A conical water tank with vertex down has a radius of 13 feet at the top and is 28 feet high. If water flows into the tank at a rate of 10 ft3/min f t 3 / m i n

Question

 A conical water tank with vertex down has a radius of 13 feet at the top and is 28 feet high. If water flows into the tank at a rate of 10 ft3/minft3/min, how fast is the depth of the water increasing when the water is 18 feet deep?

Answer 1

Let

R = 13 ft be the radius of the tank at the top,

H = 28 ft be the height of the tank,

h(t) be the height of the water at a time t,

r(t) = h(t)R/H be the radius of the top of the water at time t,

h'(t) be the rate of increase in water level (to be determined),

r'(t) = h'(t)R/H be the rate of increase in the radius of the water level at a time t,

V(t) = πr²(t)h(t)/3 be the volume of the water at time t.

For clarity, lets drop the dependence on (t) and write

V = πr²h/3

Then

V' = π/3(2rr'h + r²h') is the rate of increase in the volume, which is given as 10 ft³/min.

In the above, replace r and r' with their equivalents in terms of h:

V' = π/3(2h(R/h)h'(R/H)h + h²(R/H)²h') = π/3(2h²(R/h)² + h²(R/H)²)h' = π(hR/H)²h'

h' = V'/(π(hR/H)²)

Substituting actual values for the point in time when h=18ft

h' = 10/(π(18×13/28)²) = 0.046 ft/min