Cow Grazing Half the Circle: Newton-Raphson Method
Date: 01/18/98 at 07:53:11 From: Lord Julus Subject: Hi ! Hello ! I hail from Romania and I have this problem: Let's assume there is a perfect circle filled with grass. In this circle there is a cow tied with a rope to the fence. The cow can move inside the circle only as much the as the rope allows it to. The problem is this: how long must the rope be so that the cow can eat exactly half of the circle area? Making it more geometricaly: . - . ' ' <---- something like a circle and a rope | / | inside. What is the length of the rope '/ ' so it could balance inside the circle ` - ` on exactly half of the circle's area ? Of course we assume it's a perfect geometrical problem. The cow and grass stuff are the nice way to put it ;-) Thanks in advance, Lord Julus.
Date: 01/18/98 at 09:04:44 From: Doctor Anthony Subject: Re: Hi ! This problem comes up from time to time in the guise of the length of rope required to tether a goat on the boundary of a circular field such that the goat can eat exactly half the grass in the field. See the answer in our archives at http://mathforum.org/dr.math/problems/steele8.8.97.html -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Date: 01/19/98 at 09:29:19 From: Lord Julus Subject: Re: Hi ! Thank you so much ! This was a really neat solution. I really understood it... until the Newton-Raphson method. If it's not very hard maybe you can tell me where I can read about that. Thank you, Lord Julus
Date: 01/19/98 at 11:52:42 From: Doctor Anthony Subject: Re: Hi ! The Newton-Raphson method is used for making successive approximations to a root of f(x) = 0. The equation must be put into this form before you start; i.e. we must have all terms on the left of the equation, and zero on the right. To understand the proof draw a curve y = f(x) cutting the x axis at say x = 2 and sloping up to the right. Now we assume that our first guess x0 is at say x = 2.5. Draw a vertical line from the x axis at x = 2.5 (call this point N) to meet the curve at point P. Now draw the tangent to the curve at P and let this tangent meet the x axis at point x1 (call this point M). Then x1 is a better approximation to the root than x0. What we require now is how to calculate x1 from x0. Slope of the tangent f'(x0) = PN/MN = f(x0)/(x0-x1) x0 - x1 = f(x0)/f'(x0) x1 = x0 - f(x0)/f'(x0) This is the Newton-Raphson formula for getting a closer approximation to the required root. Having found x1, we then get an even better approximation from x2 = x1 - f(x1)/f'(x1) and this process continues for as long as you like depending on the accuracy you require in the answer. You can tell the accuracy by how how much successive answers differ from each other. If there is no change in the 5th place of decimals between successive approximations, then you are safe to assume your accuracy is at least 4 places of decimals. Example: Solve x^2 - sin(x) = 1 f(x) = x^2 - sin(x) - 1 = 0 f'(x) = 2x - cos(x) x^2-sin(x)-1 x1 = x0 - -------------- 2x-cos(x) Suppose we start at x0 = 1.35 then: -0.1532 x1 = 1.35 - ----------- = 1.4117 2.481 0.005657 x2 = 1.4117 - -------- = 1.409626 2.665 0.00000688 x3 = 1.409626 - ------------ = 1.409624 2.65878 and you could continue if you required even greater accuracy. In practice you carry out the iteration in your calculator, just keeping the last answer in memory, and there is no need to write down the intermediate values as I have shown above. When successive answers show no change you have the answer to whatever number of places your calculator can work. I find that there is no further change if we put x = 1.409624004 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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