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

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SUMMARY

The discussion centers on calculating the leak rate of water from the second container, which is a cone with a 60-degree angle, while water is poured into three containers at a constant rate of 20 cm³/sec. The first container is a cylinder with a radius of 10 cm, and the third container is a cone with a 90-degree angle. To find the leak rate from the second container, participants suggest using the volume equations for each container and comparing the rates of height increase. The key equations include V = f(h) for each container and the relationship between the rates of change of volume and height.

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  • Understanding of calculus, specifically derivatives and rates of change.
  • Familiarity with geometric volume formulas for cylinders and cones.
  • Knowledge of the relationship between volume and height in fluid dynamics.
  • Ability to set up and solve equations based on given rates and geometric properties.
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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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