# Thread: Finding final amount after X seconds

1. ## Finding final amount after X seconds

Hi
Ill try to describe my problem in a simple way:

I have a stopwatch set to X seconds.

Every second this happens:
- Y amount of water is automaticaly being put intoo a transparent ContainerA.
- I then take a bucket and fill it with the exact amount of water in ContainerA, and pour the entire bucket intoo ContainerB.

...
After X secounds, how much water is in ContainerB?
( Y can be Integer or Decimal )

2. ## Re: Finding final amount after X seconds

Originally Posted by CakeSpear
Hi
Ill try to describe my problem in a simple way:

I have a stopwatch set to X seconds.

Every second this happens:
- Y amount of water is automaticaly being put intoo a transparent ContainerA.
- I then take a bucket and fill it with the exact amount of water in ContainerA, and pour the entire bucket intoo ContainerB.

...
After X secounds, how much water is in ContainerB?
( Y can be Integer or Decimal )
Note that after $\displaystyle x$ seconds, $\displaystyle B$'s volume is given by $\displaystyle Y$ times the sum of all positive integers from 1 to $\displaystyle x.$

$\displaystyle \begin{array}{c|c|c}\mathrm{Time}&\mathrm{Containe r\ }A&\mathrm{Container\ }B\\\hline0&0&0\\1&Y&Y\\2&2Y&3Y\\3&3Y&6Y\\4&4Y&10Y \\\vdots& \vdots&\vdots\\x&xY&x(x+1)Y/2\end{array}$

Edit: I'm assuming that the containers start out empty. But if not, it would not be difficult to adjust the procedure above.

3. ## Re: Finding final amount after X seconds

Thank you Reconer!
Thats seems to be working correctly.

...
However if i add yet another Container C like this:

I have a stopwatch set to X seconds.

Every second this happens:
- Y amount of water is automaticaly being put intoo a transparent ContainerA.
- I then take a bucket and fill it with the exact amount of water in ContainerA, and pour the entire bucket intoo ContainerB.
- I then take a bucket and fill it with the exact amount of water in ContainerB, and pour the entire bucket intoo ContainerC.

...
After X secounds, how much water is in ContainerC?
( Y can be Integer or Decimal )

4. ## Re: Finding final amount after X seconds

Originally Posted by CakeSpear
Every second this happens:
- Y amount of water is automaticaly being put intoo a transparent ContainerA.
- I then take a bucket and fill it with the exact amount of water in ContainerA, and pour the entire bucket intoo ContainerB.
- I then take a bucket and fill it with the exact amount of water in ContainerB, and pour the entire bucket intoo ContainerC.
...
After X secounds, how much water is in ContainerC?
( Y can be Integer or Decimal )
$\displaystyle \begin{array}{c|c|c|c}\mathrm{Time}&\mathrm{Contai ner\ }A&\mathrm{Container\ }B&\mathrm{Container\ }C\\\hline0&0&0&0\\1&Y&Y&Y\\2&2Y&3Y&4Y\\3&3Y&6Y&10 Y\\4&4Y&10Y&20Y\\\vdots& \vdots&\vdots&\vdots\\x&xY&\frac{x(x+1)}2Y&\frac{x (x+1)(x+2)}6Y\end{array}$

And in general, if you extend the pattern to $\displaystyle n$ buckets, the volume of water in the $\displaystyle n^{\mathrm{th}}$ bucket is given by

$\displaystyle V = \frac{x(x+1)(x+2)(x+3)\cdots(x+n-1)}{n!}Y.$

5. ## Re: Finding final amount after X seconds

Thank you very much Reconer, solves all my problems!
This is very valuable to me

6. ## Re: Finding final amount after X seconds

Originally Posted by CakeSpear
Thank you very much Reconer, solves all my problems!
This is very valuable to me
Sure. If you're interested, the coefficients in each column of the table are called figurate numbers - the third column is the sequence of triangular numbers, the fourth column is the sequence of tetrahedral numbers, and the numbers produced by the generalized formula I gave are called $\displaystyle n$-simplex numbers (a simplex being a generalization of triangles and tetrahedrons to $\displaystyle n$ dimensions).