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 (i.e., in general, the tether is longer than one half the circumference of 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/(2π) yds. The involute in the fourth quadrant is a nearly circular arc. One can imagine a circular segment with the same perimeter (arc length) 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 = 3π/2; it is the arc subtended by Φ. (while the arc is minutely longer than r, the difference is negligible). So . The arc length of a circular arc is and θ here is π/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 (1018.61) 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 < 3π/2 of the involute.
If it matters, there is a constructive way to obtain a quick and very accurate estimate of : 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.
Solution by integrating with polar coordinates
Find the area between a circle and its involute over an angle of 2π to −2π 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 (z,θ). Because the "sweep" of the area under the involute is bounded by a tangent line (see diagram and derivation below) 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 (A1); the area "swept" by the tether over an angle of 2π (A2); the portion of area A2 from θ = 0 to the tangent line segment (A3); and the wedge area qFtq (A4). So, the desired area A is A1 + (A2 − A3 + A4) · 2. The area(s) 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 (see diagram, above) 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 (opposite to ) 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 (i.e. the circumference of the silo). 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, find the angle which will be used in the limits of the integrals below. 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 . So and , so . Finally, and the following equation is obtained: . That is a transcendental equation that can only be solved by trial-and-error, polynomial expansion, or an iterative procedure like Newton–Raphson. .
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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 (note, on account of the tangent tF, that this area includes the wedge section, area A4, which will have to be added back in during the final summation). 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 . Is not given, and must be determined indirectly. Unfortunately, there is no way to simplify the latter term representing the lower bound of the eval expression because is not a rational fraction of , so it may as well be substituted and evaluated at once (factoring out preemptively): which for expository reasons can be rewritten . It seems apropo to merge a factor of into the constant term to get a common denominartor for the terms, so . is dominated by a linear term from the integration, so may be written, where is a non-zero positive but negligible quantity.
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.
The final summation A1 + (A2 − A3 + A4) · 2 is . All imprecision in the calculation is now uncertainty in and the residual . . That's useful for elucidating the relationships between the parameters. is transcendental, so the definition is a recurrence relation. The initial guess is a small fraction of . The numerical answer is rounded up to the nearest square yard.[4] It is worth noting that , which is the answer given for the case where the tether length is half the circumference (or any length such that ) of the silo, or no overlap to account for. The goat can eat all but 5% of the area of the great circle defined by its tether length, and half the area it cannot eat is within the perimeter of the pond/silo. The only imprecision in the calculation is that no closed-form representation for can be derived from the geometry presented. But small inaccuracies in when don't significantly affect the final result.