Goat and SiloDate: 05/31/2001 at 11:35:32 From: David Lockhart Subject: Goat problem Our goat friend is now roped to a point A next to a silo. The rope is 15pi feet long. The diameter of the silo is 30 feet. How much area does the goat have to graze in? Date: 05/31/2001 at 14:32:17 From: Doctor Rick Subject: Re: Goat problem Hi, David. We have a similar problem in our Archives, but it doesn't do more than give some suggestions for thinking about it by starting with a square, then a hexagon, etc. Cow Grazing in Circles http://mathforum.org/dr.math/problems/cow.grazing.circles.html I would solve this problem with calculus. You didn't say what level of math you have, so I don't know if this is going over your head. I'll just say a little, and you can get back to me if you want more help. I think you're saying that point A is on the wall of the silo, that is, it is on the circle of radius R = 15 feet. If you draw the tangent to the circle at A, the goat can graze the semicircle (of radius L = 15pi) beyond this tangent with no interference from the silo. However, when the goat tries to go on the silo side of this tangent, it can't stretch quite as far because part of the rope is forced to curve along the wall of the silo. When the goat is at the end of his tether on the silo side of the tangent, the rope runs along the silo until some point B, then goes off in a straight line. The remainder of the rope lies on the tangent to the circle at B. The curve traced out by the end of the rope is called the involute of a circle. We don't have an area formula handy for the involute of a circle, but we can find the area using calculus. Consider the central angle between A and B; call this angle theta. You can calculate the length of the arc between A and B, and hence the remaining length of the rope. Let theta be your variable of integration for the area calculation. What infinitesimal area is swept out by the rope as theta increases to theta + d(theta)? What is the range of theta? Here are the details, in case you know calculus. Point A is on the wall of the silo, that is, it is on a circle of radius R feet. If you draw the tangent to the circle at A, the goat can graze a semicircle of radius L beyond this tangent with no interference from the silo. However, when the goat tries to go on the silo side of this tangent, it can't stretch quite as far because part of the rope is forced to curve along the wall of the silo. The curve traced out by the taut rope is called the involute of a circle. In consequence, the region in which the goat can graze has 3 parts: -------+----- ---- | -------- --- | \ / | \ / | \ / | \ / | \ / | \ | L| \ | | \ | | | | | | | | | | --+-- | | \ / \ | | \ / \| | \| R | | +-------+-------+A | /| \ th | | / \ \\ /| | / \ \\ / | | | ----- /* | | | / B | | | // | | | / / | | | / / | / | / / | / \/ / L| / dth\/ | / \ | / \ | / \ | / --- | / ---- | -------- -------+----- The area of the semicircle is (1/2)pi*L^2. The two regions bounded by the silo circle, the involute, and the tangent have equal areas; we will compute the area of one of them. When the goat is at the end of his tether on the silo side of the tangent, the rope runs along the silo until some point B, then goes off in a straight line. The remainder of the rope lies on the tangent to the circle at B. Consider the central angle between A and B; call this angle theta radians. The length of the arc between A and B is R*theta. Hence the remaining length of the rope is L - R*theta. The infinitesimal area swept out by the rope as theta increases to theta + d(theta) is that of a sector of a circle of radius (L - R*theta) and central angle d(theta): dA = (1/2)(L - R*theta)^2 d(theta) Of course, the infinitesimal region is not exactly a sector of a circle (because its "center" moves as theta increases), but the error is second order in d(theta). What is the range of the variable theta? When the rope is wrapped all the way against the silo so that the remaining length is zero, we have L - R*theta = 0 theta = L/R In the puzzle, the length of the rope has been chosen so that theta = pi radians (180 degrees) as shown in the figure: R = 15 ft, L = 15*pi. Let's consider only this case, and let L = R*pi dA = (1/2)R^2(pi-theta)^2 d(theta) Now we can integrate: pi INT (1/2)R^2(pi-theta)^2 d(theta) 0 Make a change of variables: x = pi-theta dx = -d(theta) Then the integral is 0 -(1/2)R^2 * INT x^2 dx pi 0 = -(1/2)R^2(1/3)x^3 ] pi = (1/6)R^2 * pi^3 The area of the semicircle is (1/2)pi*L^2 = (1/2)pi(pi*R)^2 = (1/2)pi^3*R^2 The total grazing area is (1/2)pi^3*R^2 + 2(1/6)pi^3*R^2 = (5/6)pi^3*R^2 If you're not familiar with basic integral calculus, then I can't suggest anything beyond the answer in the Archives. Whatever you do, I'd like to see your method and your results. Let me know if you have any difficulty. You might also browse the discussions of other grazing areas in the Dr. Math FAQ at: http://mathforum.org/dr.math/faq/faq.grazing.html - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/