Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Coal Consumption


Date: 01/13/97 at 01:38:37
From: Juan P. Plascencia
Subject: Decreasing Coal rate problem

I have a problem that my teacher is not sure how to answer.  I have 
found an answer but I cannot put it into equation form.

The March, 1974 edition of _Scientific American_ said that the rate of
energy consumption increased 3.1 percent per year between 1947 and 
1971.  If the coal reserves would last 600 years at the 1972 
consumption rate, how long would they last if consumption continued to 
increase at the rate of 3.1 percent per year?

My answer is rather simple for a complex question, therefore I am 
unsure as to whether it's correct or not.
 
Since the consumption rate is steadily increasing, I divided 3.1 by 
600 to get the answer of 193.54 years plus 1972, so the energy would 
last upto the year 2165.  But if that is so, how can I prove it?  It 
seems too easy of an answer for such a complex problem.

Thank you for taking the time to read this.

Thanks,
Juan P. Plascencia


Date: 01/13/97 at 04:43:03
From: Doctor Keith
Subject: Re: Decreasing Coal rate problem

Hi!

Great problem.  I am glad to see that you were not put off by it and 
that you gave it a good shot.  Since I do not know your grade or math 
background, I will try to guess where explanations are needed and 
where they would bore you.  If I write something that confuses you, do 
not get discouraged, I probably guessed you had seen some of this 
before.  Just write back and ask for further explanation on the areas 
you have problems with, and I will gladly explain further.  If in some 
areas I over-explain, please pardon it - I just want to make sure the 
idea is clear.

This is basically an interest problem that you can reference in a 
number of places in your local library (for instance, most algebra 
books will have a section on interest, as will books that discuss 
economics and finances).  The rate of increase is the interest rate 
and it tells us how much additional coal will be consumed each year 
based on the consumtion of the previous year.  

An example is in order at this point, so let's say that we consumed 
100 units of coal in 1972.  How many did units did we consume in 1973 
if we had the same 3.1 percent increase as was noted?  We would have 
consumed as follows:
 
 100 units (previous amount) + 3.1 percent of 100 units (increase)

This works out to be:

 100 + (0.031 * 100) = (1+0.031) * 100 = 103.1 units of coal

So how much did we consume in 1974 with the same rate of increase?

 103.1 units (previous amount) + 3.1 percent of 103.1 units (increase)

This is:

  103.1 + (0.031 * 103.1) = 1.031 * 103.1 = 106.296 units

As you can see it is not increasing by a constant amount but rather 
the amount gets bigger and bigger as time goes on because you are 
taking 3.1 percent of a bigger and bigger number (this is the fun part 
of what we call compound interest).

Now we can get to the heart of the problem.  Lets say the rate of coal
use in 1972 is x units per year (x is everyones favorite variable in 
algebra and we will use it since we do not know what the amount was 
specifically).  We also know we have 600 years of coal if we used it 
at x units per year.  Thus we had (I will put the unit of measure in 
[] per scientific convention):

   x [units per year] * 600 [years]  = 600x [units]

So we know we had 600x units of coal to start out with.  Now we need 
to see how much coal we will use each year with the rate of increase 
and see at what year we have used it all up.  You were right - it is 
not as easy as just dividing, but we have some tricks up our sleeves.  
Let's write out our consumption as a series, and we will agree to only 
count whole years (so we won't have to worry about interest of a 
fraction of a year at the end):

 Year        Amount of Coal
  1  -----          x
  2  -----     1.031x
  3  ----- (1.031^2)x  (the ^ means exponent so this is 1.031*1.031)
  4  ----- (1.031^3)x
  etc.

Do you see the pattern? When we have a pattern we can use it to 
simplify the equation we write. In this case we note (1.031^(n-1))x 
units are used in the year numbered n (we are using n as any whole 
number, again trying to generalize so we can write an easier 
equation).  Now we can write the total use of coal in the first n 
years as:

   x + 1.031x + (1.031^2)x + ... + (1.031^(n-1))x 

where "..." means a bunch of other terms following the same pattern.  
How does this help us, other than looking really impressive to others?  
Well, this is a standard series, which means that people have studied 
its behavior and have given us a simple formula for calculating its 
sum.  In fact this series (it is called the geometric series if you 
are interested) has a sum that can be expressed as:

     (1.031^n - 1)x/(1.031 -1)

Now that is much simpler to deal with! Many math books will have a
series like this one and what it sums to. Try some algebra books for 
starters or if you want more, let me know. All we have to do is find 
out for what value of n this simplified answer is equal to 600x.  
So we write:

     (1.031^n - 1)x/(1.031 -1) = 600x

Cancel the x's and multiply by (1.031 - 1)= 0.031:

     (1.031^n - 1) = 600(0.031)

Add one to each side:

     1.031^n = 600(0.031) +1

At this point we encounter another problem.  How do we get that n out 
of the exponent?  For this we will use another tool from our box of 
tricks, namely the logarithm.  Logarithms are the inverse of 
exponents, and thus allow us to get the n out of the exponent we have.  
We thus take the log (short for logarithm) of both sides of the 
equation to get:

     log(1.031^n) = log(600(0.031) + 1)

And using a property of log that states that log(x^y) = y*log(x):

     n*log(1.031) = log(600(0.031) + 1)

Then divide by log(1.031) to get:

     n = log(600(0.031) + 1)/log(1.031)

Now, luckily most calculators have log built in (and most computers 
have a calculator program that has this) so we can readily calculate 
the right side and get our answer.  Remember we agreed to keep n to 
whole numbers (if you don't you have to stop one year back anyway and 
figure out how to spread the increase and we don't want to deal with 
that at the moment ).  I calculate that this would give us only 97 
years of coal!  In the last year we would be using almost 19 times 
what we used in 1972!  Talk about an increase!  Thus you will want to
look out for the year 2068, if the trend continues.  Yet another 
reason for looking for alternative sources of energy.  

Note that no step that I took was a big step - the key to solving 
tough problems is to just write down what you know and look for 
patterns or similarities that you can exploit. Keep up the good work 
and let me know if I can help further. Hope the explanation helps.  

-Doctor Keith,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Exponents
High School Interest
High School Logs

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/