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?
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.