## Water leaking from a tank

Water leaks from a vertical cylindrical tank through a small hole in its base at a rate proportional to the square root of the volume remaining. The tank initially contains 175L and 15L leak out during the first day.

When will the tank be half enpty?

How much water will remain in the tank after 4 days?

So the differential equation is dV/dt=-k(sqrtV)

Differntiating I have 2(sqrtVi-sqrtVf) = -kt+c

I 'm not sure about my lower bounds for V though? I have 175 for my upper bound, but what would be the lower bound? Would it be half of 175 for the first question, and 175 -15, so 160for the second question?

I also have to find the value of c. Initially Vi is 175, but so is Vf since none has leaked out yet, but this doesn't seem right. Unless you use how much is left after one day to find c. Is this correct?

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Hi Gogsey!
 Quote by Gogsey Differntiating I have 2(sqrtVi-sqrtVf) = -kt+c
(have a square-root: √ )

(and you mean "integrating", of course)

Why are you using Vi and Vf?

Your Vi is just part of the constant of integration, c.

So just write 2√V = -kt + c …
carry on from there.

 Integrating, Yup, lol. Ok, so at t = 0, we have 175 L, so does that mean, C = 2sqrt(175)? Do we have to find the value of k?

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## Water leaking from a tank

 Quote by Gogsey Ok, so at t = 0, we have 175 L, so does that mean, C = 2sqrt(175)?
Just got up

what happened to that √ i gave you?

Yes, C = 2√175.

Now start "at t = 4, …"