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Solution to Puzzle No. 8

September 1999

For the question see Puzzle No 8 - The Gobbling Goat in issue 8.

Solution diagram 1

Let the circular field have radius $r$.

Let the length of the rope, which is anchored at point $A$ on the circumference of the field, be $R$.

Now, with the rope at full stretch, the goat will be able to move in an arc from point $B$ on the circumference to point $C$.


Solution diagram 2

Let $O$ be the centre of the field.

Clearly, the angle $OAB$ is equal to the angle $OAC$. Let the magnitude of $OAB$ be $x$ radians.

Thus, the area accessible to the goat will be a circle sector with radius $R$ and angle $2x$ (yellow), plus two circle segments (pink) from a circle of radius $r$, cut off by the chords $AB$ and $AC$ respectively.


Now, the area of the circle sector is:

  \[  {1\over 2} (R^2 . 2x) = R^2x  \]    
The area of each circle segment is:
  \[  (1/2)r^2(\pi -2x) - (1/2)r^2\sin (\pi -2x)  \]    
(because each is a sector of a circle minus a triangle) and so the total area accessible by the goat is:
  \[  R^2x + r^2[\pi - 2x - \sin (2x)]  \]    
(the yellow sector plus the two pink segments).

Solution diagram 3

Currently, we have two different variables in our area equations: $r$ and $R$. Let’s try to eliminate one.

Obviously, the length of the line segment $AO$ is $r$, the radius of the field. Similarly, the radius of the line segment $OC$ must be $r$.

Therefore, by similar triangles, if we drop a perpendicular from $O$ to the line segment $AC$, the perpendicular will bisect $AC$. Therefore the length of $AP$ (and $PC$, of course) is $R/2$.

We now have a right-angled triangle and enough information to calculate the relationship between $r$ and $R$:

  $\displaystyle  \cos x  $ $\displaystyle  =  $ $\displaystyle  R/(2r) $   (1)
  $\displaystyle R  $ $\displaystyle  =  $ $\displaystyle  2r \cos x $   (2)

So the total area accessible to the goat is:

  \[  (4r^2\cos ^2 x)x + r^2[\pi - 2x - \sin (2x)] .  \]    

We wish for this area to be half the area of the total field; therefore we have:

  $\displaystyle  4r^2 x \cos ^2 x + r^2[\pi -2x-\sin (2x)]  $ $\displaystyle  =  $ $\displaystyle  \pi r^2/2  $    
  $\displaystyle 4x \cos ^2 x + \pi - 2x - \sin (2x)  $ $\displaystyle  =  $ $\displaystyle  \pi /2 $    
  $\displaystyle 4x \cos ^2 x + \pi /2 - 2x - \sin (2x)  $ $\displaystyle  =  $ $\displaystyle  0 $    

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

$x$ is approximately $0.953$, and therefore $\cos x = 0.579$.

Now, since $R = 2r \cos x$ and $r$, the radius of the field, is 100m, we have $R = 200 \cos x$ 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:

  \[  4x \cos ^2 x + \pi /2 - 2x - \sin (2x) = 0  \]    

The Newton-Raphson method is an approximate method for finding roots of equations that are differentiable.

Let $f(x)$ be a differentiable function. Since $f(x)$ is differentiable, every point on the graph of $f(x)$ must have a gradient and a unique tangent line.

Now, the tangent at $x_0$ is an approximation to the graph of $f(x)$ near the point $(x_0,f(x_0))$.

Therefore the zero of the tangent line (the point where the tangent line crosses the $x$-axis) is an approximation (perhaps a very bad one, however!) of the zero of $f(x)$ (the point where $f(x)$ crosses the $x$-axis, i.e. the root of $f(x)$). It’s like we’re pretending that $f(x)$ is really a straight line, like the tangent line, and therefore crosses the $x$-axis at the same place the tangent does.

Newton-Raphson diagram

In the Newton-Raphson method, we start with a "best guess" $x_0$ as to the zero of $f(x)$. We then calculate the first approximation, $x_1$, as the zero of the tangent line to $f(x)$ at $x_0$.

We then calculate the second approximation, $x_2$, as the zero of the tangent line crossing the $x$-axis at $x_1$, and so forth.

The diagram above shows the initial guess $x_0$, the first approximations $x_1$ and the relevant tangents. The second approximation $x_2$ is the coordinate where the second tangent crosses the $x$-axis. As you can see, the approximations are getting closer to the actual zero point of $f(x)$. If we continue iterating like this, we will get better and better estimates for the zero point of $f(x)$.

How do we do it?

We wish to solve $4x \cos ^2 x + \pi /2 - 2x - \sin (2x) = 0$. Obviously, plotting $f(x) = 4x \cos ^2 x + \pi /2 - 2x - \sin (2x)$ and drawing tangents is not going to be very much fun! However, we can perform Newton-Raphson numerically.

Our initial point is $x_0$. The gradient of $f(x)$ at $x_0$ is given by $f’(x_0)$, and the tangent line to $f(x)$ at $x_0$ is therefore given by:

  \[  y - f(x_0) = f’(x_0) (x - x_0)  \]    

To find $x_1$, we must find the point where this tangent crosses the $x$-axis, i.e. to let:

  \[  0 - f(x_0) = f’(x_0) (x_1 - x_0)  \]    

and therefore

  \[  x_1 - x_0 = \frac{-f(x_0)}{f'(x_0)}  \]    

so that

  \[  x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}  \]    

Similarly, in the general case we obtain:

  \[  x_{n+1} = x_ n - \frac{f(x_ n)}{f'(x_ n)}  \]    

Now, our function is $f(x) =4x \cos ^2 x + \pi /2 - 2x - \sin (2x)$. Via standard differential calculus, the gradient $f’(x)$ of this function is

  \[  4 \cos ^2 x - 8x \cos x \sin x - 2 - 2 \sin (2x) \cos (2x).  \]    

Therefore, to find the approximate root of $f(x)$ we can use the following:

  \[  x_{n+1} = {x_ n} - \frac{4x_ n \cos ^2 x_ n + \pi /2 - 2x_ n - \sin (2x_ n)}{4\cos ^2 x_ n - 8x_ n \cos x_ n \sin x_ n - 2 - 2 \sin (2x_ n) \cos (2x_ n)}  \]    

So, we know how to calculate $x_{n+1}$ from $x_ n$. But how do we find our starting value, $x_0$? Well, in this particular case we know that the magnitude of $x$ must be between 0 and $\pi /2$ 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) $\pi /4$.

As it turns out, all sorts of values will do: here’s a table of the iterative steps of Newton-Raphson on our function $f(x)$ for a range of initial values of $x_0$. As you can see, they all converge quite rapidly to the same twelve-significant-digit approximation.

$x_0 = \pi /4$ $x_0 = \pi /6$ $x_0 = \pi /3$ $x_0 = \pi /2$
$x_0$ 0.785398163397 0.523598775598 1.047197551200 1.570796326790
$x_1$ 0.967088277214 1.254847487960 0.956164730983 0.785398163397
$x_2$ 0.953058379193 0.966611488070 0.952884951928 0.967088277214
$x_3$ 0.952849994306 0.953049230238 0.952848237620 0.953058379193
$x_4$ 0.952847886046 0.952849901115 0.952847868401 0.952849994306
$x_5$ 0.952847864870 0.952847885110 0.952847864693 0.952847886046
$x_6$ 0.952847864657 0.952847864860 0.952847864655 0.952847864870
$x_7$ 0.952847864655 0.952847864657 0.952847864655 0.952847864657
$x_8$ 0.952847864655 0.952847864655 0.952847864655 0.952847864655
$x_9$ 0.952847864655 0.952847864655 0.952847864655 0.952847864655

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