# Math Help - grazing goat

1. ## grazing goat

This problem may be a little cliche, but I will post it anyway for those who have not seen it.

A goat is tied to a silo of radius 12 feet by a tether 36 feet long. Calculate the area the goat can graze.

Here is a hackneyed diagram. It kind of resembles a limacon.

2. If I remember correctly it is a cardioid.

I think Stuart called it
"Clara the calculus cow"

This problem made me crazy for about a week.

A cool more advanced way to solve this is to use Greene's thorem.

3. The length of the tether from where the goat is tied to the end of its rope is $36-12{\theta}$.

That's because the rope wraps around the silo before it becomes tangent to the silo at that point.

The total angle the goat grazes is 3 because $12{\theta}=36$

There is a small piece it can not reach because the semi-diamter of the silo is slightly longer than 36 feet. Therefore, we don't have to integrate to Pi, but slightly less than that.

Limits of integration are ${\theta}=[0,3]$.

The semicircular region is easy $\frac{36^{2}\pi}{2}=648{\pi}$

The region we are interested in is the second and fourth quadrants where the goat goes around the silo.

$2\int_{0}^{3}\frac{1}{2}(36-2{\theta})^{2}d{\theta}=1296$

So, the total area is $648{\pi}=1296=3331.75 \;\ ft^{2}$

I'd like to see the Green's theorem method. I will look into it.

4. I will work a variation of the problem.

Suppose that a cow is tied to a silo of radius r with a rope of length $\pi r$

See the diagram below

using trig and the fact that the rope is always at a right angle to the silo we can find $d_1,d_2,d_3,d_4$ and parametric equations of the curve.

Note the length of the hypotenuse of the small triangle is $r\theta$
becuase it have the same length as the arc of the circle subtended by the angle theta.

$d_1=r\cos(\theta)$
$d_1=r\theta\sin(\theta)$

so x as a function of theta is $x=d_1+d_2$ so

$x(\theta)=r(\cos(\theta+\theta\sin(\theta))$

using a similar argument for y we get

$
y(\theta)=r(\sin(\theta)-\theta\cos(\theta))
$

if we graph this from $-\pi \mbox{ to } \pi$ we get
Note for the graph I will use r=12

Now for the other half of the equation we want the half circle
centered at (-r,0) with a radius of $\pi r$

$x(\theta)=-r +\pi r \cos(theta)$
$y(\theta)=\pi r \sin(theta)$

graphed in green theta $\frac{\pi}{2}\mbox{ to } \frac{3\pi}{2}$

These two curves enclose the total area for the cow, but this is too much area we need to remove the silo.

lets add one more line

let $c_1$ be the blue curve
let $c_2$ be the purple curve
let $c_3$ be the green curve

Now for Green's theorem

$\iint_A1dA=-\int_{\partial A}ydx$

$-\int_{c_1 \cup c_2}ydx=-\int_{c_1}ydx-\int_{c_2}ydx$

but dx=0 on the purple curve so we get...

$-\int_{-pi}^{pi}r(\sin(\theta)-\theta\cos(\theta))r\theta\cos(theta)d\theta=\frac {1}{3}r^2\pi^3+r^2\pi$

The area of the other half circle is $\frac{1}{2}r^2\pi^3$

Then we subtact off the area of the silo to get

$A=\frac{5}{6}\pi^3r^2$

whew that was longer than I remember.