THE TETHERED GOAT { A CALCULUS
APPROACH
Michael Brennan
The problem of the goat and the eld was relayed to me by a friend
who had been asked by a quantity surveyor to solve this seemingly
simple problem of determining the rope length for a grazing goat
so as the goat could graze half of a given eld. The problem that
I was given involved a circular eld and at rst seems to involve
little more than doing some algebra on a standard area formula.
Some of my colleagues have heard a similar problem concerning grazing goats, except with regards to rectangular, square and
even triangular elds. A reference to the triangular eld dates
back to 1917 and appears in [1]. The problem with the circular eld is well known and recently appeared as a problem in the
Mathematical Spectrum , [2, p.23]. In it, Andrew Lobb used a mix
of geometry and trigonometry to solve the problem. Intuitively,
even though these problems are clear, the mathematics is more
involved than one would initially expect. Ultimately the problem
reduces down to solving a transcendental equation which is best
tackled by a Computer Algebra System like Maple or Mathematica. In our solution below, we use integral calculus to obtain an
expression for the grazing area of the goat. Then by forming the
appropriate equation, the problem is next to nd the root of this
equation.
1. Setup
Given a circular eld of grass of diameter 100m. A farmer ties a
goat to a pole on the circumference. When the rope is taut, the
goat path sweeps out an arc of a circle of radius r. The problem is
to determine the value of r so that the area of the ground accessible
to the goat is half the area of the circular eld. First we need to
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IMS Bulletin 40, 1998
72
nd an expression in r that gives the area of grass accessible to
the goat. This area is the region enclosed by two circles, one of
radius r and the other of radius 50.
The equation of the goat's circle (assuming that the rope pole
is at (0,0) and we measure the x{coordinate along the radial line
to the centre of the circular eld and the y{coordinate perpendicular to this line) is given by x2 + y2 = r2 , whereas the equation
of the xed eld will be (x 50)2 + y2 = 502. Using some elementary algebra we can nd the intersection of these curves. The
r2 .
x-coordinate is all that is needed and this is found to be at x = 100
The accessible area available to the goat can be found using integration. First we need to integrate over the region of the eld under
the curve dening the eld from the pole, i.e. from x = 0 to
r2 . Then we integrate over the region under the curve that
x = 100
r2 to x = r, which is the length
denes the goat's path from x = 100
of the rope. This area is the given by the following sum of two
integrals
Z
r2
100
p
0
502 (x 50)2 dx +
Z
r
r2
100
p
r2 x2 dx:
2. Calculation
Using the Computer Algebra System Maple V R4 [3], we can evaluate this integral sum to give an expression for the area. Note
that due to symmetry the sum of the two integrals above needs to
be multiplied by two to obtain an expression that represents the
area that is accessible to the goat.
For Maple to evaluate the denite integrals correctly, we need
rst to make an assumption on the range of the variable r (to avoid
the incorrect introduction of complex numbers).
> assume(r<100);
> A:=2*(int(sqrt(50^2-(x-50)^2,x=0..r^2/100)
+int(sqrt(r^2-x^2),x=r^2/100..r));
The Tethered Goat
73
p
r~2 1) + 1250
A := 12 r~4 + 10000r~2 + 2500 arcsin( 5000
r
r
1 r~2 ):
1
+ 12 r~3 12 r~2 arcsin( 100
r~
r~2
The tilde (~) on r indicates that r carries an assumption. Now we
2
set this equal to one half of the area of eld, a = 50
2 . Dene a
new function B to be the dierence between A and a.
> B:=A-50^2*Pi/2
p
r~2 1)
B := 12 r~4 + 10000r~2 + 2500 arcsin( 5000
r
r
1
1 r~2 ):
+ 21 r~3 12 r~2 arcsin( 100
r~
r~2
3. Answer
All that is left to do is to nd where B has a root and this value will
be the required length of rope. Intuitively we would expect this
value to be greater than 50m. To nd this value, we can use the
Maple solve command. To obtain a oating point approximation,
the numerical rootnder fsolve will approximate the root.
> solve(B=0,r);
100 sin(RootOf(4 Z sin( Z )2 arcsin( 1 + 2 sin( Z )2 )
p
+ 2 sin( Z )4 + sin( Z )2 2 sin( Z )2 ))
> fsolve(B=0,r);
57.93642365
We conclude that the length of the rope is r = 57:9364m,
correct to four decimal places. A good homework exercise would
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IMS Bulletin 40, 1998
be to solve the problem where the eld is of a more conventional
shape like a rectangle or a square. Note that for these problems
the rope pole is located at one of the corners of the eld.
4. Acknowledgement
The author would like to acknowledge the valuable assistance of Dr
Tom Carroll, resulting in the necessary introduction of the assume
command in the work above.
References
[1] H. E. Dudeney, Amusements in Mathematics. Dover, 1958.
[2] Andrew Lobb, Problem 29.7, Mathematical Spectrum 30 1997/8.
[3] K. M. Heal, M. L. Hansen, K. M. Rikard, Maple V, Learning Guide.
Springer-Verlag, 1996.
Michael Brennan
Department of Mathematics
University College, Cork.