Goat grazing problem
The goat grazing problem is either of two related problems in recreational mathematics involving a tethered goat grazing a circular area: the interior grazing problem and the exterior grazing problem. The former involves grazing the interior of a circular area, and the latter, grazing an exterior of a circular area. For the exterior problem, the constraint that the rope can not enter the circular area dictates that the grazing area forms an involute. If the goat were instead tethered to a post on the edge of a circular path of pavement that did not obstruct the goat, the interior and exterior problem would be complements of a simple circular area.
The original problem was the exterior grazing problem and appeared in the 1748 edition of the English annual journal The Ladies' Diary: or, the Woman's Almanack, designated as Question attributed to Upnorensis, stated thus:
Observing a horse tied to feed in a gentlemen's park, with one end of a rope to his fore foot, and the other end to one of the circular iron rails, enclosing a pond, the circumference of which rails being 160 yards, equal to the length of the rope, what quantity of ground at most, could the horse feed?
The related problem involving area in the interior of a circle without reference to barnyard animals first appeared in 1894 in the first edition of the renown journal American Mathematical Monthly. Attributed to Charles E. Myers, it was stated as:
A circle containing one acre is cut by another whose center is on the circumference of the given circle, and the area common to both is one-half acre. Find the radius of the cutting circle.
The solutions in both cases are non-trivial but yield to straightforward application of trigonometry, analytical geometry or integral calculus. Both problems are intrinsically transcendental – they do not have closed-form analytical solutions in the Euclidean plane. The numerical answers must be obtained by an iterative approximation procedure. The goat problems do not yield any new mathematical insights; rather they are primarily exercises in how to artfully deconstruct problems in order to facilitate solution.
Three-dimensional analogues and planar boundary/area problems on other shapes, including the obvious rectangular barn and/or field, have been proposed and solved. A generalized solution for any smooth convex curve like an ellipse, and even unclosed curves, has been formulated.
Exterior grazing problem
The question about the grazable area outside a circle is considered. This concerns a situation where the animal is tethered to a silo. The complication here is that the grazing area overlaps around the silo : the goat can only eat the grass once, he can't eat it twice. The answer to the problem as proposed was given in the 1749 issue of the magazine by a Mr. Heath, and stated as 76,257.86 sq.yds. which was arrived at partly by "trial and a table of logarithms". The answer is not so accurate as the number of digits of precision would suggest. No analytical solution was provided.A useful approximation
Let tether length R = 160 yds. and silo radius r = R/ yds. The involute in the fourth quadrant is a nearly circular arc. One can imagine a circular segment with the same perimeter would enclose nearly the same area; the radius and therefore the area of that segment could be readily computed. The arc length of an involute is given by so the arc length |FG| of the involute in the fourth quadrant is. Let c be the length of an arc segment of the involute between the y-axis and a vertical line tangent to the silo at θ = 3/2; it is the arc subtended by Φ. . So. The arc length of a circular arc is and θ here is /2 radians of the fourth quadrant, so, r the radius of the circular arc is and the area of the circular segment bounded by it is. The area of the involute excludes half the area of the silo in the fourth quadrant, so its approximate area is 18146, and the grazable area including the half circle of radius R, totals. That is 249 sq.yds. greater than the correct area of 76256, an error of just 0.33%. This method of approximating may not be quite so good for angles < 3/2 of the involute.Solution by integrating with polar coordinates
Find the area between a circle and its involute over an angle of 2 to −2 excluding any overlap. In Cartesian coordinates, the equation of the involute is transcendental; doing a line integral there is hardly feasible. A more felicitous approach is to use polar coordinates . Because the "sweep" of the area under the involute is bounded by a tangent line which is not the boundary between overlapping areas, the decomposition of the problem results in four computable areas: a half circle whose radius is the tether length ; the area "swept" by the tether over an angle of 2 ; the portion of area A2 from θ = 0 to the tangent line segment ; and the wedge area qFtq. So, the desired area A is A1 + · 2. The area required to be computed are between two quadratic curves, and will necessarily be an integral or difference of integrals.The primary parameters of the problem are, the tether length defined to be 160yds, and, the radius of the silo. There is no necessary relationship between and, but here is the radius of the circle whose circumference is. If one defines the point of tethering as the origin with the circle representing the circumference of the pond below the x-axis, and on the y-axis below the circle representing the point of intersection of the tether when wound clockwise and counterclockwise, let be a point on the circle such that the tangent at intersects, and + Is the length of the tether. Let be the point of intersection of the circumference of the pond on the y-axis below the origin. Then let acute be.
The area under the involute is a function of because it is an integral over a quadratic curve. The area has a fixed boundary defined by the parameter . In this case the area is inversely proportional to, i.e. the larger, the smaller the area of the integral, and the circumference is a linear function of . So we seek an expression for the area under the involute.
First, the area A1 is a half circle of radius so
Next compute the area between the circumference of the pond and involute. Compute the area in the tapering "tail" of the involute, i.e. the overlapped area. Recall that the area of a circular sector is if the angle is in radians. Imagine an infinitely thin circular sector from to subtended by an infinitely small angle. Tangent to, there is a corresponding infinitely thin sector of the involute from to subtending the same infinitely small angle. The area of this sector is where is the radius at some angle, which is, the arc length of the circle so far "unwrapped" at angle. The area under the involute is the sum of all the infinitely many infinitely thin sectors through some angle. This sum is
The bounds of the integral represent the area under the involute in the fourth quadrant between and. The angle is measured on the circle, not on the involute, so it is less than by some angle designated.
Next, find the angle. Let. is complementary to the opposite angle of the triangle whose right angle is at point t; and also complementary to that angle in the third quadrant of the circle. is the unrolled arc, so its arclength is times the central angle. So. is also a leg of the triangle vFtv so and. Equating the 2 expressions for and solving for, the following equation is obtained:. That is a transcendental equation that can be solved approximately by iterative substitution of, polynomial expansion of, or an approximation method like Newton–Raphson. A constructive way to obtain a quick and very accurate estimate of is: draw a diagonal from point on the circumference of the pond to its intersection on the y-axis. The length of the diagonal is 120yds. because it is of the tether. So the other leg of the triangle, the hypotenuse as drawn, is yds. So radians, rounded to three places.. Small inaccuracies in when don't significantly affect the final result.
A4 is the area of the peculiar wedge. That area is the area of a right triangle with vertex t, minus the area of a sector bounded by. where x is |tF| and θ is the angle opposite to Φ in the right angle triangle. So,. If, then the area of the wedge is by reduction.
Algebraic combination of the pieces does not reduce to anything useful because of the differing geometries of the pieces, and is not a rational fraction of so it doesn't combine with the other angle. So evaluate the pieces at once then combine them numerically. Since the total area is a bit less than the area of the great circle whose radius is , keep 6 digits of precision :
Thus is the area bounded by the involute and pond in the 4th quadrant.
; the grazing area
The numerical answer is sq.yds.rounded up to the nearest square yard. It is worth noting that, which is the answer given for the case where the tether length is half the circumference of the silo, or no overlap to account for. The goat can eat all but a bit more than 5% of the area of the great circle defined by its tether length, and almost half the area it cannot eat is within the perimeter of the pond/silo.