Solution to Puzzle No. 8
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 :
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.
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:
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.