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Vi Nguyen · Jul 19, 2009, 02:02 PM

If water was leaking from an inverted conical tank at a rate of 10 000 cm³/min and at the same time water form another source is entering the tank at a constant rate Q. The tank is 6m high and the diameter at the top is 4m (r=2m=200cm). If the water level is rising at a rate of 20cm/min when the hight of water is 2m (200cm), find Q (rate going in the tank).

That is the question, I have converted all the numerics to cm, volume of cone as (1/3)pi.r²h and differentiated this with respect to time using the product rule to get:

dV/dt= (2/3)pi.r.(dr/dt)h+(1/3)pi.r².(dh/dt)

with r=(1/3)h=200/3 (this is from drawing diagram and letting (r/200)=(h/600))
h=200
(dr/dt)=(dr/dh)x(dh/dt)=(1/3)x20=20/3
(dh/dt)=20

and ended with an answer of (dV/dt), then reasoning that Q=(dV/dt)-10 000(rate going out of tank) to get 269252.6803cm³/min but the answer should have been (10000+800000.pi)/9=289252.6803 cm³/min, where have I gone wrong or am I totally off track?

galactus · Jul 19, 2009, 04:16 PM

If I follow your work correctly, it looks find. You probably just have an arithmetic error since your solution is so similar.

The volume of a cone is given by $V=\frac{\pi}{3}r^{2}h$ ... [1]

By similar triangles, we have $\frac{200}{600}=\frac{r}{h}$

$r=\frac{h}{3}$

Sub into [1]:

$V=\frac{\pi}{3}(\frac{h}{2})^{2}h=\frac{{\pi}h^{3} }{27}$

Differentiate implicitly:

$\frac{dV}{dt}=\frac{{\pi}h^{2}}{9}\cdot\frac{dh}{d t}$

But, $\frac{dV}{dt}=Q-10000$

$Q-10000=\frac{{\pi}h^{2}}{9}\cdot\frac{dh}{dt}$

dh/dt=20 and h=200 cm:

$Q-10000=\frac{(200)^{2}{\pi}}{9}(20)$

Solving for Q gives $Q=289252.68 \;\ cm^{3}/min$

Unknown008 · Jul 20, 2009, 07:40 AM

You error comes in here:

and ended with an answer of (dV/dt), then reasoning that Q=(dV/dt)-10 000(rate going out of tank) to get 269252.6803cm³/min but the answer should have been (10000+800000.pi)/9=289252.6803 cm³/min, where have I gone wrong or am I totally off track?

It is Q=(dV/dt)+10 000

Vi Nguyen · Jul 21, 2009, 06:04 AM

Hey your really good! I figured this out after sitting down and playing around a bit and writing the DE again. :)

Originally Posted by Unknown008

You error comes in here:

It is Q=(dV/dt)+10 000

Unknown008 · Jul 21, 2009, 06:24 AM

Originally Posted by Vi Nguyen

Hey ur really good! I figured this out after sitting down and playing around a bit and writing the DE again. :)

You're welcomed! :)