Relationship between Circumference and Area

Date: 11/08/2007 at 09:41:13
From: Channel
Subject: shortcut between area and circumference and back

Hello,

My daughter (grade 6) asked me to find a shortcut for converting from 
Area to Circumference and back.  It seems to me that if you have a 
round ordinary circle and know the area then radius is a component of 
that number.  I know just enough to be dangerous and she was curious 
(hooray) so we set about it.

Most confusing is that it seems intuitive that there should be a
relationship between the two that could be easily defined in an 
elegant equation without resorting to radius.  Also frustrating, I'm 
questioning my math; algebra, order of operations, canceling terms 
and such.

We calculated circumference and area from 1 - 20 in a spreadsheet 
to test our theories against.  I looked for patterns and found that:
Circumference increases at a rate of 6.28 per unit of radius.
Area increases at an increasing rate of which 6.28 is a component.
I found that A=1/2C*R and C=A/r*2 (not helpful because one needs the 
radius to solve.

My daughter went this way:

A = Pi * r * r 
C = 2 * Pi * r
so 
A/r = Pi * r
C/2 = Pi * r
so
A/r = C/2

Also not helpful because radius is required.

I realized that there's an extra squared in the Area formula that we 
don't use for calculation only for presentation so I thought that 
might be my problem.  The result of A = Pi r r is a number like 3.14 
if the radius is 1.  However, my answer to the area question for 
this circle has to be 3.14 cm^2 ... Still haven't figured out how 
this fits in.

Date: 11/08/2007 at 09:59:43
From: Doctor Ian
Subject: Re: shortcut between area and circumference and back

Hi Channel,

>My daughter (grade 6) asked me to find a shortcut for converting 
>from Area to Circumference and back.  

Nice!  She's thinking like a mathematician.  Why keep solving special
cases of the same problem over and over, if you can just solve the
general case one time and be done with it?

>It seems to me that if you have a round ordinary circle and know the
>area then radius is a component of that number.  I know just enough
>to be dangerous and she was curious (hooray) so we set about it.

Good for you.  In the long run, she'll probably learn more from your
willingness to jump in and play around and look for the answer, than
from actually finding the answer itself.  It's that kind of playing
around that kids are supposed to be learning in their math classes,
but it hardly ever works out that way. 

>Most confusing is that it seems intuitive that there should be a 
>relationship between the two that could be easily defined in an 
>elegant equation without resorting to radius. 

That's a good intuition. 

>We calculated circumference and area from 1 - 20 in a spreadsheet 
>to test our theories against.  I looked for patterns and found that:
>Circumference increases at a rate of 6.28 per unit of radius.

And 6.28 is about twice pi.  Interesting. 

>Area increases at an increasing rate of which 6.28 is a component.

Again, interesting. 

>I found that A=1/2C*R and C=A/r*2 (not helpful because one needs the 
>radius to solve.

It might be more helpful than you think.  But let's set that aside for
now. 

>My daughter went this way:
>
>A = Pi * r * r 
>C = 2 * Pi * r
>so 
>A/r = Pi * r
>C/2 = Pi * r
>so
>A/r = C/2
>
>Also not helpful because radius is required.

Again, it might be more helpful than you think.  Let's go back to
where your daughter started:

  A = pi * r * r 
 
  C = 2 * pi * r

But before we start pushing symbols around, let's think about what
we'd like to end up with.  We want a formula where we can input an
area and get a circumference, or input a circumference and get an
area, right?  That is, we want an equation that contains ONLY an area
and a circumference, with no mention of the radius.  

Suppose, then, that we solve the first equation FOR the radius:

            A = pi * r^2

        A/pi = r^2

  sqrt(A/pi) = r

And suppose we solve the second equation, also FOR the radius:

          C = 2 * pi * r

   C/(2*pi) = r

Now we have two expressions equal to r, neither of which involves r
itself.  Which is almost what we want. 

To get what we DO want, we just set those expressions equal to each other:

         r = r

  C/(2*pi) = sqrt(A/pi)

And this is what we were looking for: an equation that relates A and C
directly, with no mention of r.  

Of course, it's kind of ugly, with that square root in there.  So we
can square both sides, to get

    C^2     A
  ------ = --
  4*pi^2   pi

which looks a little better.  If we cross-multiply, we get

  C^2 * pi = 4 * A * pi * pi

and since multiplying by pi on both sides doesn't accomplish anything,
we can cancel that:

  C^2 = 4 * A * pi 

This makes sense, that C would have to be squared, since C is going to
have units of length, while A will have units of area, or length^2. 

Of course, we'll want to check it, with some simple cases, to make
sure we haven't made some careless error.  If a circle has a radius of
1, it has an area of 

  A = pi * 1^2 = pi

and a circumference of

  C = 2 * pi * 1

    = 2*pi

So let's substitute those into our equation:

       C^2 = 4 * A * pi 

  (2*pi)^2 = 4 * pi * pi

  4 * pi^2 = 4 * pi^2

So that works, at least for one case.  I'll leave it to you to try
some others.  Note that this doesn't PROVE anything!  We might just 
get lucky (or unlucky!) in choosing just some cases that happen to
work.  But it's a good way to get some confidence that we're right. 

All of which is to say, you were really on the right track, but just
missing the insight that the way to 'get rid' of r was to solve for it!

Does this help? 

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