Solving the Interesting Problem of the Last Bit of Water in a Bottle

[ioscope]

There is a popular myth that after drinking a bottle of water, the last bit is mostly backwash. Well I decided to try and test it, but got stumped.

Lets call y the amount of backwash in the bottle
Lets call x the number of sips taken
The volume of the bottle will be 1000mL
Assume each sip is 20mL
Assume that each sip backwashes 0.05mL into the bottle

dy/dx= 0.05 -20( y / (1000 -(20-0.05) x ) )

I can't separate variables here, so I do not know what to do. This is not a homework problem, I was just wondering if anyone could help me solve this differential equation. At 51 sips there will be nothing left in the bottle.

 

[diazona]

I tinkered with it a bit... try making the change of variables $x' = 1000 - (20 - 0.05)x$, then look at http://en.wikibooks.org/wiki/Differential_Equations/Exact_1 . I didn't take the calculation all the way through but it looks solvable that way.

 

[sir-pinski]

First of all, it's great that you're trying to test this popular myth! It's always important to question and experiment with things we hear or assume to be true. Now, let's dive into solving this interesting problem of the last bit of water in a bottle.

From the given information, we can set up the following differential equation: dy/dx = 0.05 - 20(y/(1000 - (20 - 0.05)x)). This equation represents the rate of change of backwash in the bottle with respect to the number of sips taken.

To solve this equation, we can use the method of separation of variables. First, we can multiply both sides by (1000 - (20 - 0.05)x) to get rid of the denominator on the right side. This gives us: (1000 - (20 - 0.05)x) dy/dx = 0.05(1000 - (20 - 0.05)x) - 20y.

Next, we can integrate both sides with respect to x. This gives us: ∫(1000 - (20 - 0.05)x) dy = ∫(0.05(1000 - (20 - 0.05)x) - 20y) dx.

The left side integrates to 1000y - (10 - 0.025)x^2 + C, where C is the constant of integration. The right side integrates to 50x - 0.025x^2 + C. Thus, our equation becomes: 1000y - (10 - 0.025)x^2 + C = 50x - 0.025x^2 + C.

Now, we can solve for y in terms of x: y = (50x - 0.025x^2 + C + (10 - 0.025)x^2 - C)/1000. Simplifying this, we get: y = (50x - 0.025x^2 + 10x^2)/1000. This can be further simplified to y = (50x + 9.975x^2)/1000.

Now, to find the amount of backwash left in the bottle after 51 sips, we can plug in x = 51 into our equation. This gives us: y = (50(51) + 9.975(