[QUOTE=xilman]I'd like to introduce now the featured artist of this forum. I'm sure that you'll all agree without any hesitation that Xilman is the most brilliant creative genius that Britain has produced in almost 200 years, so perhaps a few words of biographical background might not be amiss.
...........
Paul:/UnQuote:
:unsure: Bravo! A brilliant piece of literature! Its even humorous (or less?)
But as the bard says..........:sad:
Mally :coffee:

I bet I could make even a wise man believe a lie that he wanted to believe.
[spoiler] But that would be evil. <devilish grin> [/spoiler]

help!-tethered cow problem
 

[QUOTE=nibble4bits]I bet I could make even a wise man believe a lie that he wanted to believe.
[spoiler] But that would be evil. <devilish grin> [/spoiler][/QUOTE]
:smile:
To all the posters who commented on this thread and even traded virtual verbal blows and to Drew in particular who took it very seriously!
I have found the exact problem stated in my book 'Ingenious Mathematical Problems and Methods' by L.A. Graham. I cannot draw the diagram here but the correct answer he works out is that the total grazing area is
(5*pi*L^2)/6 where the length of the rope, L, = semi perimeter of the silo. So that puts an End to all speculation and spurious calculations. :showoff:
It will be interesting to note that this area is not much different if the cow had to be tethered in the centre of the field and had the whole circle to graze in. It will be only 0.833% greater! :flex:
Mally :coffee:

[QUOTE=mfgoode]:smile:
To all the posters who commented on this thread and even traded virtual verbal blows and to Drew in particular who took it very seriously!
I have found the exact problem stated in my book 'Ingenious Mathematical Problems and Methods' by L.A. Graham. I cannot draw the diagram here but the correct answer he works out is that the total grazing area is
(5*pi*L^2)/6 where the length of the rope, L, = semi perimeter of the silo. So that puts an End to all speculation and spurious calculations. :showoff:
It will be interesting to note that this area is not much different if the cow had to be tethered in the centre of the field and had the whole circle to graze in. It will be only 0.833% greater! :flex:
Mally :coffee:[/QUOTE]
pi*L[sup]2[/sup] / (5/6 pi*L[sup]2[/sup]) = 6/5 = 1.2

That's 20%, not .833%

I think you meant to say that the grazing area *with* the silo present is 83.3% of what it would be without.

I'm curious. Does the solution in your book mention Pascals Limacon or the Cardioid?

Drew

[QUOTE=mfgoode]It will be interesting to note that this area is not much different if the cow had to be tethered in the centre of the field and had the whole circle to graze in. It will be only 0.833% greater![/QUOTE]

Mally,

Did you misplace a decimal point?

I believe that the correct ratio of the areas which you mention (whether they are correct, or not) is that the full circle is 20% greater.

[QUOTE=mfgoode]I have found the exact problem stated in my book 'Ingenious Mathematical Problems and Methods' by L.A. Graham. I cannot draw the diagram here but the correct answer he works out is that the total grazing area is (5*pi*L^2)/6 where the length of the rope, L, = semi perimeter of the silo. So that puts an End to all speculation and spurious calculations.[/QUOTE]

Mally,
Your reasoning is faulty. Just because you found something in some book does not necessarily imply that the information is correct. Even the best of authors have, on occasion, been found to have made a mistake.

Therefore, finding the exact problem in the book may support a particular answer, but it cannot be considered a "final authority" as you imply. It still remains for each individual to examine the evidence, including your latest reference, and attempt to resolve, for themself, the correct answer.

Mally, see post #14, [I]supra[/I].

Also note that the original puzzle was for the case when the rope length is less than half of the circumference of the circle (rope= 2 meters, circumference = 6.28 meters); the original poster notes that he had found the problem (and presumably, the solution ) for the situation where the rope length = 1/2 circumference.

help!-tethered cow problem
 

Drew[/QUOTE]
[QUOTE=drew]pi*L[sup]2[/sup] / (5/6 pi*L[sup]2[/sup]) = 6/5 = 1.2

That's 20%, not .833%

I think you meant to say that the grazing area *with* the silo present is 83.3% of what it would be without.

I'm curious. Does the solution in your book mention Pascals Limacon or the Cardioid?

Drew[/QUOTE]
:redface:
Im sorry that I gave some misinformation in my post to the silo problem.

Actually he gives two goats one (George) tethered to the silo,at half circumference length ( call it L) This rope is 11 feet long. The other goat (Bill) grazes freely in the field tied with a rope of 10' length.

I will give a partial solution restricted to my presentation of no diagram.
The Integral of one side of the curve is of [(pi*x^2 dx)/2L] from 0 to L.
The one side area is thus (pi*l^2)/6

So for both sides, which are equal in this case, the area is twice the single area. i.e.(2Pi*L^2)/6. So the total of George's grazing area is the area of the 2 symetrical curves on either side of the ring attached to the silo. plus the area of the semi circle which is (pi*l^2)/2.

Adding these up we get (5 Pi*l^2)/6 which if compared to Bills area of 100*pi gives the fraction 121/120 as much, or just 0.833% greater

I have put it as clear as I can and hope there are no mistakes

No he doesnt mention Pascals Limacon or the cardioid. I wouldnt expect that, in a book for the lay reader, altho he uses calculus and admits he gives an easier but correct version by using similar triangles and infinitesimals.

It will be interesting to compare this area to the cardioid and find out if its the same area or not. :unsure:
I leave that up to you Drew as its for your benefit that I have given a partial solution. If you still want to go further I will describe the diagram used by Graham but only if necessary as it will entail a lot of trouble on my part.
Mally :coffee:

[QUOTE=mfgoode]

:redface:
Im sorry that I gave some misinformation in my post to the silo problem.

Actually he gives two goats one (George) tethered to the silo,at half circumference length ( call it L) This rope is 11 feet long. The other goat (Bill) grazes freely in the field tied with a rope of 10' length.

I will give a partial solution restricted to my presentation of no diagram.
The Integral of one side of the curve is of [(pi*x^2 dx)/2L] from 0 to L.
The one side area is thus (pi*l^2)/6

So for both sides, which are equal in this case, the area is twice the single area. i.e.(2Pi*L^2)/6. So the total of George's grazing area is the area of the 2 symetrical curves on either side of the ring attached to the silo. plus the area of the semi circle which is (pi*l^2)/2.

Adding these up we get (5 Pi*l^2)/6 which if compared to Bills area of 100*pi gives the fraction 121/120 as much, or just 0.833% greater

I have put it as clear as I can and hope there are no mistakes

No he doesnt mention Pascals Limacon or the cardioid. I wouldnt expect that, in a book for the lay reader, altho he uses calculus and admits he gives an easier but correct version by using similar triangles and infinitesimals.

It will be interesting to compare this area to the cardioid and find out if its the same area or not. :unsure:
I leave that up to you Drew as its for your benefit that I have given a partial solution. If you still want to go further I will describe the diagram used by Graham but only if necessary as it will entail a lot of trouble on my part.
Mally :coffee:[/QUOTE]
No, that's alright. I'm usually more intetested in knowing how to solve these things than actually going through the trouble...I can think of a few good approaches.

Anyway, the .833% comes from the fact that one rope is 11/10 as long as the other, which when squared yields an area 121% as much as the other rope would. Since the ratio we found earlier is 120, then you can see how the answer becomes 121/120 as large.

Drew

help!-tethered cow problem
 

[QUOTE=drew]No, that's alright. I'm usually more intetested in knowing how to solve these things than actually going through the trouble...I can think of a few good approaches.

Anyway, the .833% comes from the fact that one rope is 11/10 as long as the other, which when squared yields an area 121% as much as the other rope would. Since the ratio we found earlier is 120, then you can see how the answer becomes 121/120 as large.

Drew[/QUOTE]
Well drew, without using the integral of (2 pi*x^2)dx/6 for one side how will you arrive at the denominator of the fraction 121/120 ? You assume it is worked out, but how?
Mally :coffee:

[QUOTE=mfgoode]Well drew, without using the integral of (2 pi*x^2)dx/6 for one side how will you arrive at the denominator of the fraction 121/120 ? You assume it is worked out, but how?
Mally :coffee:[/QUOTE]
Read post number 48. Assuming the expression you provided is correct (5/6 pi L[sup]2[/sup]), that's the ratio I had determined for one answer over the other if the ropes were equal length. But since one rope is 11/10 as long as the first, the area for that rope is 121/100 as large as it would be. 121% / 120% = 1.00833.

Drew