# An inverted conical water tank has a radius of 10m at the top and 24m high. If water flows into the tank at a rate of 20m^3/min, how fast is the...depth of the water increasing when water is 16m deep?

### 1 Answer | Add Yours

We'll recall the formula that gives the volume of a right circular cone:

V = pi*h*r^2/3, where h is the height of the cone and r is the radius of the base.

To determine the depth of increasing water, at the given rate, we'll have to find out dh/dt.

First, we'll determine r, using similar triangles:

r/h = 10/24

r = 5h/12

V = pi*h*25h^2/3*144

We'll differentiate both sides with respect to t:

dV/dt = 3*25*pi*h^2/3*144

dV/dt = (25pi*h^2/144)*(dh/dt)

Since dV/dt = 20m^3/min, when h = 16, we'll get:

20 = (25pi*16^2/144)*(dh/dt)

(dh/dt) = 144*20/25*256*pi

(dh/dt) = 144/5*64*pi

(dh/dt) = 36/5*16*pi

(dh/dt) = 9/5*4*pi

(dh/dt) = 0.1433

**The depth of water is increasing at a rate of about 0.1433 m/min.**