# Math Help - Cow tied outside a circular silo

1. ## Cow tied outside a circular silo

Problem: A cow is tied to the outside of a circular silo of diameter 30 ft with a rope that has length L = 10*Pi ft.Find the area of the region that is within the cow's reach for grazing.

How would I go about solving this one? I'm guessing I have to do something with polar coordinates and the area formula, but I couldn't figure out how to get an equation for the cow's grazing range. Thanks for the help!

2. A clasic problem - one which I've always known as "the goat problem" (not cow), and it was a goat on the outside of a circular corral, not silo, but who cares.

You break up the grazing area into 2 parts - call part (A) the area that is covered when the rope is wound part way around the outside of the silo (there are two area A's), and part (b) for the semi-circle of area where the cow's leash is not constrained by the silo. Part B is easy - the area of the semicircle is simply $1/2 \pi L^2$. Part (A) is the hard part - and yes, use polar coordinates. Set $\theta=0$ at the point of attachment of rope to silo. When the cow's rope is stretched around the silo to angle $\theta$, the amount of rope that is wound against the silo is $R\theta$. For each $\Delta \theta$ that the rope is wound around the silo, the area that the cow has to graze is:
$
dA = \frac 1 2 (L-R \theta) d\theta
$

So the total area to graze is:

$
A = \int_0 ^{\alpha} \frac 1 2 (L-R\theta) d\theta
$

where $\alpha$ is the max angle that the rope can strech around the silo:
$
\alpha = \frac L R
$

Can you take it from here?

3. Think I got it, thanks so much for the help =)

4. Just checking, the (L-Rθ) inside the integral is the r(θ), correct? So I would actually have to square it to get the correct value because the polar area formula says $

\frac{1}{2}\int_{\theta 1}^{\theta 2}{r}^{2}(\theta )d\theta
$

5. Crysolice - You are correct - I realize now that I made an error in the integral. The incremental area $dA$ is found by approximating the area swept by $d\theta$ as a triangle of length $L-r\theta$ and base $(l-r\theta)d\theta$. So the incremental area is:
$
dA = \frac 1 2 (L-R \theta)^2 d\theta
$

And the total area to graze should be:

$
A = \int_0 ^{\alpha} \frac 1 2 (L-R\theta)^2 d\theta
$

Sorry for the confusion.

6. ## Re: Cow tied outside a circular silo

Hi there,
This problem is treated at least six times on the internet, but each time the length of the rope is equal to radius x pi, which makes it easy to solve. Your solutions for the involute portion of the grazing area are correct as I understand them. But what about a rope that is longer than radius x pi. I propose a radius of 10 and a rope of 50. Wit the standard formul there is an overlap. So the question is: When will the tip of the rope reach the center line? I have a way of doing this which I will post later, but I cannot seem to do it purely with math. Has anyone found a way to do this problem with a longer rope? I have been working with a radius of 10 and a rope of 50.

7. ## Re: Cow tied outside a circular silo

Has anyone found a way to do this problem with a longer rope?

well,
given the nature of the problem,
I'd say a partial solution is "zero",
as it may happen that the cow is not hungry... :P