Question

What is the rate at which water is being pumped into the tank in cubic centimeters per minute, If the water level is rising at a rate of 19 centimeters per minute when the height of the water is 4.0?

Water is leaking out of an inverted conical tank at a rate of 7400 cubic centimeters per minute at the same time that water is being pumped into the tank at a constant rate. The tank has height 9 meters and the diameter at the top is 4.0meters. If the water level is rising at a rate of 19 centimeters per minute when the height of the water is 4.0
find the rate at which water is being pumped into the tank in cubic centimeters per minute.

Answer 1

`479026.749\ \ text{cubic cm./min}`

Explanation:

Let `h` be the height & `r` be the radius of water level in inverted cone tank of total height `H=9\ m` & radius of base `R=4/2=2\ m` at any time `t` then by property of similar triangles we have

`r/h=R/H`

`r={hR}/H`

`r={2h}/9`

At any instant of time `t`, if height `h` is the height of water level in the tank then at the same time, the radius of water surface will be `r={2h}/9`. Hence, the volume `V` of water filled at any time `t` is given as

`V=1/3\pi r^2h`

`V=1/3\pi ({2h}/9)^2h`

`V=\frac{4\pi}{243}h^3`

Differentiating the volume `V` w.r.t. time `t` as follows

`{dV}/{dt}=\frac{4\pi}{243}d/dt (h^3)`

`{dV}/{dt}=\frac{4\pi}{243}(3h^2)\ {dh}/dt`

`{dV}/{dt}=\frac{4\pi h^2}{81} {dh}/dt`

Above expression shows that net rate of filling the tank is `{dV}/{dt}` & `{dh}/{dt}` is the rate of change of height/level when height of water level is `h`

Now, setting `h=4\ m=400\ \text{cm` & `{dh}/dt=19\ \text{cm/min` in above expression, we get net rate of filling the tank

`{dV}/{dt}=\frac{4\pi (400)^2}{81} (19)`

`{dV}/{dt}=\frac{12160000\pi}{81} \ \ text{cubic cm./min}`

Now, the total rate of pumping the water into the tank

`=\text{net rate of increase of water }+\text{rate of leakage of water}`

`=\frac{12160000\pi}{81} +7400`

`=479026.749\ \ text{cubic cm./min}`