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Equations for the Paths of Four Lady Bugs


Date: 12/02/98 at 15:06:54
From: Corey Morcombe
Subject: Calculus and Polar Equation

Four ladybugs are placed at the four corners of a square with sides of 
length S. The bugs crawl counterclockwise at the same speed and each 
bug crawls directly toward the next bug at all times. They approach the 
center of the square along spiral paths. Find the polar equation of a 
bug's path assuming that the pole is at the center of the square. Also 
find the distance traveled by the bug by the time it meets the others 
in the center.

I have no idea even where to get started on this problem, and could use 
some help.  Thank you.


Date: 12/02/98 at 17:17:08
From: Doctor Anthony
Subject: Re: Calculus and Polar Equation

I shall be using the 'pedal' equation of the curve to answer this.

Pedal Equation of a Curve
--------------------------
Let O be the usual pole of co-ordinates with OX the initial line, and p 
the perpendicular ON to the tangent at P (r, theta) to the curve. The 
relation between p and r only is known as the 'pedal equation' of the 
given curve.

If phi is the angle that the tangent at P makes with OP, then (if you 
draw a diagram) it is easy to see that

   p = r.sin(phi)

Also by consideration of elementary triangle formed by neighbouring 
points on the curve, (r, theta) and (r+dr, theta+d(theta)) we get

   sin(phi) = r.d(theta)/ds     
   cos(phi) = dr/ds

   ds^2 = (r.d(theta))^2 + (dr)^2     (Pythagoas)

  [ds/d(theta)]^2 = r^2 + [dr/d(theta)]^2

      ds/d(theta) = sqrt[r^2 + (dr/d(theta))^2]

Now turning to the question of the ladybugs, we take the origin at the 
centre of the square and OX the initial line from O to one of the 
corners.

By symmetry it is easy to see that as the bugs spiral in they continue 
to form a square (decreasing in size) and the direction of the path 
keeps a constant angle of 45 degrees with the radius vector r joining 
O to one of the bugs. This means that the angle phi (as described 
above) is a constant 45 degrees. So  

   p = r.sin(45) = r/sqrt(2) 

We have also  

   r.d(theta)/ds = sin(45)  and  dr/ds = -cos(45)

From the second of these  
 
   ds = -dr/cos(45)  

      = -dr.sqrt(2)

From  r.d(theta)/ds = sin(45)

      r.d(theta)/[-dr.sqrt(2)] = sin(45)

     -r. d(theta)/dr = sqrt(2).sin(45)

     -r. d(theta)/dr = 1   and separating the variables;

           -d(theta) = dr/r

Integrating   -theta = ln(r) + constant

and then, a = 0  when  r = S.cos(45) = S/sqrt(2)  so 

          0 = ln[S/sqrt(2)] + constant

   constant = -ln[S/sqrt(2)]

The polar equation is

     -theta = ln[r/(S/sqrt(2))]

     -theta = ln[r.sqrt(2)/S]

which can be written:

   r.sqrt(2)/S = e^(-theta)

             r = S.e^(-theta)/sqrt(2)

and this is the required polar equation of the curve. 

As theta -> infinity, r -> 0 

To find the distance travelled by a bug we can use  

   dr/ds = -cos(45)

  So  ds = -dr/cos(45)     

      ds = -sqrt(2).dr  

 INT[ds] = -sqrt(2).INT(S/sqrt(2) to 0)[dr]

       s = -sqrt(2)[0 - S/sqrt(2)]

       s = S
 
So each bug walks a distance equal to the side S of the square.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus
High School Trigonometry

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