./min when the height level of the water is 4 ft., what is rate at which water is flowing into the tank?

B) If water flows into the tank at a rate of 3ft^3/min, what is the radius when the water level is rising at a rate of 2ft/min?

Answer 1

The volume of water in the conical tank is:

V = r²*h*π/3, with r the radius of the water level and h its height.

Using similar triangles to get the relationship between the radius and the height:

16/h = 12/r

h = 4*r/3

Substituting this in V:

V = 4*r³*π/9

The rate the volume increases (= the rate of water flowing in) is:

dV/dt = 4/3*r²*π * (dr/dt)

A)

We know that the rate of the radius is changing with 3 ft/min (= dr/dt) when the height h = 4 ft.

For that height, the radius r = 3 ft.

The amount of water flowing in at that moment is

dV/dt (h = 4) = 27*π ft³/min

B)

Now we know that dV/dt = 3 ft³/min, and the height is changing by 2 ft/min (= dh/dt)

dV/dt = 4/3*r²*π * (dr/dt),

dr/dt = (dr/dh)*(dh/dt) = 3/4 * dh/dt (since r = 3/4 *h)

dV/dt = r²*π*dh/dt

r = 0.7 ft (rounded to 1 SF)