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Compactness in Town Planning


Date: 07/22/99 at 02:15:41
From: Sandi
Subject: Geometry

Hi Doctors,

I'm working on a large project related to "compactness" of various 
areas in town planning. I've been investigating squares, rectangles, 
circles, spheres, triangles etc. using various functions for 
compactness. A problem I'm having at the moment is a small but 
challenging part relating to area of a polygon. I'm really enjoying it 
and I hope that you can help me with it.

The particular function for compactness I'm working with is:

     K = (4/pi)(A/D^2)

where

     A = area
     D = the distance between the two points in the region 
         that are the farthest apart.

First I had to show that when n is even, the compactness K for a 
regular n-sided polygonal region is given by K = n/(2pi) sin(2pi/n). 
I've done this using area = .5bcsin theta, with b and c equaling r 
(which cancels out eventually) and theta = 2pi/n.

The area is 

     A = n[(r^2/2)sin(2pi/n)]
     D = 2r

so   K = (4/pi)(n[(r^2/2)sin(2pi/n)])(1/(4r^2)

which cancels out to give

     K = (n/2pi)sin (2pi/n).

My problem is that I now need to find a formula for K in terms of n 
for a regular n-sided polygonal region when n is odd.

The area would be the same, but I'm stumped on finding D. For an 
odd-sided polygon, the two points farthest apart are not easy to find. 
I'm about to use Excel to experiment with various values, to see where 
that leads me, but was wondering if you would be able to help because 
I will still need to understand what's happening and have to solve the 
question algebraically in terms of n using D.

Thank you
Sandi


Date: 07/22/99 at 03:33:49
From: Doctor Floor
Subject: Re: Geometry

Dear Sandi,

You did a good job and I am glad you are enjoying the problem.

To calculate D when n is odd is indeed a little more difficult. In my 
calculations I will use a regular n-gon circumscribed by a circle (C) 
with center C.

Suppose we take one of the vertices, say A, of the regular polygon. 
Two "opposite" vertices, say B1 and B2, are the points that create the 
greatest distance. We know that angle B1CB2 = 2pi/n. Since an 
inscribed angle is half the associated central angle, we can conclude 
that angle B1AB2 = pi/n.

See, from the archives:

   Inscribed Angle Theorem
   http://mathforum.org/library/drmath/view/55073.html   

Now make a diameter in (C) from point A and meeting (C) at a second 
point E. We will consider triangle AB1E, which is a right triangle 
with a right angle at B1. (You can see this from the fact that 'angle' 
ACE is pi, so angle AB1E is half of it: pi/2). We find that 
angle A = pi/(2n)

In triangle AB1E we have:

     cos(pi/(2n)) = AB1 / AE = D / 2r

so   D = 2r cos(pi/(2n)).

I hope this helps you to finish your calculations. If you need more 
help, feel free to write back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   


Date: 07/24/99 at 20:08:17
From: Sandi
Subject: Re: Geometry

I've looked at ellipses and triangles since the earlier message and 
I am now looking at a housing estate bounded on three sides by 
existing roads. The southern boundary is a road running east-west 
(along the x-axis) and the other two parallel roads run north-south 
(the y-axis is placed symmetrically between the north-south roads so 
that the north-south roads are 2 units apart. The curved boundary of 
the enclosed region representing the housing estate has equation 
y = a+sqrt(1-x^2).

I've looked at a = 1, but also need to look at the effect on K of 
varying a. The problem is that I can't decide whether d runs through 
(0,a) or not, and even if it does, how to formulate an expression for 
d that doesn't look like a quadratic or make the formula for K awfully 
messy. Do you think that this case warrants me just writing a 
spreadsheet giving various numbers to a and showing what happens to K 
and leaving it at that, or is there a way to actually find a neat 
formula for d? Does D go through (0,a) every time? It seems when a = 2 
that it doesn't. But when I looked at a = 4, I couldn't decide whether 
it did or not.

Also, (with thanks to Dr. Floor) I found D for the odd-sided polygon a 
different way (achieving the same answer) without having to use point 
E. It was a little less complicated.

Thank you.


Date: 07/28/99 at 03:49:46
From: Doctor Floor
Subject: Re: Geometry

Dear Sandi,

Thanks for your answer.

The formula: 

     y = a + sqrt(1-x^2) 

describes a semicircle that passes through (-1,a) as well as (1,a) and 
has the segment between these two as a diameter.

I hope you can handle the rest yourself now.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   


Date: 07/28/99 at 02:47:54
From: Sandi
Subject: Re: Geometry

Hi Dr. Floor,

If a function K(x) = K(1/x) what does this actually mean? I have 
graphs of the two functions and I have proved it algebraically but I 
can't at the moment see any actual connection between the graphs of 
the two functions.

Thank you,
Sandi


Date: 07/28/99 at 03:54:34
From: Doctor Floor
Subject: Re: Geometry

Dear Sandi,

Thanks for this reaction. I am running out of time today. But I'll do 
my best:

If a function satisfies K(x) = K(1/x), then it means that there is a 
local maximum or minimum for x = 1. You can see this in the following 
way: If, say, K(0.9) < K(1), then also K(1/0.9) = K(1 1/9) < K(1).
Often in practical situations this will mean that x = 1 yields an 
optimum.

I hope this helps for now.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   


Date: 07/28/99 at 04:05:19
From: Sandi
Subject: Re: Geometry

Thank you, Dr. Floor, for taking the time to answer my question when 
you are very busy. I really appreciate it and it has helped me a lot. 
Thanks!

Sandi


Date: 07/28/99 at 04:08:51
From: Sandi
Subject: Re: Geometry

Dr. Floor,

Does D as earlier defined in my message pass through (0,a)? That is 
what I actually need to know.

Sandi


Date: 07/28/99 at 18:00:19
From: Doctor Floor
Subject: Re: Geometry

Dear Sandi,

It is a bit unclear to me what you mean by 'D' in your question, which 
you defined by the longest possible distance in a certain figure, 
which is a number. I suppose you want to know whether the longest 
possible segment runs through (0,a), when you have

     y = a + sqrt(1-x^2).

It indeed does, because, as I wrote this morning, y = a + sqrt(1-x^2) 
forms a semicircle with a diameter from (-1,a) to (1,a), and the 
longest possible segment there is the diameter, passing through (0,a).

I hope this clears all up. Good luck!

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   


Date: 07/28/99 at 11:41:44
From: Sandi
Subject: Re: Geometry

Hi, I'm hoping that someone can help me find the second derivative of 
a function

     y = 4x/[pi(1+x^2)]

I have found first derivative:

     y' = (4pi - 4pi x^2)/[(pi + pi x^2)^2]

and have tried unsuccessfully several times to differentiate this 
function. I need to prove that the point (1,0.6366) is a maximum of 
the first function and I am supposed to prove it with the second 
derivative. I can see it's a maximum easily enough from the graph, but 
unfortunately that's not good enough.

Thank you,
Sandi


Date: 07/29/99 at 03:35:23
From: Doctor Floor
Subject: Re: Geometry

Dear Sandi,

Thanks for your question!

I don't see why you should need a 2nd derivative for the maximum of a 
function. You can use the 1st derivative, look for the zeroes, and see 
whether there is a change of sign.

The 1st derivative is given by y' = (4pi - 4pi x^2)/[(pi + pi x^2)^2. 
The denominator is always > 0, so the zeroes of the derivative are 
found when the numerator equals 0, and changes of sign of the 
numerator are also changes of sign of the derivative.

Now, 4pi - 4pi x^2 = 0 gives x^2 = 1 and thus x = +/- 1.

We can now show the signs of the derivative is a number line:

  ____neg____0____pos____0____neg____
            x=-1        x=1

When the derivative is positive, the function is ascending, and when 
the derivative is negative, the function is descending. We find a 
maximum when x = 1.

When you really want the 2nd derivative, you should use the chain rule 
for the derivative of the denominator:

     (pi + pi x^2)^2 ---derivative--> 2*(pi + pi x^2) * 2 pi x

(Here 2 pi x is the derivative of pi + pi x^2). I suppose you will 
manage to do the rest.

I hope this helped. If you need more help, just write us back!

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   


Date: 07/29/99 at 15:52:51
From: Sandi
Subject: Re: Geometry

Dr. Floor,

I have a function K(x) = 4x/[pi(1 + x^2)] and K(1/x) = K(x). I've been 
looking as you know at a rectangular area bordered by a semicircle 
that has equation a + sqrt(1-x^2). As a approaches 0, the height of 
the rectangle becomes smaller and D (the distance between two points 
in the region the farthest apart) approaches 2. But also, the area 
approaches pi/2, which is the area of the semicircle. What I have 
found, however, is that the maximum of K(x) occurs at (1,0.6366) and 
the reciprocal of 0.6366 is rather close to pi/2. Is this a 
coincidence or is it something to do with it being a reciprocal 
function? 

Incidentally, is it a reciprocal function when K(1/x) = K(x) or is it 
called something else?

Thank you, 
Sandi


Date: 07/29/99 at 16:35:12
From: Doctor Floor
Subject: Re: Geometry

Dear Sandi,

In my last messages I missed the rectangle part, which I suppose you 
noticed. I hope and trust you have found a good approach yourself 
already.

By the way, the longest possible segment still passes through the 
point (0,a), where the rectangle is bordered by (-1,0), (1,0) (-1,a) 
and (1,a).

You can see this in the following way. Consider the point not on the 
semicircle where the longest segment starts. The circle with this 
starting point as center and the longest segment as radius must be 
tangent to the semicircle. The direction of tangency must be 
perpendicular to the radius of both (semi-)circles, so the radii are 
in fact one line. And of course this line passes through (0,a). That 
brings D = sqrt(1+a^2) + 1 (diagonal in half rectangle plus radius).

About the maximum of K(x): this is indeed equal to 2/pi, as you can 
see by inserting x = 1 without calculator or computer.

But I can not think of any connection with the semicircle/rectangle. 
K(x) seems to me to belong to a 1 by x rectangle, and the 2/pi drops 
in from the definition of K = 4/pi  A/D^2.

As far as I know, functions K(x) = K(1/x) do not have a special name. 
I hope this helps quickly enough. 

Good luck! Let me now how it went.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   


Date: 07/29/99 at 15:55:33
From: Sandi
Subject: Re: Geometry

Thank you,

Yes, I can do it the way you propose, it's just that this year as part 
of the course we learned how to tell a maximum or a minimum by using 
the 2nd derivative. It is expected that I show it in that way.

Thanks for your reply.


Date: 07/29/99 at 16:38:28
From: Doctor Floor
Subject: Re: Geometry

Dear Sandi,

Sorry to ask, but HOW do you use the 2nd derivative for finding maxima 
and minima? I don't see it. I teach my pupils to use the 2nd 
derivative to find points of inflection...

Just curious.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   


Date: 08/02/99 at 06:05:15
From: Sandi
Subject: Re: Geometry

Hi Dr. Floor,

This year we were taught how to use the 2nd derivative to prove that 
we have either a maximum or a minimum.

In general,

If f(x) has a local minimum at x = a, then f"(a) < 0
If f(x) has a local maximum at x = a, then f"(a) > 0

So, in my report I was supposed to double differentiate and then say 
that it was a maximum because f"(x) was > 0, but I was unable to carry 
out the second differentiation. Instead I just wrote that before the 
turning point the derivative is positive and after the turning point 
the derivative is negative and this means that it has to be a maximum. 
I ran out of time to do anything more than that.

Thank you for all of your help,
Sandi


Date: 08/02/99 at 17:00:37
From: Doctor Floor
Subject: Re: Geometry

Dear Sandi,

I see now how you use the 2nd derivative to see if there is a maximum 
or minimum. I think however it is needlessly difficult, and I am 
afraid you agree. I hope my help was sufficient.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   


Date: 08/02/99 at 20:30:49
From: Sandi
Subject: Re: Geometry

Thank you, Dr. Floor,

Unfortunately, I may be penalized for not using the second derivative. 
Although you and I both agree that it is the hardest way to prove a 
maximum or minimum point, it is part of the requirement of my course 
that I show that I understand the theory and can use it. It would have 
been best to have included the 2nd derivative in my report and I knew 
that I could be penalized for not including it. However, I hope my 
report as a whole will compensate - I think that I did a fantastic job 
(even if I say so myself!) - and as I wrote below, I did prove that it 
was a maximum.

For the exam, though, I am going to have to use it. I should have 
asked you this earlier, but could you please help me to find the 
second derivative for the function? This is my original question:

     y = 4x/[pi(1+x^2)]

I have found first derivative:

     y' = (4pi - 4pi x^2)/[(pi + pi x^2)^2]

and have tried unsuccessfully several times to differentiate this 
function. I need to prove that the point (1,0.6366) is a maximum of 
the first function and I am supposed to prove it with the second 
derivative. I can see it's a maximum easily enough from the graph but 
unfortunately that's not good enough.

Your help is, as always, very much appreciated Dr. Floor.

Thank you.
Sandi


Date: 08/03/99 at 02:38:02
From: Doctor Floor
Subject: Re: Geometry

Dear Sandi,

Thanks for your kind words.

Let me slightly rewrite the first derivative:

     f'(x) = (4pi - 4pi x^2)/[(pi + pi x^2)^2]
           = [4pi(1-x^2)]/[pi^2(1+x^2)^2]
           = [4(1-x^2)]/[pi(1+x^2)^2]

Let's write for the numerator

      n(x) = 4(1-x^2), then n'(x) = -8x. 

And for the denominator 

      d(x) = pi(1+x^2)^2, then, using the chain rule, 
     d'(x) = pi * 2 (1 + x^2) *2x.

Here 2x serves as the derivative of 1 + x^2. So 

     d'(x) = 4 pi x(1+x^2).

We find 

     f"(x) = [d(x)n'(x) - n(x)d'(x)]/[d(x)^2], so 

     f"(x) = [pi(1+x^2)^2*(-8x)-4(1-x^2)*4 pi (1+x^2)] / 
               [pi^2(1+x^2)^4]
           = [-8pi x (1+x^2)^2-16 pi x (1-x^2)(1+x^2)]/[pi^2(1+x^2)^4]
           = [-8pi x (1+x^2) {(1+x^2) - 2(1-x^2)}] / [pi^2(1+x^2)^4]
           = [-8pi x (1+x^2)(3x^2 -1)] / [pi^2(1+x^2)^4]
           = [-8x (1+x^2)(3x^2 - 1)] / [pi(1+x^2)^4]

We see that f"(1) < 0, so indeed x = 1 yields a local maximum.

Here, I think you made an error:

>>>If f(x) has a local minimum at x = a, then f"(a) < 0
>>>If f(x) has a local maximum at x = a, then f"(a) > 0

It should be the other way around.

I hope this helps!

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   


Date: 08/03/99 at 22:08:24
From: Sandi
Subject: Re: Geometry

Hi Dr. Floor,

I'm just checking the finding of the second derivative that you have 
set out here and was wondering why in the 3rd line of your work you 
have factorized the 2(1-x^2) to be negative, i.e., -2(1-x^2). 
Shouldn't it be positive so that when it is multiplied out it gives 
the negative 16 pi in the line above?

Thanks heaps.
Sandi


Date: 08/04/99 at 02:12:48
From: Doctor Floor
Subject: Re: Geometry

Dear Sandi,

You are right, I made a mistake in that line. Luckily, it doesn't 
change the sign of f"(1), so still x = 1 yields a local maximum.

Well done!

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   
    
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