At what rate is the water leaking out of the second container?

AI Thread Summary
The problem involves three containers being filled with water at a rate of 20 cm³/sec, with the second container leaking water at an unknown rate. To determine the leak rate, the volume equations for each container must be established, focusing on the relationship between volume and height. The key is to find the time when the water levels in all three containers rise at the same rate, which can be achieved by comparing the first and third containers. By calculating the height and volume of the third container, the leak rate from the second container can be derived. Further assistance is requested to clarify the setup and calculations involved.
ziddy83
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HELP me please!

Hi, I am having trouble figuring out how to tackle this problem, it would be great if anyone can help! :eek:

Here is the problem...
The three containers shown below (sorry, i can only describe them and not draw them here, i'll describe em at the bottom) are all initially empty. Water is simultaneously poured into the containers at the constant rate of 20 cm^3/ sec. Water leaks out of a hole in the bottom of the second container at a constant rate. at some point in time, the water levels in the three containers are all rising at the same time. At what rate is the water leaking out of the second container?

Container 1: A cylinder with radius of 10cm.
Container 2: Cone shape with 60 degrees point
Container 3: Another cone with a 90 degrees point.

It would be awesome if i could get any help! THANKS!
 
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Hint : Write V = f(h), (where V : volume, h : height) for all three containers. Find the time at which the levels rise at the same rate by comparing 1 and 3. From this determine tha leak rate of 2.
 
can you give me another hint?
 
Ok, do i set this up by using the volume equation, f(V)= 1/3 *pi *r^2* h...

this is also given in the problem..

\frac {dV_3} {dt} = 20, V_3 = 20t, and since \frac {dV_2}{dt} = r(in) - r(out) = 20 - r (out), V_2 = (20-r (out))t

sorry i didnt include this before...but yeah, another push in the right direction would be great, thanks.
 
can anyone help me on this problem?
 
V_1 = \pi R^2 h_1 => \frac{dV_1}{dt} = \pi R^2 \frac{dh_1}{dt} = 20

=> \frac{dh_1}{dt} = \frac {20}{100 \pi} = \frac{1}{5 \pi}

V_3 = \frac{1}{3} \pi r_3^2 h_3

But~\frac{r_3}{h_3} = tan 45 = \frac{1}{\sqrt{2}} => V_3 = \frac{1}{6} \pi h_3^3

So~ \frac{dV_3}{dt} = \frac {\pi}{2} h_3^2 \frac {dh_3}{dt} = 20

=> \frac{dh_3}{dt} = \frac {40}{\pi h_3^2}

But~at~some~t,~ \frac{dh_3}{dt} = \frac{dh_1}{dt} = \frac{1}{5 \pi}

From this, you can get h3, and from that, V3. Dividing V3 by 20 gives you the time in seconds. Can you take it from there ?
 
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