This week's CR4 Challenge Question:
A goat is placed in a fenced-in circular field with radius r. It is tethered to a point on the perimeter of the field by a rope which allows it to graze as far as the radius R. What is the value of R to the nearest cm, if the goat can graze exactly ½ the circular field area, and r = 10m
Thanks to SlideRuler for this question!
And the Answer Is....
The grazing area is segment ABC + segment ABD
Therefore ½*r²*(2θ-sin(2θ))+½*R²*(2Φ-sin(2Φ))=½*π*r². ... eq(1)
also we have θ+Φ+Φ = π (isosceles Δ origin,C,A) ... eq(2)
and r*sin(θ) = R*sin(Φ) (both equal to ½*AB) ... eq(3)
Replacing 2Φ with (θ-π); and replacing R with r*sin(Φ)/sin(θ); and using the relationship sin²(Φ) = ½*(1-cos(2Φ)) = ½*(1-cos(π-θ)) = ½*(1+cos(θ)) in equation (1) gives
(π-θ-sin(θ))+(2θ-sin(2θ)-π)*(1+cos(θ)/(2*sin²(θ))=0 ... eq(4)
This has only one variable θ, and the root can be found using Newton's approximation. xn+1 = xn - yn/y'n
Using a starting value of θ=1 gives successively
θ =1 --> 1.2143 --> 1.2327 --> 1.2354 --> 1.2358 --> 1.2359
and the corresponding values of R by using eq(2) to get Φ and eq(3) to get R; (remembering r=10m) are
R= 10m --> 11.41m -->11.56m --> 11.58m --> 11.59m --> 11.59m
The answer to the given problem is therefore 11.59m
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