Expert: Ahmed Salami - 1/7/2010

Question
62. Draining Water- water drains from the conical tank shown in the figure at a rate of 5 ft^3/min. (There is a pic with a large cone that a smaller cone inside it. The height of the larger cone is 10 in. with a radius of 4 in. The smaller cone inside it has a hight of h and a radius of r.)
a. what is the relationship between the variables of h and r.
b. how fast is the water level dropping when h= 6ft?

Answer
Hi Ellen,
The volume of a conical tank with radius R and height H is
V = πR²H/3
If water is draining from the tank, the volume of water at any point where the radius at the surface of the water is r and with height h is given by
v = πr²h/3
The smaller cone containing all the water at this point is similar to the larger cone and so the ratio of the dimensions is constant i.e
r/h = R/H
   = 4/10
   = 2/5
r = 2h/5
Now, writing v in terms of h only
v = π(2h/5)².h/3
 = π(4h²/25).h/3
 = 4πh³/75
Differentiating with respect to time,
dv/dt = (4π/75)(3h²)dh/dt
     = (4πh²/25)dh/dt
dh/dt = (25/4πh²)dv/dt
but dv/dt = 5 ft³/min
So, at h = 6
dh/dt = (25/4π.6²).5
     = 125/144π ft/min
The water level is dropping at about 0.28 ft/min. Actually, both dv/dt and dh/dt are negative since v and h are decreasing with time.

Regards