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Cow Grazing Half the Circle: Newton-Raphson Method

Date: 01/18/98 at 07:53:11
From: Lord Julus
Subject: Hi !

Hello !

I hail from Romania and I have this problem:

Let's assume there is a perfect circle filled with grass. In this
circle there is a cow tied with a rope to the fence. The cow can move 
inside the circle only as much the as the rope allows it to. The
problem is this: how long must the rope be so that the cow can eat
exactly half of the circle area?

Making it more geometricaly:

          . - .
         '     '  <---- something like a circle and a rope
        |   /   |       inside. What is the length of the rope
         '/    '        so it could balance inside the circle
          ` - `         on exactly half of the circle's area ?

Of course we assume it's a perfect geometrical problem. The cow and 
grass stuff are the nice way to put it ;-)

Thanks in advance,
Lord Julus.

Date: 01/18/98 at 09:04:44
From: Doctor Anthony
Subject: Re: Hi !

This problem comes up from time to time in the guise of the length of 
rope required to tether a goat on the boundary of a circular field 
such that the goat can eat exactly half the grass in the field. See 
the answer in our archives at   

-Doctor Anthony,  The Math Forum
 Check out our web site!   

Date: 01/19/98 at 09:29:19
From: Lord Julus
Subject: Re: Hi !

Thank you so much ! This was a really neat solution. I really
understood it... until the Newton-Raphson method. If it's not very 
hard maybe you can tell me where I can read about that.

Thank you,
Lord Julus

Date: 01/19/98 at 11:52:42
From: Doctor Anthony
Subject: Re: Hi !

The Newton-Raphson method is used for making successive approximations 
to a root of f(x) = 0. The equation must be put into this form before 
you start; i.e. we must have all terms on the left of the equation, 
and zero on the right.

To understand the proof draw a curve y = f(x) cutting the x axis at 
say x = 2 and sloping up to the right.  Now we assume that our first 
guess x0 is at say x = 2.5. Draw a vertical line from the x axis at 
x = 2.5 (call this point N) to meet the curve at point P. Now draw the 
tangent to the curve at P and let this tangent meet the x axis at 
point x1 (call this point M). Then x1 is a better approximation to the 
root than x0. What we require now is how to calculate x1 from x0.

Slope of the tangent f'(x0) = PN/MN  =  f(x0)/(x0-x1)

                    x0 - x1 =  f(x0)/f'(x0)

                         x1 = x0 - f(x0)/f'(x0)

This is the Newton-Raphson formula for getting a closer approximation 
to the required root. Having found x1, we then get an even better 
approximation from

             x2 = x1 - f(x1)/f'(x1)

and this process continues for as long as you like depending on the 
accuracy you require in the answer. You can tell the accuracy by how 
how much successive answers differ from each other. If there is no 
change in the 5th place of decimals between successive approximations, 
then you are safe to assume your accuracy is at least 4 places of 

Example:  Solve x^2 - sin(x) = 1

     f(x) = x^2 - sin(x) - 1 = 0

    f'(x) = 2x - cos(x)

       x1 = x0 - --------------

Suppose we start at x0 = 1.35  then:

          x1 = 1.35 - -----------   =  1.4117

          x2 = 1.4117 -   --------  =  1.409626

       x3 = 1.409626 - ------------ =  1.409624

and you could continue if you required even greater accuracy.  In 
practice you carry out the iteration in your calculator, just keeping 
the last answer in memory, and there is no need to write down the 
intermediate values as I have shown above. When successive answers 
show no change you have the answer to whatever number of places your 
calculator can work. I find that there is no further change if we put 
x = 1.409624004

-Doctor Anthony,  The Math Forum
 Check out our web site!   
Associated Topics:
High School Conic Sections/Circles
High School Geometry

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