Related rates (conical tank)

[jwpaine]

Water is leaking out of an inverted conical tank at a rate of 10900 cubic centimeters per
minute at the same time water is being pumped into the tank at a constant rate. The tank
has height 9 meters and diameter at the top is 3.5 meters. If the water level is rising at a
rate of 22 centimeters per minute when the height of the water is 1.5 meters, find the rate
at which water is being pumped into the tank in cubic centimeters per minute.

This is what I have so far:

V = (1/3)(pi)(r^2)h
V' = (1/3)(pi)2rr'h'

Vout = 10900cc/min
V' = [ Vin - 10900cc/min ]

h' = 22cm/min when h = 3/2 meters

not sure where to go from here.

 

[galactus]

Hey JW:

\(\displaystyle \frac{dV}{dt}=\text{rate in - rate out}\)

Let's say the rate in is x and out is 10900. Therefore, \(\displaystyle \frac{dV}{dt}=x-10900\)

Now, by similar triangles and consistent units.

The radius of the tank is 175 cm, height is 900 cm. We have dh/dt=22 when h=150 cm.

Similar triangles:

\(\displaystyle \frac{r}{h}=\frac{175}{900}=\frac{7}{36}\)

So, \(\displaystyle r=\frac{7}{36}h\)

Sub into the volume of a cone formula:

\(\displaystyle V=\frac{1}{3}{\pi}(\frac{7}{36}h)^{2}h=\frac{49\pi}{3888}h^{3}\)

\(\displaystyle \frac{dV}{dt}=\frac{49\pi}{1296}h^{2}\frac{dh}{dt}\)

Now, you have everything you need. You have dV/dt, dh/dt, h. Sub them in and solve for x (the rate in).

 

[jwpaine]

Thanks.

I came up with \(\displaystyle x = \frac{69695.8^{3}cm}{min}\)

Correcto?

 

[galactus]

Yep, that's right. If you want it in m^3/min, divide by 1,000,000

That would be 0.0697 m^3/min