A conical water tank with vertex down has a radius of 12 feet at the top and is 23 feet high. If water flows into the tank at a rate of 20 {\rm ft}^3{\rm /min}, how fast is the depth of the water increasing when the water is 18 feet deep?
Make your sketch of the cone
At a given time of t minutes,
let the height of the water be h ft
let the radius of the water level be r ft
by ratios:
r/h = 12/23
23r = 12h
given : dV/dt = 20 ft^3/min
find : dh/dt , when h = 18
so I will need an equation containing V , for volume, and h
V = (1/3)π r^2 h , but from 23r = 12h , r = 12h/23
V = (1/3)π(144h^2/529)(h)
= (48/529)π h^3
dV = (144/529)π h^2 dh/dt
You have dV/dt and h
sub in, simplify and solve for dh/dt