Question

A 1000 gallon tank has a leak. 500 gallons drained the first 30 minutes. Find A(t), amount of water left in tank t minutes. Drains at rate proportional to the product of the time elapsed and the square root of the amount of water present.

Answer 1

I am just really confused on what to do with the rate... I have a few examples and just dont see how they are getting these equations.

For example... one says the same 1000 tank. 200 gallons leaked in first 10 minutes. The rate proportional to the product of the time elapsed and the amount of water present.

and the answer is

A(t) = 1000(4/5)^(t^2/100)

Answer 2

Well you're looking at the `answer`, not the differential equation they started with. So that's a little more confusing.

This one will setup similarly to that population problem.
If I'm interpretting the problem correctly, I think this is our differential equation,
\[\large \frac{dA}{dt}=k t \sqrt{A}\]
Although maybe dA/dt should be negative since the rate of change is decreasing.. I'm not sure though.

Answer 3

@zepdrix , i saw the same thing but the solution to that doesn't look like the solution to the example?

@zonazoo , is that answer?
\[1000(\frac{4}{5})^{t^{2}/1000}\]

Answer 4

that is the answer to the example I gave yes... not my initial problem though.

Answer 5

well should be t^2/100 though

Answer 6

Hmm I dunno :( Yah my equation didn't get me anything near that.
Blah I gotta get to bed D:

Answer 7

well thanks for the help. i will probably do the same. try again tomorrow.

Answer 8

The equation I got was A(t) = 1000 + Ce^(-kt^2/2)

But I really don't know if that seems right or not.

Answer 9

my example that had 200 gallons in 10 minutes I now solved, so I think I can get this

Answer 10

@zepdrix

I think I have solved it

And when I put in 30 minutes for t i get 500... yay...

Thanks for the help.

Answer 11

yay \:D/