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# Circular field problem.

Fri 4 Mar, 2011 05:39 pm
A donkey is tethered to the side of a circular field. Express the length of its tethering rope as a fraction of the diameter of the field if the donkey's movement is confined to exactly half of the area of the field.

(I have tried this several times myself with no satisfactory answer as yet)
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View best answer, chosen by fresco
rosborne979

1
Fri 4 Mar, 2011 05:47 pm
@fresco,
Sounds like an interesting problem. Unfortunately I have no math skills, so someone else will have to help you out.
fresco

1
Fri 4 Mar, 2011 05:59 pm
@rosborne979,
It looks like a calculus problem involving the area bounded by two curves, and the problem lies in specifying the co-ordinates of the intersection.
laughoutlood

1
Fri 4 Mar, 2011 07:22 pm
@fresco,
579 / 1000
High Seas

1
Fri 4 Mar, 2011 07:37 pm
@fresco,
It's not a calculus problem, it's ole Plato's geometry; the chord (donkey's rope) is known to be longer than r, the circle radius, since a shorter chord will not cover half the circle area, and half of that area is already known to equal {pi r^2~~3.14159 r^2}/ 2. Visualize 2 overlapping circles, or draw them:

http://reference.wolfram.com/mathematica/tutorial/InteractiveGraphicsPalette.html

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High Seas

1
Fri 4 Mar, 2011 07:39 pm
@laughoutlood,
Trig?
0 Replies

laughoutlood

2
Fri 4 Mar, 2011 07:40 pm
@laughoutlood,
I stubbornly refuse to entertain providing any details without certain guarantees as to how tight my tether will be. And don't call me trig.
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engineer

1
Fri 4 Mar, 2011 08:04 pm
@fresco,
If the radius of the field is R then the length of the line is 2RF where F is the desired factor. If a circle of radius 2RF is centered at (R,0), then the two circles will intersect when x=R(1-2F^2). At this point, y = +/- 2RF(sqrt(1-F^2)) From this point, you need to calculate the circular segment of the field circle from the x value above to R and of the donkey circle from x=R(1-2F) to R(1-2F^2). Find the value of F that makes that sum equal to half the area of the original circle (piR^2). The circular segment formula is not very simple so good luck with the calculation.
engineer

2
Fri 4 Mar, 2011 08:28 pm
@engineer,
I refuse to do the math, but I can put it into Excel. I got F=0.5794 so laughoutlood gets the last laugh.

R 10
F 0.579364162
R2 11.58728324
Intersection 3.286743351
h1 6.713256649
h2 4.874026595
Circular Segment 1 92.54824624
Circular Segment 2 64.53135364
Sum 157.0795999
1/2 Circle 157.0796327
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fresco

1
Fri 4 Mar, 2011 11:51 pm
Thank you all.

I am wondering if there is an iterative procedure.
fresco

1
Sat 5 Mar, 2011 12:13 am
@fresco,
Quote:
This cannot be solved analytically, but the solution can be found numerically

Quote following engineer's links for "quarter tank problem"
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laughoutlood

1
Sat 5 Mar, 2011 12:23 am
@fresco,
Newton Raphson

here's a spoiler

http://mathforum.org/library/drmath/view/54871.html
fresco

1
Sat 5 Mar, 2011 01:15 am
@laughoutlood,
Aha !
laughoutlood

1
Sat 5 Mar, 2011 04:30 am
@fresco,
Yeah, lovely question.

Even more beautiful was seeing engineers brain kick in from first principles which meant he could then switch to the suite of excel routines to obviate the crunching.
0 Replies

rosborne979

1
Sat 5 Mar, 2011 08:51 am
@fresco,
fresco wrote:
It looks like a calculus problem involving the area bounded by two curves, and the problem lies in specifying the co-ordinates of the intersection.

I was thinking it would be calculus as well, but I can barely do arithmetic, so I stopped there
0 Replies

amyjohn

0
Tue 17 Apr, 2012 12:58 am
I have similar kind of question http://www.youtube.com/watch?v=0V2Hq8iQys4 related to circular field. Can anyone help me in understanding the problem
laughoutlood

1
Tue 17 Apr, 2012 08:30 pm
@amyjohn,
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

It is a question on how to find the LCM (Lowest Common Multiple) of the two numbers 12 and 18. You might be able to guess that the LCM is 36 , that is (12 x 3) and (18 x 2 ) both equal 36, or in other words Sonia drives around 2 times in 36 minutes and Ravi drives around 3 times in 36 minutes before they are both exactly at the same place on the circular track again.

Or you could try this type of analysis involving identifying the factors of the numbers:

12 = 2 x 2 x 3 ( thats's 2 squared times 3 )

18 = 2 x 3 x3 ( that's 2 times 3 squared )

The lowest number that is a multiple of 2 and 2 squared is 2 squared
The lowest number that is a multiple of 3 and 3 squared is 3 squared
So the lowest common multiple of both expressions is

2 squared times 3 squared = 36
0 Replies

markr

1
Tue 17 Apr, 2012 08:30 pm
@amyjohn,
The solution is given in the video. Are you wondering why the answer is the LCM of 12 and 18?
0 Replies

sandy111

1
Mon 30 Apr, 2012 08:01 am
@fresco,
does it mean the area covered by donkey is limited to the half of area of circular field......?reply
fresco

1
Mon 30 Apr, 2012 08:26 am
@sandy111,
Yes.

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