Grazing Half of a Square FieldDate: 08/30/2001 at 12:36:07 From: Philip Roche Subject: Grazing Animal My puzzle consists of a cow tied to the middle of one side of a SQUARE field. Apart from that all the details shown in the Dr. Math Grazing Animals FAQ are the same. http://mathforum.org/dr.math/faq/faq.grazing.html Can you tell me the length the rope should be to enable the cow to eat half the grass? I tried solving the problem using geometry and integration but could never work it out. I did get to a point where I had an equation containing x and sin x but had no method of solving such an equation. Now I have forgotten most of what I once knew but I would be very interested in seeing a solution using whatever methods are required. Thanks for your time. Date: 08/30/2001 at 14:10:54 From: Doctor Rob Subject: Re: Grazing Animal Thanks for writing to Ask Dr. Math, Philip. Let s be the length of the side of the square field. Let the length of the rope be r. Set up a coordinate system with origin at the point of tethering, and with the x-axis along one side of the square, so that the corners of the square are at (0,s/2), (-0,-s/2), (s,s/2), and (s,-s/2). We are interested in finding the ratio t = r/s, which is independent of r or s. Half the area of the square is s^2/2, and the area the cow can graze is the area within the square but below the circle y = sqrt(r^2-x^2). Thus you get the equation s/2 s^2/2 = INTEGRAL sqrt(r^2-x^2) dx, -s/2 s/2 s^2/2 = [x*sqrt(r^2-x^2)/2+r^2*arcsin(x/r)/2] x=-s/2 If you put r = s*t, you end up with the following equation to solve: 4*t^2*arcsin(1/[2*t]) + sqrt(4*t^2-1) - 2 = 0, 1 = 2*t*sin([2-sqrt(4*t^2-1)]/[4*t^2]). This can only be solved numerically, not algebraically. The answer is r/s = t = 0.5828221624459..., r = s*0.5828221624459..., approximately. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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