Formulas and Missing Quantities

Date: 10/09/2004 at 04:45:34
From: Pilar
Subject: Drinking a swimming pool

Could you drink enough water in a life time to empty a pool?  My math 
teacher told us to solve this problem and I understand it but I really 
don't get how we have to solve it!

I thought of using a pool that is 4m by 8m, and depth 2m.  But I don't 
know how to find out how much water it would contain. 

Also I figured that our lifetime on average is about 80 years.  But 
I'm not sure what to do next. 

Pilar

Date: 10/09/2004 at 08:57:18
From: Doctor Ian
Subject: Re: Drinking a swimming pool

Hi Pilar,

If you assume that the pool is shaped like a box, you can compute the
volume by multiplying the length, width, and depth.  What do you get
when you do that? 

How many days are in 80 years?  How much water can you drink in a 
single day?  If you multiply those together, you get the total amount
of water you would drink in a lifetime. 

Is this enough to get started?  Write back if you need more help.

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 10/09/2004 at 11:03:16
From: Pilar
Subject: Drinking a swimming pool

Hi, it's me again. 

I thought about what you told me and the volume I came up with was 
120,000 liters. 

Next, I found out that 80 x 365 = 29,200. So in 80 years there are
29,200 days. And if in one day we drink on average 2 liters, in a
lifetime we drink 58,400 literes. 

But now I would like to know what is the method to find out how many
years I would have to live to drink all the water in my swimming pool
if I drink on average 2 liters per day.

Thank you so much for your help.
Pilar

Date: 10/09/2004 at 12:09:13
From: Doctor Ian
Subject: Re: Drinking a swimming pool

Hi Pilar,

Thanks for writing back. 

I get a slightly different volume for the pool:

  4 m * 8 m * 2 m = 64 m^3

and there are 1000 liters in a cubic meter.  So I get 64,000 liters. 
But it's in the same ballpark, which is okay, since the question isn't
asking about a specific swimming pool. 

Similarly, the number of days you computed doesn't take into account
leap days, but for purposes of this question, it's okay to ignore them. 

As for figuring out how long you'd have to live to drink 120,000 (or
any other quantity) of water...  In fact, there are a number of
quantities here:

  the length, width, and depth of the pool, in meters;

  the length of a lifetime, in years;

  the amount of water drunk in a day, in liters;

and we'd like to relate them all in one equation:

   volume of pool = volume drunk in a lifetime

The volume of the pool is

  length * width * depth = volume drunk in a lifetime

but that's in cubic meters.  To convert it to liters, we have to
multiply by 1000:

  L * W * D * 1000 = volume drunk in a lifetime

(I've abbreviated length as L, width as W, and depth as D to save
space and typing.) 

The volume drunk in a lifetime is

  L * W * D * 1000 liters = years * days per year * liters per day

We know how many days are in a year:

  L * W * D * 1000 liters = years * 365 * liters per day

So here's the thing:  If we know any 4 of the named quantities, we can
get the 5th.  That means we can ask a number of different questions. 
For example:

  1) If you live 80 years, and drink 3 liters of water a day,
     how deep a 5m by 8m pool could you drink?

        8 * 5 * ? * 1000 = 80 * 365 * 3

  2) How long would you have to live, drinking 1 liter per day,
     to drink a pool that is 4m by 7m by 2m?

        7 * 4 * 2 * 1000 = ? * 365 * 1

  3) If you live to be 75, how many liters would you have to 
     drink each day to drink a pool that is 8m by 3m by 1.5m?

        8 * 3 * 1.5 * 1000 = 75 * 365 * ?

In fact, for any quantity that you can choose from the equation, you
can make up a question where you supply all the other quantities. 

Does this make sense?  You'll see this kind of thing often for various
geometric shapes.  For example, the volume of a cylinder is 

  volume = pi * radius^2 * height

So if you know the radius and height, you can find the volume; if you
know the volume and radius, you can find the height; and if you know
the volume and height, you can find the radius.  It's all the same
basic idea:  If you have an equation that relates N quantities, and
you know N-1 of those quantities, you can find the remaining one. 

Does this help? 
     
- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 10/10/2004 at 06:19:47
From: Pilar
Subject: Thank you (Drinking a swimming pool)

Thank you so much. You were a really good help and if I need any more
help I will surely get back to you. But you explained it very well for
me to understand it. Thanks so much...

Pilar