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Goat and Silo


Date: 05/31/2001 at 11:35:32
From: David Lockhart
Subject: Goat problem

Our goat friend is now roped to a point A next to a silo. The rope is 
15pi feet long. The diameter of the silo is 30 feet. How much area 
does the goat have to graze in?


Date: 05/31/2001 at 14:32:17
From: Doctor Rick
Subject: Re: Goat problem

Hi, David.

We have a similar problem in our Archives, but it doesn't do more than 
give some suggestions for thinking about it by starting with a square, 
then a hexagon, etc.

  Cow Grazing in Circles
  http://mathforum.org/dr.math/problems/cow.grazing.circles.html   

I would solve this problem with calculus. You didn't say what level of 
math you have, so I don't know if this is going over your head. I'll 
just say a little, and you can get back to me if you want more help.

I think you're saying that point A is on the wall of the silo, that 
is, it is on the circle of radius R = 15 feet.

If you draw the tangent to the circle at A, the goat can graze the 
semicircle (of radius L = 15pi) beyond this tangent with no 
interference from the silo. However, when the goat tries to go on the 
silo side of this tangent, it can't stretch quite as far because part 
of the rope is forced to curve along the wall of the silo. 

When the goat is at the end of his tether on the silo side of the 
tangent, the rope runs along the silo until some point B, then goes 
off in a straight line. The remainder of the rope lies on the tangent 
to the circle at B. The curve traced out by the end of the rope is 
called the involute of a circle. We don't have an area formula handy 
for the involute of a circle, but we can find the area using calculus.

Consider the central angle between A and B; call this angle theta. You 
can calculate the length of the arc between A and B, and hence the 
remaining length of the rope. Let theta be your variable of 
integration for the area calculation. What infinitesimal area is swept 
out by the rope as theta increases to theta + d(theta)? What is the 
range of theta?

Here are the details, in case you know calculus.

Point A is on the wall of the silo, that is, it is on a circle of 
radius R feet. If you draw the tangent to the circle at A, the goat 
can graze a semicircle of radius L beyond this tangent with no 
interference from the silo. However, when the goat tries to go on the 
silo side of this tangent, it can't stretch quite as far because part 
of the rope is forced to curve along the wall of the silo. The curve 
traced out by the taut rope is called the involute of a circle. In 
consequence, the region in which the goat can graze has 3 parts:

                   -------+-----
               ----       |     --------
            ---           |               \
           /              |                  \
          /               |                     \
         /                |                        \
        /                 |                         \
       /                  |                           \
      |                  L|                            \
      |                   |                             \
      |                   |                              |
      |                   |                               |
      |                   |                                |
      |         --+--     |                                |
       \    /           \ |                                |
        \  /             \|                                 |
         \|           R   |                                 |
          +-------+-------+A                                |
         /|        \ th   |                                 |
        /  \        \\   /|                                 |
       /    \        \\ / |                                |
      |         ----- /*  |                                |
      |              /  B |                                |
      |           //      |                               |
      |        / /        |                              |
      |     /  /          |                             /
      |   /  /            |                            /
       \/  /             L|                           /
     dth\/                |                         /
         \                |                        /
          \               |                     /
           \              |                  /
            ---           |               /
               ----       |     --------
                   -------+-----

The area of the semicircle is (1/2)pi*L^2. The two regions bounded by 
the silo circle, the involute, and the tangent have equal areas; we 
will compute the area of one of them.

When the goat is at the end of his tether on the silo side of the 
tangent, the rope runs along the silo until some point B, then goes 
off in a straight line. The remainder of the rope lies on the tangent 
to the circle at B.

Consider the central angle between A and B; call this angle theta 
radians. The length of the arc between A and B is R*theta. Hence the 
remaining length of the rope is L - R*theta. The infinitesimal area 
swept out by the rope as theta increases to theta + d(theta) is that 
of a sector of a circle of radius (L - R*theta) and central angle 
d(theta):

  dA = (1/2)(L - R*theta)^2 d(theta)

Of course, the infinitesimal region is not exactly a sector of a 
circle (because its "center" moves as theta increases), but the error 
is second order in d(theta).

What is the range of the variable theta? When the rope is wrapped all 
the way against the silo so that the remaining length is zero, we have

  L - R*theta = 0

  theta = L/R

In the puzzle, the length of the rope has been chosen so that theta = 
pi radians (180 degrees) as shown in the figure: R = 15 ft, L = 15*pi. 
Let's consider only this case, and let

  L = R*pi

  dA = (1/2)R^2(pi-theta)^2 d(theta)

Now we can integrate:
     pi
  INT  (1/2)R^2(pi-theta)^2 d(theta)
     0

Make a change of variables:

  x = pi-theta

  dx = -d(theta)

Then the integral is

                 0
  -(1/2)R^2 * INT  x^2 dx
                 pi

                       0
  = -(1/2)R^2(1/3)x^3 ]
                       pi

  = (1/6)R^2 * pi^3

The area of the semicircle is

  (1/2)pi*L^2 = (1/2)pi(pi*R)^2 = (1/2)pi^3*R^2

The total grazing area is

  (1/2)pi^3*R^2 + 2(1/6)pi^3*R^2 = (5/6)pi^3*R^2


If you're not familiar with basic integral calculus, then I can't 
suggest anything beyond the answer in the Archives. Whatever you do, 
I'd like to see your method and your results. Let me know if you have 
any difficulty. You might also browse the discussions of other grazing 
areas in the Dr. Math FAQ at:

  http://mathforum.org/dr.math/faq/faq.grazing.html   
  
- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Calculus
High School Euclidean/Plane Geometry
High School Geometry

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