# Problem 93 Solution

## Solution

Let p = volume of pee in the pool, in cubic inches.

Let t = time since the kid peed in the pool, in seconds.

dp/dt = (-30/7000000)×p

dp = (-30/7000000)×p dt

dp (7000000/30p)= dt

Taking the integral of each side...

(1) -(7000000/30)×ln(p)= t + c, where c is the constant of integration.

We're given that when t=0, v=8. Plug that into the equation to find c.

-(7000000/30)×ln(8)= 0 + c

c = -(7000000/30)×ln(8)

Putting that value of c in equation (1) ...

-(7000000/30)×ln(p)= t -(7000000/30)×ln(8)

t = -(7000000/30)×ln(p) + (7000000/30)×ln(8)

t = (7000000/30)×(ln(8)-ln(p))

t = (7000000/30)×ln(8/p)

The question at hand is what is t when p = 1

t = (7000000/30)×ln(8/1)

t = 485,203 seconds = 8,087 minutes = 134.8 hours = 5.6158 days.

The general formula is t=(w/r)×ln(p_{1}/p_{2}), where

t = time

w = total volume in pool.

r = rate water flows in and out of pool.

p_{1}=Initial volume of pee.

p_{2}=Volume of pee at time t.

Michael Shackleford, ASA — Nov. 4, 2013