Grazing Animals

Classic Problems || Dr. Math FAQ || About Dr. Math || Search Dr. Math || Dr. Math Home

A cow/goat/donkey is tethered to a building/fence/post in a field...
Answers to a number of similar questions can be found in the Dr. Math archives (see below for a list). We'll show you a few of them and we encourage you to explore the others.

1. A donkey is attached by a rope to a point on the perimeter of a circular field. How long should the rope be in terms of the radius of the field so that the donkey can reach exactly half the field and eat half the grass?

From: Doctor Anthony
Subject: Donkey in Field

This problem comes up from time to time in the guise of the length of the rope required to tether a goat (instead of a donkey) on the boundary of a circular field such that the goat can eat exactly half the grass in the field.

Here is a diagram to refer to while we go through the working:

Draw a circle with suitable radius r.

Now take a point C on the circumference and with a slightly larger radius R draw an arc of a circle to cut the first circle in points A and B. Join AC and BC.

Let O be the centre of the first circle of radius r. Let angle OCA = x (radians). This will also be equal to angle OCB.

The area we require is made up of a sector of a circle radius R with angle 2x at the centre, C, of this circle, plus two small segments of the first circle of radius r cut off by the chords AC and BC.

The area of the sector of circle R is (1/2)R^2*2x = R^2*x

The area of the two segments

= 2[(1/2)r^2(pi-2x) - (1/2)r^2sin(pi-2x)]
= r^2[pi - 2x - sin(2x)]

We also have R = 2rcos(x)   so R^2*x = 4r^2*x*cos^2(x)

We add the two elements of area and equate to (1/2)pi*r^2

4r^2*x*cos^2(x) + r^2[pi-2x-sin(2x)] = (1/2)pi*r^2   divide out r^2

4x*cos^2(x) + pi - 2x - sin(2x) = (1/2)pi

4x*cos^2(x) + (1/2)pi - 2x - sin(2x) = 0

We must solve this for x and we can then find R/r from R/r = 2cos(x)

Newton-Raphson is a suitable method for solving this equation, using a starting value for x at about 0.7 radians.

The solution I get is x = 0.95284786466 and from this cos(x) = 0.579364236509

and so finally R/r = 2cos(x) = 1.15872847

- Doctor Anthony, The Math Forum

2. A cow is tethered in a field using a 50-ft. rope tied to one corner of the outside of a 20 ft. by 10 ft. barn. What is the total area that the cow is capable of grazing?

From: Dr. Ken
Subject: Grazing Cow

I have made a sketch of this problem using The Geometer's Sketchpad:

Basically, the shape of the grazing area is circular with a little chunk out of it. In my drawing, the barn looks like this:

```                                             COW
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
____________/
|            |
|            |
|            |
|            |
|            |
|            |
|            |
|            |
|            |
|            |
|            |
|____________|

```
With this setup, the chunk that's missing from the grazing circle is in the lower lefthand corner. The right half of the grazing area, then, is just the area of the semicircle, which is 1250. To find the area of the remaining grazing area (including the area of the barn, which you subtract later), you use integration.

Do you see how the grazing area in the lower lefthand corner is the intersection of two circles? You need to find that intersection point, integrate to find the area of the grazing area to the left of the intersection point, and then do the same to the area to the right of the intersection point.

- Doctor Ken, The Math Forum

3. We can generalize this problem slightly by allowing the cow to be tethered at any point along the longer wall of the barn.

From: Dr. Peterson
Subject: Grazing Cow

Let's say that the cow is x feet from one end of the longer wall of the barn and y = 20-x feet from the other:

The heavy curve in the figure shows the maximum extent of the cow's range, which consists of several arcs of circles. First there is an arc of radius 50 centered at the tether point P; then when the rope wraps around either corner of the barn, it follows an arc of 50-x or 50-y feet respectively, centered at that corner (A or D); when it wraps around another corner (B or C) the radius becomes 50-x-10 or 50-y-10 respectively.

We don't need to worry about all these arcs, because the last few overlap and do not add any area to the region that can be grazed. (You'll have to be careful to determine which arcs intersect for a given location of P.) The total area consists of:

• a semicircle (red) with radius 50;
• a quarter-circle (magenta) with radius 50-x;
• two sectors(yellow) with radii 50-y and 40-x that meet at point Q where two circles intersect; and
• a quadrilateral BCDQ (green) determined by the intersection point Q and two walls of the barn.

The hard part is to find where the circles intersect, so we can get the angles of the yellow sectors. The area of the quadrilateral is easy, if we think of it as the difference between the triangle BQD determined by the intersection point and the diagonal of the barn, and the right triangle BCD forming half of the barn itself. You can use Heron's formula to find the area of BQD.

So how do we find where the circles intersect? Applying the Law of Cosines to triangle BQD, we can find angles BDQ and DBQ; and since we know angles ADB and ABD, we can determine the angles we need for the two sectors. There's some work left to do, but it's all doable, and it will be easier if x is a constant rather than a variable.

Finally, if the rope is short enough (less than half the perimeter of the barn), the problem simplifies immensely. With no overlap to consider, the area is merely the sum of half- and quarter-circles:

- Doctor Peterson, The Math Forum