Donkey Grazing Half a FieldDate: 08/08/97 at 09:27:13 From: Tim Steele Subject: Donkey in Field Can you solve this for me? A (zero-dimensional) donkey is attached by a (one-dimensional) rope to a point on the perimeter of a (two-dimensional) circular field. How long should the rope be (in terms of the radius of the field) so that the donkey can reach exactly half the field (eat half the grass)? Date: 08/08/97 at 12:38:24 From: Doctor Anthony Subject: Re: Donkey in Field This problem comes up from time to time in the guise of the length of rope required to tether a goat (instead of a donkey) on the boundary of a circular field such that the goat can eat exactly half the grass in the field. Here is a diagram you should refer to while I go through the working: Draw a circle with suitable radius r. Now take a point C on the circumference and with a slightly larger radius R draw an arc of a circle to cut the first circle in points A and B. Join AC and BC. Let O be the centre of the first circle of radius r. Let angle OCA = x (radians). This will also be equal to angle OCB. The area we require is made up of a sector of a circle radius R with angle 2x at the centre, C, of this circle, plus two small segments of the first circle of radius r cut off by the chords AC and BC. Area of sector of circle R is (1/2)R^2*2x = R^2*x Area of two segments = 2[(1/2)r^2(pi-2x) - (1/2)r^2sin(pi-2x)] = r^2[pi - 2x - sin(2x)] We also have R = 2rcos(x) so R^2*x = 4r^2*x*cos^2(x) We add the two elements of area and equate to (1/2)pi*r^2 4r^2*x*cos^2(x) + r^2[pi-2x-sin(2x)] = (1/2)pi*r^2 divide out r^2 4x*cos^2(x) + pi - 2x - sin(2x) = (1/2)pi 4x*cos^2(x) + (1/2)pi - 2x - sin(2x) = 0 We must solve this for x and we can then find R/r from R/r = 2cos(x) Newton-Raphson is a suitable method for solving this equation, using a starting value for x at about 0.7 radians The solution I get is x = 0.95284786466 and from this cos(x) = 0.579364236509 and so finally R/r = 2cos(x) = 1.15872847 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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