Conference rusure::math

    This seems easy enough and I have an answer.  But, by all proof
    do you mean prove that Area = Pi * radius� also?   I used it as
    an assuption.
    
    Martin H.

    Re .1
    Martin H.
    Assuming     Area = Pi * R^2     is o.k.
    Ed.
    p.s. Lets have it.

    Ok I'll admit it I got a bit cocky.  I thought I had an answer
    but I now realize I made one of those mistakes which made some
    problems go away.  Here is what I have so far though.      
          
    Find Rg the radius of a goat's teather when it is at the edge
    a circular field of radius 20 (Rf) and you only wish to allow
    the goat to use half of the area of the field (Af).        
          
    
    1) Af = Pi * Rf�                 ! The area of the field   
    
    2) Ag = � Af                     ! Area of the goat's teather
    
                 ________                                                  
    3) F1(x) = \/Rg� - x�            ! The arc of the goats teather
                      ________
    4) F2(x) = 20 - \/Rf� - x�       ! The lower half of the field
    
    5) Ag = Integral (F1(x) - F2(x)) ! From the intersection points.
    
    6) Limits are when F1(x) = F2(x)                           
          
         ________          ________                            
       \/Rg� - x� = Rf - \/Rf� - x�  ! square and solve for x  
          
       x = �(Rg� - 2Rf� + 2Rf�)/2Rf  ! The Limits of integreation
                              ________          ________
    7) Pi * Rf� = Integral [\/Rg� - x� - 20 + \/Rf� - x�]
    
          
    I'm currently working in the integral and Using the limits
    to find Rg.  Is my basic approach right?                
          
    Martin H.

    Martin,
    Basic approach seems right, I'll check it at the house tonight.
    I think that Integrating will cause mega problems.
    Ed.

    I knew I recognized that equation.  It's one of the standard forms
    which requires trig substitution to integrate.
               _______          _______
    integral \/a� - x� dx = �x\/a� - x� + �a�arcsin(x/a) + C
    
    In order to atone for my looking up the integral I set up the
    problem in polar form.  It's much simpler.
    
    r = Rg  0 < theta < Pi
    
    r = Rf*sin(theta) -Pi/4 < theta < Pi/4
    
    I think this will work.
    
    Martin H.

    One could have computed the area directly, not using integration,
    but the end result is effectively the same. So, we got an equation
    with arcsin, and possibly arccos or arctan, and we want to find
    its roots. Personally I have no idea how to solve that equation,
    except by numerical methods. But I don't think that is what the
    person who posted it wants.
    
    A friend of mine said he had read something similar in the Mathematics
    Magazine or something like that a few years back that deal with
    the ratio of (Rg/Rf) in a n-dimensional setting as n -> infinity,
    and it had something to do with pi, but he couldn't remember. Is
    that where the problem originated from?
    
    					David Ng
                                                   

    RE .6
    I had heard of this problem some time ago and remember putting in
    very much time.  All I came up with was an approximation, never an
    exact answer.  Through integration I got an answer, but with arc's
    in it, so not a definite number.  Through geometry I got an
    approximation plus/minus 0.5*10^-8.  Never an exact answer so that's
    why I asked the Digital Math world if anybody could do better than
    I.                       Regards, The Originator.

    I found this answer in an OU text book;
    
    The solution says forget the length of the teather work with the
    following angle:-
    
    Let O be the centre of the circular field
    Let R = radius of this field (20 in this case)
    Let A be the point on the edge of the field that the teather is
    tied
    Let a = length of the teather
    
    Let angle x be the angle of the arc drawn by the teather fully extended
    as it moves from distance R from the origin O to the other point
    on the other side of the field at distance R from O.
    
    
    Then (skipping a hell of a lot - its very difficult not being able
    to draw a circle):-
                   2
    Area grazed = R (xcosx+pi-sinx)
    
             x
    a = 2Rcos-
             2
                          1   2
    put the area grazed = -piR  = half the field
                          2
    
    Does this help.
    
    Tony.