Or use some pencil on a string. :showoff:

There's a very similar problem on one of my calculus textbooks. (Multivariable Calculus/5E by Stewart, ch. 10.2, problem 74) I also have the solutions manual, but I don't have it with me right now. If I find the manual, I will post the solution.

All right, here's the problem from my textbook:

[quote]74. A cow is tied to a silo with radius [b]r[/b] by a rope just long enough to reach the opposite side of the silo. Find the area available for grazing by the cow.[/quote]

Here's the solution (in .png):

help!-tethered cow problem
 

:smile:
Its surprising how this problem has crept in disguise, and not been recognized as Pascals Limacon as the general case and the Cardioid as the particular depending on the radii being equal or different as in the problem.

For the cardioid the Area is A = 3/2 pi r^2 which is obtained from a simple integration of 1/2 r^2 d(theta) where r = a (1 + cos (theta) ) theta ranging from 0 to 2 pi. I could help you in the integration but make sure what a is.
In some derivations its the radius and in the cardioid the diameter.
Subtract the area of the silo and you have your problem solved.

URL/ [url]http://mathworld.wolfram.com/Cardioid.html[/url]
Also see the same for Limacon.
Mally :coffee:

[QUOTE=mfgoode]:smile:
Its surprising how this problem has crept in disguise, and not been recognized as Pascals Limacon as the general case and the Cardioid as the particular depending on the radii being equal or different as in the problem.[/QUOTE]
That's because it's not, although it looks very close.

Drew

help!-tethered cow problem
 

[QUOTE=drew]That's because it's not, although it looks very close.

Drew[/QUOTE]
As before, have it your way Drew but TRUTH WILL TRIUMPH! :flex:
Ask yourself by a downright denial how have you contributed positively to the thread ?. Is your answer helpful in solving this problem?
How about a concrete solution? :question: :question:
Mally :coffee:

[url]http://www.mersenneforum.org/attachment.php?attachmentid=922[/url] <--- I've seen that also in Mandelbrot fractals with a constant number of cycles of the inner loop instead of a maximium radius. The problem is that there's a different length of the rope in these different versions of the same problem - areas covered/touched by the rope. What's interesting is to do a thought experiment where the rope stays the same length and the silo's size changes. :)

For the picture you provide there is a length f(Pi/2)=0 where f(x) is the length of rope remaining after touching the furtherest north point on the silo. Note that if you have the rope longer then this, then you can _ignore_ any further length upto a max, then add a circle with a center at the north tip of the silo. As the silo's radius divided by the the length of the rope approaches infinite, the rope's path becomes closer to a perfect circle.

I'll draw a picture and post it soon because this is a fun bit of math. :)

You'll notice that I made the angle of the lines parallel (actually on top of) a line from the center of the silo's area. At the top (north end) of the silo the leftover length is 0 because it's the end of the rope (duh!). At the bottem of the circle (south end of the silo) is where the rope can go the furthest. There's a reason I kept the angles equal.

Once you know the lengths and angles then you can plug them into a formula using a limit. I'll let you try this before writing the solution. Hint: This is a triangles problem as well as a circle problem.

[QUOTE=mfgoode]As before, have it your way Drew but TRUTH WILL TRIUMPH! :flex:
Ask yourself by a downright denial how have you contributed positively to the thread ?. Is your answer helpful in solving this problem?
How about a concrete solution? :question: :question:
Mally :coffee:[/QUOTE]
Interesting that I didn't see an original argument to support your claim but you demand one from me. At least I had the sense to prove it to myself before challenging youj. I'd expect the same from you before receiving such a criticism.

However, for the sake of argument, I'll humor you. This one's easy:

I'll refer you to [url=http://mathworld.wolfram.com/Cardioid.html]this[/url] page for the governing equations of a cardioid. Let's pay particular attention to equations 4 and 5 and assume the amplitude 'a' is 1 for the sake of simplicity.

We know that with the silo problem, the tether should form a simicircle to the left of the attachment point as it is drawn on the MathWorld page. The x value of the tether attachment point can be assumed to be the point at which y has a maximum and minimum as t varies in equation 4. Differentiating equation 4, I get the following:

dy/dt = sin[sup]2[/sup](t) + [1-cos(t)]cos(t)

For values of t=2*pi/3 and -2*pi/3, it can be shown that the above derivative is equal to 0, which represent a local maximum and minimum of the equation. Plugging in these values of t in the original equations yields a y value of 3*sqrt(3)/4 and -3*sqrt(3)/4, and an x value of -1/4.

So, we should expect all values of our cartioid with x < -1/4 to coincide with a circle of radius 3*sqrt(3)/4 centered at the point (-1/4, 0).

Therefore, the leftmost point on the cartioid should be the point (-1/4-sqrt(3)/4, 0). Plugging a value of t=pi into equation 4 of the above referenced web site yields the point (-2, 0).

Therefore, the left part of our cartioid is not a semicircle, and therefore can not be considered the solution to the tethered cow problem.

I've proven a counter to the specific case of a cartioid, and I'll leave it as an exercise to the reader to prove the same for the limacon. The ball's in your court.

By the way, I'm still waiting for a response to the last point I made in the coin problem thread. Or are you the type that avoids admitting so when he realizes he's wrong?

Drew

[QUOTE=drew]Interesting that I didn't see an original argument to support ....
snip~
By the way, I'm still waiting for a response to the last point I made in the coin problem thread. Or are you the type that avoids admitting so when he realizes he's wrong?

Drew[/QUOTE]
:smile:
Compliments of the season and peace and good will to you Drew.

In the Coin problem with all your meandering circumlocution I think you have overlooked this final line of mine regards my stance in the solution of it.
This was made in post number 28. To save you the trouble I c+ p it here

"In short please don't waste my time and I will not reply, as to me, the problem is hooked cooked booked. and closed. "
Mally

Regards the 'cow' problem I'm still pondering over it.
You may be right after all.
Mally :coffee:

[QUOTE=mfgoode]"In short please don't waste my time and I will not reply, as to me, the problem is hooked cooked booked. and closed. "
Mally [/QUOTE]
Mally,

Not to belabor the issue, but I feel you were rude to me in your replies to me in the coin thread. No circumlocution, there. If you read my posts and attempted to understand them, you'd realize that I have a point. Wacky followed it and now agrees with me. You tried reducing a 3-D problem with a 2-d perspective...incorrectly.

I suppose I could have posted the solution, but I feel Wacky had already described it well. It's a single integral of one variable. High-school math. I posted the final answer as well as described the means to solve it, and explained the error in your logic. I don't know what else you expect from me.

You may consider the problem closed, but it is not solved according to your solution. You may continue in your ignorance if you so choose, I can't stop you. But it's a shame that you can not accept a bit of a challenge to your discussion in a math forum, of all things.

Drew