For the question see Puzzle No 8 - The Gobbling Goat in issue 8.
Let the circular field have radius .
Let the length of the rope, which is anchored at point on the circumference of the field, be .
Now, with the rope at full stretch, the goat will be able to move in an arc from point on the circumference to point .
Let be the centre of the field.
Clearly, the angle is equal to the angle . Let the magnitude of be radians.
Thus, the area accessible to the goat will be a circle sector with radius and angle (yellow), plus two circle segments (pink) from a circle of radius , cut off by the chords and respectively.
Now, the area of the circle sector is:
Currently, we have two different variables in our area equations: and . Let’s try to eliminate one.
Obviously, the length of the line segment is , the radius of the field. Similarly, the radius of the line segment must be .
Therefore, by similar triangles, if we drop a perpendicular from to the line segment , the perpendicular will bisect . Therefore the length of (and , of course) is .
We now have a right-angled triangle and enough information to calculate the relationship between and :
(1) | |||||
(2) |
So the total area accessible to the goat is:
We wish for this area to be half the area of the total field; therefore we have:
We can't easily solve the equation but we can use a graphical calculator or numerical method such as Newton-Raphson to find an approximate solution.
Using the Newton-Raphson method as described in the Coda, we find that
is approximately , and therefore .
Now, since and , the radius of the field, is 100m, we have and thus the required length of rope is approximately 116m.
Coda: Solving the equation using Newton-Raphson
The basic idea
In the goat puzzle, we were left with the following equation to solve:
The Newton-Raphson method is an approximate method for finding roots of equations that are differentiable.
Let be a differentiable function. Since is differentiable, every point on the graph of must have a gradient and a unique tangent line.
Now, the tangent at is an approximation to the graph of near the point .
Therefore the zero of the tangent line (the point where the tangent line crosses the -axis) is an approximation (perhaps a very bad one, however!) of the zero of (the point where crosses the -axis, i.e. the root of ). It’s like we’re pretending that is really a straight line, like the tangent line, and therefore crosses the -axis at the same place the tangent does.
In the Newton-Raphson method, we start with a "best guess" as to the zero of . We then calculate the first approximation, , as the zero of the tangent line to at .
We then calculate the second approximation, , as the zero of the tangent line crossing the -axis at , and so forth.
The diagram above shows the initial guess , the first approximations and the relevant tangents. The second approximation is the coordinate where the second tangent crosses the -axis. As you can see, the approximations are getting closer to the actual zero point of . If we continue iterating like this, we will get better and better estimates for the zero point of .
How do we do it?
We wish to solve . Obviously, plotting and drawing tangents is not going to be very much fun! However, we can perform Newton-Raphson numerically.
Our initial point is . The gradient of at is given by , and the tangent line to at is therefore given by:
To find , we must find the point where this tangent crosses the -axis, i.e. to let:
and therefore
so that
Similarly, in the general case we obtain:
Now, our function is . Via standard differential calculus, the gradient of this function is
Therefore, to find the approximate root of we can use the following:
So, we know how to calculate from . But how do we find our starting value, ? Well, in this particular case we know that the magnitude of must be between 0 and radians (go back to the second diagram and think about it if you’re not sure why!). So a good initial guess might be (for example) .
As it turns out, all sorts of values will do: here’s a table of the iterative steps of Newton-Raphson on our function for a range of initial values of . As you can see, they all converge quite rapidly to the same twelve-significant-digit approximation.
0.785398163397 | 0.523598775598 | 1.047197551200 | 1.570796326790 | |
0.967088277214 | 1.254847487960 | 0.956164730983 | 0.785398163397 | |
0.953058379193 | 0.966611488070 | 0.952884951928 | 0.967088277214 | |
0.952849994306 | 0.953049230238 | 0.952848237620 | 0.953058379193 | |
0.952847886046 | 0.952849901115 | 0.952847868401 | 0.952849994306 | |
0.952847864870 | 0.952847885110 | 0.952847864693 | 0.952847886046 | |
0.952847864657 | 0.952847864860 | 0.952847864655 | 0.952847864870 | |
0.952847864655 | 0.952847864657 | 0.952847864655 | 0.952847864657 | |
0.952847864655 | 0.952847864655 | 0.952847864655 | 0.952847864655 | |
0.952847864655 | 0.952847864655 | 0.952847864655 | 0.952847864655 |
You can simplify the equation by using the cos(2x) trig identity
4 x cos^2(x) + pi/2 - 2x - sin(2x) = 0
can be simplified a little using cos(2x) = 2 cos^2(x) - 1
You get:
2 x cos(2x) + pi/2 - sin(2x) = 0
or replacing variables (y = 2x):
y cos(y) + pi/2 - sin(y) = 0