help! - tethered cow problem
 

I need to know how to solve a problem about a cow tethered to one point on a silo. If the length of the rope is two meters and the radius of the silo is one meter, how much area can the cow graze on around the silo? I found the problem where the length of rope is half the circumference, but I'd rather not use polar coordinates if it's possible. I'm pretty sure this problem can be solved with more simple calculus, but any solution will help.

Thanks for your time and help!

are you stupid? You have two circles.


Take the area of the larger one

subtract the area of the smaller

Done.



No calculus, no polar mathematics...

What about the wall of the silo if he is teatherd to one side it will affect the distance you would end up with a half moon shape

My understanding is the Cow was tethered outside the silo, and it would be full two meters to one side, and the silo to the other... so the circles dont line up perfectly, but its still larger minus smaller.

[QUOTE=Unregistered]My understanding is the Cow was tethered outside the silo, and it would be full two meters to one side, and the silo to the other... so the circles dont line up perfectly, but its still larger minus smaller.[/QUOTE]The effective length of the rope changes as the rope wraps around the silo.

[QUOTE=Unregistered]are you stupid? You have two circles.[/QUOTE]

Except for the fact that you do NOT have two circles, your approach might work.

Consider tearing down the silo. Since the silo covered the area, there is no grass inside.

In this case, your approach is correct. The cow can reach any point in its circle but you must subtract the silo area because it is void of grass.

But look at the tether when the cow is grazing near the far side of the silo. Consider the tether stretched tightly from the anchor point. This is the farthermost point that the cow could reach. The tether stretches across the silo floor and back into the grass on the far side.

But if the silo is in the way, the cow would not be able reach that far. The tether will have to wrap around the outside wall. Since the resulting tether position is not a straight line, it cannot reach as far in that direction. So the perimeter is not a circle.

[QUOTE=Unregistered]are you stupid?
[/QUOTE]
You have no right to insult other people, even if you think (wrongly) that they ask silly questions. Please learn to behave, or stop posting here.

Sorry if I didn't explain the problem fully the first time. It's not two circles and subtracting them. I don't know where you get that idea. The silo is tall and the rope has to wrap around it. Say the diameter of the silo is 1M and the length of the rope is 2M. The rope is tethered to one point on the silo, so there is a half circle of grazing area that can be calculated very easily. Once the cow goes beyond this half circle into the the half that has the silo in it, the line of the circle begins to approach the silo resulting in the half moon shape that someone mentioned earlier. The cow will continue to curve into the silo and eventually hit the wall. Since the circumference is pi, and the length of the rope is 2, the wrapped around grazing areas overlap. So, once you find the area of the half moon shaped grazing area, you have to subtract the areas that overlap. So, I hope that clears is up a little bit. Sorry about the first post, i meant to put diameter instead of radius. Can anyone think of a way to find the area? Thanks!

Have you tryed geometric tools?
A compass and ruler may help you come up with more elegant ideas. :)

The calculus solution is to just use the limit series definition of area. For angles not involving the silo you can just use (Pi*r^2)*(angle in degrees/360). The limit is used to find an approximation for angles involving the silo.

If it was me, then I'd just set the angle of the first equation to 180 degrees. Write this as a constant since the limit of a constant is still a constant. It's nice to eliminate as many variable terms as possible. The area of half a circle is 0.5*Pi*r^2 of course.

The second part is to find the critical points on the silo's perimeter. The point where the cow is no able to move (y=0) is easy to calculate. A full circle is Pi*r^2 so if r=1/2, then the circumference is Pi/4. It turns out that for a 2-meter-long cord you can go around the silo more then once.
This doesn't sound right. If the radius is 1 then it makes more sense so that's what I'll use. :) The unit circle has a radius of 1 and a circumference of Pi. A 2-meter cord would go to 2/Pi meters around the silo on either side. This means there's an overlap that will have to be considered as another section. Once you have the point for y=0, then you need the point for y=2 which is not too hard to get using the same method. Sorry if I'm a little tired - lator I'll probally get the solution into 3 steps since it's not that hard. :P

<2nd try> The easiest way using a compass is to start with half a circle, then at every point on the silo, calculate the length (or use a piece of string and a coin with the same diameter as my drawing) of each point and draw circles. You end up with a group of circles. Hehe, get ready for working with some sets or summation notation no matter what you do. You have to find just the areas of the _extensions_ of the previously enclosed area and add those to the final sum. After a couple trys I'd come up with some formula that I could plug into a "E" type function on my calculator.

[QUOTE=Marc Bacani]I need to know how to solve a problem about a cow tethered to one point on a silo. If the length of the rope is two meters and the radius of the silo is one meter, how much area can the cow graze on around the silo? I found the problem where the length of rope is half the circumference, but I'd rather not use polar coordinates if it's possible. I'm pretty sure this problem can be solved with more simple calculus, but any solution will help.

Thanks for your time and help![/QUOTE]
The best way to approach such a problem is to be systematic about it.

If it were me, I'd define a polar coordinate system centered about the tie-down point. Find a function, r(theta) describing the farthest distance the cow can reach at every angle theta. Because of the point at which the chain begins to wrap around the silo, this will be a piecewise function. Then integrate the area r*dr*d(theta) from -pi to pi (pi being the 'back side' of the silo). When you're done, subtract the area of the silo.

If you do it this way, you won't have to subtract out any overlap areas either, because you'll only account for each region once.

Drew

I would adopt a similar approach except that I would avoid worrying about the polar co-ordinates, as such.

If we draw a line which passes through both the center of the silo and the attachment point, we will have an axis of symmetry. Therefore, we can compute the grasing area on one side of that line and double it. This will resolve the overlap that might occur when the tether is longer than the half-circumference of the silo. Assume that the origin is at the center of the silo and that the attachment point is to the right (0 degrees).

We now wish to define an orientation and for each orientation the amount of grazing area that is assigned to it. Then we can simply integrate over all orientations to determine the area.

Note that for any given orientation, we can stretch the tether in that direction. In the general case, a portion of the tether will wrap around the silo and the remainder will be along a tangent in that direction.

Now, draw the radius of the silo that passes through that point of tangency.
Let's call the angle that that radius makes with the axis of symmetry "theta". The tether will be stretched at an angle of (theta + pi/2).

Note that the area grazed is just the area swept by the portion of the tether that is not against the silo. We need to determine L(theta) for each angle and integrate to find the total area grazed. The incremental area grazed is 1/2 L^2 dtheta.

Now for the endpoints in the interation.
When theta < -pi/2, the cow would be grazing on the wrong side of the axis.
For -pi/2 < theta < 0, the tether is stretched to its maximum length. Call it "T". This grazing area is a quadrant of a circle and the area is pi*(T^2)/4.

Call the radius of the silo "R". If T/R is sufficiently small, the cow would graze around the silo until the tether is completely against the silo.

Note that for an angle theta, R*theta of the tether is against the silo.
The remainder of the tether (T-R*theta) is the length to be used in the integration. If T = pi*R, the cow would reach the end of its rope just as it gets to the intersection of the silo and the axis opposite the attachment point.

For any longer tether, at some angle (less than pi), the cow would reach the axis. Determining this angle is left as an exercise for the student.

If we stop the integration before theta = pi, we must also add the additional area. This incremental area can be determined by expressing the remaining length as a function of theta. However, it is probably easier to observe that there is a triangle formed by the tangency point where we stopped the integration, the location of the cow on the axis at that time and the center of the silo. Calculate the area of the triangle and subtract the portion that is within the silo.

Now, add up the three parts and you have your answer in terms of T and R.
Substituting the values given produces the final result.

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

help!-tethered cow problem
 

[QUOTE=drew]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.
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[/QUOTE]
:smile:
Drew: To settle the problem once and for all for you and the other posters, who contributed, and gave their opinions Please check out this URL from the TOI which has a readership of over a million readers in India and the Gulf countries. If the answer (two identical ones given) was wrong its strange there were no objections in any follow up.

[url]http://www.mindsport.org/archives/2005/november05/20november05.htm[/url]

You will find it in 'Googled-Coins Dept'. and a little before that you will find the question asked (to remind one) in a previous article of Mind Sport.
So Drew you and the rest are not infallible in Mersenneforum and its best we bury the hatchet.

The rest of your post is so full of holes in your reasoning it is worse than Eratosthenes' Sieve to say the least. I am tempted to tackle each point, one by one but I leave it to you to sort yourself out. You are your your best teacher in faith and morals but take my advice-don't be so dogmatic and a faithful follower of your mentors. Remember they may be wrong in many matters, and probably in most, and then its the case of the blind leading the blind. Where do you expect them to land you but in a ditch ?

Its an interesting study to me to study the psychology of the posters in this forum.

Some are genuine seekers of truth, others are know alls, and consider themselves infallible, others, thrive on the the name they have built up by demeaning others, and so on and on. I lack the time to give my full opinion as I have other more enjoyable social commitments on an evening such as this, to be stuck behind an in human computer, communicating to someone who has a coded name, with no idea of how old he is, where does he stay, and what is his nationality, to mention a few unknowns.

But I hope to send in a thread asking the powers that be to remove this cloak of secrecy and the unknowns in a poster and lets all see openly who they are and their weapons.

Im sorry to say there are genuine posters I recommended to this forum who have been hounded out of it, so much so they have joined other organisations and tell me they are happy.

Well to cut this post short, I'm very sorry Drew, if I have been rude in my comments to you and the others, and I admit I may have. Its the hectic pace in a mega city like Mumbai which perhaps I could not have taken in my stride at the time and sesaon.
So please over look my remarks and get on with it.

Im off to rock and roll and roll the evening away with loads of JW down my gullet and a lavish sumptious dinner to bless a couple in newly wedded bliss!
May God Bless you too. :rolleyes:
Mally :coffee:

[QUOTE=mfgoode] and gave their opinions … there were no objections [/QUOTE]

Mally,

You should know that just because something is "popular" does not mean that it is correct. Therefore, this portion of your posting fails the test for valid argument.

[QUOTE] you and the rest are not infallible [/QUOTE] Kettle, meet Pot.

[QUOTE] someone who has a coded name, with no idea of how old he is, where does he stay, and what is his nationality, to mention a few unknowns. … this cloak of secrecy [/QUOTE]
I, for one, take exception to your attitude in the matter. It is very common practice for individuals to be known by nicknames. It in no way implies that they are attempting to hide their identity or other attributes.

Would you propose that William J. Clinton was attempting to hide his identity by using the name Bill? What about Richard Bernard Skelton? Most people know of him by the name "Red". Have you heard of Leonard Slye or William Sidney Porter?

In my own case, I have used the nickname "Wacky" for well over 50 years and have been published under it. I have also published under my legal name, Richard Wackerbarth. Anyone who has been reading this forum for such time as yourself should recognize my postings and, without much effort, be able to ascertain a reasonable estimate of both my age and location. I have left many clues about the Internet (ne: ARPANET) in the past 30 years.

However, I really consider your complaint to be in bad faith. For any legitimate purpose, have you ask anyone for such details and been refused?

[QUOTE]I'm off to rock and roll and roll the evening away with loads of JW down my gullet and a lavish sumptuous dinner to bless a couple in newly wedded bliss![/QUOTE]

I hope that you enjoyed the Wedding Feast and I wish you a Prosperous New Year.

Richard

Mally,
Your claim that the curve described by the cow is a cardioid is false.

Take a look at the diagram of a cardioid at the link you provided. Note that in the left-hand diagram the x and y axes are marked. Note also in the java applet on the right that the two points where the red spot is closest to and farthest from the silo are both on the x-axis. Since the cow will be closest to the silo when it has used up all the rope, it suffices to show that when the rope has wrapped completely around the silo the cow is not on the x-axis.

[quote=Snorko] Say the diameter of the silo is 1M and the length of the rope is 2M.[/quote]

The circumference of the silo is 2*pi*0.5 = pi metres.

The length of the rope is 1 metre, which is not pi/2 metres. QED

Drew has already provided a rigorous proof of the same thing, which you ignored and changed the subject to that of another thread.

[quote=mfgoode] have it your way Drew but TRUTH WILL TRIUMPH![/quote]

Indeed it will.

[QUOTE=mfgoode]:smile:
Drew: To settle the problem once and for all for you and the other posters, who contributed, and gave their opinions Please check out this URL from the TOI which has a readership of over a million readers in India and the Gulf countries. If the answer (two identical ones given) was wrong its strange there were no objections in any follow up.

[url]http://www.mindsport.org/archives/2005/november05/20november05.htm[/url]

You will find it in 'Googled-Coins Dept'. and a little before that you will find the question asked (to remind one) in a previous article of Mind Sport.
So Drew you and the rest are not infallible in Mersenneforum and its best we bury the hatchet.

The rest of your post is so full of holes in your reasoning it is worse than Eratosthenes' Sieve to say the least. I am tempted to tackle each point, one by one but I leave it to you to sort yourself out. You are your your best teacher in faith and morals but take my advice-don't be so dogmatic and a faithful follower of your mentors. Remember they may be wrong in many matters, and probably in most, and then its the case of the blind leading the blind. Where do you expect them to land you but in a ditch ?

Its an interesting study to me to study the psychology of the posters in this forum.

Some are genuine seekers of truth, others are know alls, and consider themselves infallible, others, thrive on the the name they have built up by demeaning others, and so on and on. I lack the time to give my full opinion as I have other more enjoyable social commitments on an evening such as this, to be stuck behind an in human computer, communicating to someone who has a coded name, with no idea of how old he is, where does he stay, and what is his nationality, to mention a few unknowns.

But I hope to send in a thread asking the powers that be to remove this cloak of secrecy and the unknowns in a poster and lets all see openly who they are and their weapons.

Im sorry to say there are genuine posters I recommended to this forum who have been hounded out of it, so much so they have joined other organisations and tell me they are happy.

Well to cut this post short, I'm very sorry Drew, if I have been rude in my comments to you and the others, and I admit I may have. Its the hectic pace in a mega city like Mumbai which perhaps I could not have taken in my stride at the time and sesaon.
So please over look my remarks and get on with it.

Im off to rock and roll and roll the evening away with loads of JW down my gullet and a lavish sumptious dinner to bless a couple in newly wedded bliss!
May God Bless you too. :rolleyes:
Mally :coffee:[/QUOTE]
Mally,

It disappoints me to see you so eager to accuse me of not providing sound arguments, and then see you refer to a popular consensus rather than provide a sound defense of your own. If you had bothered to spend 10 minutes working out the integral (per Wacky's suggestion) on your own for the coin problem, you'd have discovered a discrepancy between that result and your own. It would then be necessary to resolve this discrepancy and prove to yourself which result is correct and why, at which point you could enumerate the differences so we may all benefit, or have the opportunity to identify where you are mislead.

However, that's not the case. I have done this myself, and have attempted to communicate this to you. You have dismissed this as 'circumlocution' or 'full of holes', but you have yet to explain yourself.

It is not my intention to use insulting philosophical rhetoric about the psychology of posters and so forth. My interest is only in a mathematical discussion about the problem at hand, and I'm not about to take up this debate with millions of posters in India. I would expect, however, that had I posted in that forum, I'd have been received with common courtesy rather than condescending remarks about how you're so sure you're right that you refuse to entertain my argument to the contrary. That reeks of hypocracy considering your recent posts.

With regard to anonymity on the internet, Drew is short for Andrew, which is my first name. I don't really see the need to disclose my complete identity, because I don't think it matters in the least. My arguments need only stand on their own merit, not be appreciated or refuted based on superficial facts such as age, race, or career choice. Albert Einstein was a patent clerk. Would you have taken him seriously if you encountered him on an on-line forum?

I appreciate the fact that I can participate in on-line discussions without having to disclose my identity. Not so much because I don't want you to know who I am, but to keep my on-line communications private from people I know personally. I'd rather not have anyone be able to Google my name and read hundreds of conversations I've had with complete strangers. I assure you, though, that I have nothing to hide with regard to my qualifications to participate in discussions such as this. I just ask that you address my arguments with the respect and consideration you would any colleague or other acquaintance.

Drew

Wacky makes some good points, mathematically and otherwise. I particularly agree with the comments regarding Mally's concern about the "unknowns" surrounding the posters. Age, location, race, religion- though interesting on a personal level, these certainly shouldn't play much of a part in deciding who is correct in a mathematical argument. Some may be biased toward believing older people because they believe that a larger volume of experience is important, while others may feel the younger, fresher insights are more important. Shall we give greater deference to male arguments, because statistically there are more male mathematicians than female mathematicians? Or should we defer to the female mathematicians because they have overcome more prejudice to reach their position, and therefore must be very good?

This is the second time in recent weeks I've seen postings commenting on the use of nicknames for identification, as if people are hiding deliberately. If there is any arena where characteristics such as race, gender, age, name, etc. are unimportant, it is math. Base your decisions on the math, not on subtle biases you may feel (but not even know that you feel) about the posters here. The traditional representation of "Justice" is blind; that's how decisions should be made in law, as well as in math.

Norm

Very well put Spherical Cow.

help!-tethered cow problem
 

[QUOTE=Numbers]Very well put Spherical Cow.[/QUOTE]
:smile:
Happy New year to all you posters and hope you add more humour in your posts. As it is, you sound as MORONS.

Well after last nights binge here am I again replying by parody.
Please take it in the spirit(pardon the pun) that it is given !
:smile:
Reply by Parody.
There was ‘Mally’ and his mallet,
Who requested to bury the hatchet,
Do you think they’d leave him alone?
No, they bit him deep to the bone !.

There was ‘Wackerbarth’ ashamed of his real name
He blamed it on receiving too much fame;
Insisting on pricking the thoughtful bubble,
He was the one who started all the trouble!

On probability; commented ‘Drew’ on integration;
Of 3D to 2D was not permitted separation;
He possibly drew on a joint,
And never came actually to the point!

Then there was ‘Numbers’, of which he knew nothing,
Who raved and he ranted and had his fling;

To join him came ‘Spherical cow’,
His ego made him take the bow, :bow:
He spoke of blind justice in law
And then finally called it all a draw:! :surprised

Refrain:
Heaven only knows, what these programmers are up too? :question:
It’s a pity original mathematicians in this forum are few; :sad:
All praise to Ewmayer, Silverman, Akruppa and Xilman,!
They alone can evaluate a post within!

Pythagoras must have turned sev’ral times in his grave,
But there was no one to save Mally the brave.
Wolves hunt in packs it is known
But the Lion fights his battle alone
Refrain: :grin: Bravo
Heaven only knows………
Mally :coffee:

[QUOTE=mfgoode]Please take it in the spirit(pardon the pun) that it is given ![/QUOTE]
[QUOTE]There was ‘Wackerbarth’ ashamed of his real name
He blamed it on receiving too much fame;[/QUOTE]

Now, I REALLY take offense!

I am certainly not ashamed of my surname. I will have you know that it is a very old, and respected, name. It was only a couple generations ago that my ancestors dropped the "von".

However, it still contains more characters than some systems allocate. Further, numerous persons manage to do a horrible job when attempting to spell or pronounce it. I've had to put up with that for as long as I can remember.

To borrow a phrase from Dr. Silverman, "You are a CRANK. Go away!"

help!-tethered cow problem
 

:grin: Hey Wacky. Don't be so touchy!
Haven't you heard the bard say
"Whats in a Name?
"A rose by any other name will smell just as sweet" :rolleyes:
Mally. :coffee:

Well my request to be treated with respect has once again fallen on deaf ears. That's okay. Mally can continue in his delusion as a mathematical God who cannot be questioned, and the rest of us will just know him for the crank that he is.

Drew

[QUOTE=mfgoode]There was ‘Mally’ and his mallet,
Who requested to bury the hatchet,
Do you think they’d leave him alone?
No, they bit him deep to the bone !.

...
[/QUOTE]
On the whole, I think I prefer William Topaz McGonagall. He had a better grasp of metre and a more skillful use of vocabulary.

Paul

help!-tethered cow problem
 

:rolleyes:
Drew My boy, respect has to be earned and not demanded.
This is lesson No.1 in the 'School of Hard Knocks'
Mally :coffee:

help!-tethered cow problem
 

[QUOTE=xilman]On the whole, I think I prefer William Topaz McGonagall. He had a better grasp of metre and a more skillful use of vocabulary.

Paul[/QUOTE]
:smile:
Its plain Iambic metre.

Try Lord Byron for size.

'Tell me not of a name great in Story,
The days of youth are the days of glory
And the mirth and Ivy of two and twenty
Are worth all the laurels,
Be it ever so plenty'
Can you analyse the metre and is the vocab okay? Try getting this verse on the net and its title?
Mally :coffee:

[QUOTE=mfgoode]:rolleyes:
Drew My boy, respect has to be earned and not demanded.[/QUOTE]
Interesting. Most people tend to give me the benefit of the doubt. Or at least refrain from ridiculing me at the first encounter. Especially when I tactfully point out an error in their analysis...most actually try to resolve the issue instead of slinging insults. :smile:

Drew

help!-tethered cow problem
 

:smile: Drew you are like a tortoise. Its only when I will tap you on the shell will you stick your head out.
I wish you all the best and may you prove the RH with your mastery in analysis.
Mally :coffee:

Since I've been named in this thread I suppose I ought to say something... but I just don't figure it's worth it. Poorly posed problem leads to different solutions, flamewar. More at 11. Granted, the poetry (if you want to call it that) bit was a new one.

I'd be tempted to move this thread into Miscellaneous, but there's some decent math in here so I'm hesitant. Opinions?

Alex

Edit: wait, this is the wrong thread. My bad. The problem here was well posed. The coin thing was ill-posed and where the bickering started.

[QUOTE=akruppa]Since I've been named in this thread I suppose I ought to say something... but I just don't figure it's worth it. Poorly posed problem leads to different solutions, flame war. More at 11. Granted, the poetry (if you want to call it that) bit was a new one.
Alex
Edit: wait, this is the wrong thread. My bad. The problem here was well posed. The coin thing was ill-posed and where the bickering started.[/QUOTE]
:smile:
Thank you akruppa for being impartial but frank. :bow:

Yes the coin problem started the flame war particularly when I decided not to
reply any more. :bounce:
That would be a good point from which to move the thread and relegate it to 'Misc'
Posts not containing single math, numbers or equations should be moved off.
Circumlocution at all costs should not be tolerated. This is more a display of language skills than math skills. We want MATH, MATH AND Nothing but MATH!
in Puzzles and Math. :banana:
Also since evidently the pollster has not even entered a law court it should be a mandatory requirement to state Name, age, and state, and qualifications in every ones profile. There is no need for a separate PM asking for one's credentials.Its only then, when flaming will stop and respect or disregard can be judicially given. Nick names will have more meaning then, as Dr, Silverman has also mentioned in a recent post.
Mally :coffee:

Or we could all be mature and accept that even if we don't know who another person is, we can just freaking accept them as having a chance of being right or at least interesting. I'm not sure if I'd rather see 1 thinking about 100 or 100 thinking about 1. :) Of course if someone comes off as being crude, rude, hypercritical, sycophantic and toadyish, mean, on an ivory tower, or just plain A retentive then I can (try to, atleast) ignore them.
It's not so much that an idea is right, but that it makes the poster think.

So to post on this site, we're all supposed to provide our SSN, DL, and military ID? :P Funny, but I thought the point of forums was that you have no idea who a person really is - so you don't know if they're black, young, silver haired, white, left handed, female, northern hemisphere, or whatever. Thier behaviour will tell you what you need to know. If you want to make a game of trying to determine who someone "really is", then remember not to be an online stalker. LOL :bounce:

[QUOTE=nibble4bits]Or we could all be to,....... atleast) ignore them.
It's not so much that an idea is right, but that it makes the poster think.

So to post on this site, we're all supposed to provide our SSN, DL, and military ID? :P Funny, but I thought the point of forums was that you have no idea who a person really is - so you don't know if they're black, young, silver haired, white, left handed, female, northern hemisphere, or whatever. Thier behaviour will tell you what you need to know. If you want to make a game of trying to determine who someone "really is", then remember not to be an online stalker. LOL :bounce:[/QUOTE]
:smile:
SSN,DL and military Id: This is a disguised form of circumlocution IMO. of what is known as exageration
Its going beyond the domain required and permitted and certainly not mathematical.
By stating age and qualification one can still be incognito, yet give one the level of understanding in, a debate of the other person.
Mally :coffee:

Ah, but they can lie.
Hahahahahaha Most illogical I guess.

[QUOTE=mfgoode]By stating age and qualification one can still be incognito, yet give one the level of understanding in, a debate of the other person.[/QUOTE]
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. Endowed by nature with perhaps the most glorious baritone voice to be heard on the global stage since the memorable concert debut in 1835 of Millard Fillmore; endowed also with twelve incredibly agile fingers; Xilman has had a long and varied career in the field of entertainment starting with nine years Oxford University... where it was that he first decided to devote his life to what has since become a rather successful scientific project -- namely, the attempt to prolong adolescence beyond all previous limits.
Even before he came to Oxford, however, he was well known in academic circles for his masterly translation into Latin of The Wizard of Oz, which remains even today the standard Latin version of that work. A few years ago he was inducted... forcibly... into Her Majesty's Army and spent most of his indenture in London as sort of Army liaison to the Office of Naval Contemplation. About his service record he is justifiably modest, but it is known that in a short time he rose to the rank of brigadier general. However, before he could acquire tenure, he was discharged and, owing to nepotism and intrigue, he emerged with only the rank of specialist 3rd class, which was roughly equivalent to the rank of corporal without portfolio.
But to return to his career in show business: for several years he toured vaudeville theatres with an act consisting of impressions of people in the last throes of various diseases. I'm sure that many of you here tonight still recall with pleasure his memorable diphtheria imitation. He is generally acknowledged to be the dean of living British composers, and is currently working on a musical comedy based on the life of Adolf Hitler.

Paul

[QUOTE=nibble4bits]Ah, but they can lie.
Hahahahahaha Most illogical I guess.[/QUOTE]
:rolleyes: Not so nibble4tits! They can!
Short of a lie detector test their level of reasoning will give them away definitely.
The bard has this to say
"A tale told by an idiot full of sound and fury,
Signifying nothing!" :smile:
Mally :coffee:

[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

help! -tethered cow problem.
 

[QUOTE=drew]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[/QUOTE]
:smile: Fair enough Andy, as long as you establish that the expression is mine.
The rest a school boy coud work out. Thanks
Mally :coffee:
P.S. BTW the curve we discussed is NOT a Cartoid in your spelling but a CARDIOID.

[QUOTE=Wacky]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.[/QUOTE]
:smile: Well Richard, how do you feel about the problem now?
As an expert do you still say the reasoning is faulty? Have you bothered to test out the solution or are you only interested in giving totally negative comments?
Mally :coffee:

[QUOTE=mfgoode]Well Richard, how do you feel about the problem now?
As an expert do you still say the reasoning is faulty? Have you bothered to test out the solution or are you only interested in giving totally negative comments?[/QUOTE]

I am only pointing out that your "argument" for the correctness (that it is published somewhere) is lacking from a mathematical perspective. Had you given the publication as a reference which shows the derivation of the answer, then said publication might well be of some value. However, I reitterate that it is not proper math, or debate, to simply state that someone else said so, and therefore it MUST be so.

help! -tethered cow problem.
 

[QUOTE=Wacky]I am only pointing out that your "argument" for the correctness (that it is published somewhere) is lacking from a mathematical perspective. Had you given the publication as a reference which shows the derivation of the answer, then said publication might well be of some value. However, I reitterate that it is not proper math, or debate, to simply state that someone else said so, and therefore it MUST be so.[/QUOTE]
:sad: Von wackerbarth there is no 'argument' about the matter. You simply want to make one out of it.
I have clearly stated the author and the name of the book and can give the publishers name if required. I have also lucidly given the solution STRAIGHT from the book albeit a diagram which any high school student can follow.
And still you stubbornly refuse to accept it simply because it is from Mally??? :censored:
Your post analysis is that you are terribly confused within your self. And by the way I have not come across a single post of yours with a correct solution to ANY of the puzzles. :furious:
You simply hitch your wagon to the side which has the fastest horses!!!! :confused:
Mally :coffee:

[QUOTE=mfgoode]:smile: Fair enough Andy, as long as you establish that the expression is mine.
The rest a school boy coud work out. Thanks
Mally :coffee:
P.S. BTW the curve we discussed is NOT a Cartoid in your spelling but a CARDIOID.[/QUOTE]

Call it what you want. Neither cardioid nor cartioid is a solution to this problem. :wink:

help! -tethered cow problem.
 

[QUOTE=drew]Call it what you want. Neither cardioid nor cartioid is a solution to this problem. :wink:[/QUOTE]
:surprised Well, Well, quite a nonchalant attitude in mathematics!
You know that Mathematical symbolism is very important and there is no dearth of it in our world today. BY adding one more like Cartoid will confuse the issue as this is a medical term familiar to doctors.
FYI: when calculus was jointly invented by Newton and Leibnitz at much the same time, Germany and France stole an advance over England mainly because of the symbol for differentiation. Leibnitz's was simpler and more clear on understanding infinitesimals. Not to sound partial they also had the better Mathematicians, at the time, who took up the challenge!
The nearest I could get to cartoid was this
"Did you mean: carotid
Cartoid
Cartoid, Carotid artery stenosis angiogram. Cartoid, Carotid body tumour (angiogram).
Cartoid, Carotid body tumour (MRA)
[url]www.freevas.demon.co.uk/cartoid.htm[/url] - 2k - Cached - Similar pages
Cartoid angioplasty and stenting"
Mally :coffee:

[QUOTE=mfgoode]:surprised Well, Well, quite a nonchalant attitude in mathematics!
You know that Mathematical symbolism is very important and there is no dearth of it in our world today. BY adding one more like Cartoid will confuse the issue as this is a medical term familiar to doctors.
FYI: when calculus was jointly invented by Newton and Leibnitz at much the same time, Germany and France stole an advance over England mainly because of the symbol for differentiation. Leibnitz's was simpler and more clear on understanding infinitesimals. Not to sound partial they also had the better Mathematicians, at the time, who took up the challenge!
The nearest I could get to cartoid was this
"Did you mean: carotid
Cartoid
Cartoid, Carotid artery stenosis angiogram. Cartoid, Carotid body tumour (angiogram).
Cartoid, Carotid body tumour (MRA)
[url]www.freevas.demon.co.uk/cartoid.htm[/url] - 2k - Cached - Similar pages
Cartoid angioplasty and stenting"
Mally :coffee:[/QUOTE]
Well Mally, I was just pointing out your Red Herring. Rather than admit you were wrong about something, you criticize my spelling. It's a very transparent tactic.

FWIW, I spelled it 'cartioid', not 'cartoid'. With that correction you'll be relieved to find that Google returns "Did you mean cardioid?". And I later realized my error and began spelling it correctly, which you'll see if you read subsequent posts in this thread. I'm not sure how you pronounce English words over in India, but in the U.S. we'd pronounce them both the same. It's a simple mistake and absolutely irrelevant to the discussion at hand.

Drew

help! -tethered cow problem.
 

[QUOTE=drew]Well Mally, I was just pointing out your Red Herring. Rather than admit you were wrong about something, you criticize my spelling. It's a very transparent tactic.

FWIW, I spelled it 'cartioid', not 'cartoid'. With that correction you'll be relieved to find that Google returns "Did you mean cardioid?". And I later realized my error and began spelling it correctly, which you'll see if you read subsequent posts in this thread. I'm not sure how you pronounce English words over in India, but in the U.S. we'd pronounce them both the same. It's a simple mistake and absolutely irrelevant to the discussion at hand.

Drew[/QUOTE]
:smile: Well Drew we in India follow the 'Queen's English' with BBC accents.
Pronunciation: 'kär-dE-"oid Function: noun Date: 1753
Mally :coffee:

help! -tethered cow problem.
 

[QUOTE=mfgoode]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
Mally :coffee:[/QUOTE]

:rolleyes:
I have now established that the grazing area is definitely not a cardioid, the area of which is (3*Pi *l^2)/2
Mally :coffee: