SOLUTION: Water is being poured at the rate of 2pi cubic meter per minute into an inverted conical tank that is 12 m deep with the radius of 6 m at the top. If the water level is rising at

Algebra ->  Test -> SOLUTION: Water is being poured at the rate of 2pi cubic meter per minute into an inverted conical tank that is 12 m deep with the radius of 6 m at the top. If the water level is rising at       Log On


   



Question 1117080: Water is being poured at the rate of 2pi cubic meter per minute into an inverted conical tank that is 12 m deep with the radius of 6 m at the top. If the water level is rising at the rate of 1/6 m/min and there is a leak at the bottom of the tank, how fast is the water leaking when the water is 6-m deep?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
For the conical tank, with radius R=6 and height h=12 ,
R%2Fh=1%2F2 <--> R=%281%2F2%29h .
Whatever the water level, the water in the tank is a cone
similar in shape to the tank itself, with R=%281%2F2%29h .

The volume of water in the tank when the water id h m deep is
,
with h in m and V in m%5E3 .
The volume in the tank increases at a rate, in m%5E3%2Fmin of
dV%2Fdt=%28pi%2F12%293h%5E2%28dh%2Fdt%29=%28pi%2F4%29h%5E2%28dh%2Fdt%29 .
When the water depth is h=6 , dh%2Fdt%29=1%2F6 ,
and the rate of water volume increase, in m%5E3%2Fmin , is
dV%2Fdt=%28pi%2F4%296%5E2%281%2F6%29=3pi%2F2 .

With water entering the tank at 2pi m%5E3%2Fmin ,
and water leaking out at an unknown rate of x m%5E3%2Fmin ,
the water volume is increasing at a rate of 3pi%2F2 m%5E3%2Fmin .
That means that
3pi%2F2=2pi-x --> x=2pi-3pi%2F2=%282-3%2F2%29pi=%281%2F2%29pi=pi%2F2

When the water depth is 6m, water is leaking at highlight%28pi%2F2%29 cubic meters per minute.