Expert: Ahmed Salami - 4/13/2010

Question
a conical tan with vertex down is 20ft across the topa nd 24ft deep. if the wateris flowing inot the tank at a rate of 14 cubic ft per minute, find the rate of change of the depth of water when the water is 10 ft deep.

Answer
Hi Amanda,
The volume of a conical tank with radius R and height H is
V = πR²H/3
Let r and h be the instantaneous radius and height of the water filled section of the tank. The cone formed is similar to the conical tank in dimensions and so
r/h = R/H
r = hR/H
 = 10h/24 = 5h/12
The volume of water at point in time is
v = πr²h/3
 = π(5h/12)².h/3
 = π(25h²/144).h/3
 = 25πh³/432
The rate of change of this volume with respect to height of water is
dv/dh = 3.25πh²/432
     = 25πh²/144
Also, the rate of changes with respect to time are related by
dv/dt = dv/dh . dh/dt
14 = (25πh²/144)dh/dt
dh/dt = 14.144/25πh²
     = 2016/25π(10)²
     = 2016/2500π
     = 0.26 ft/min

Regards