Goat Tied by a 10-Meter RopeDate: 11/28/2001 at 06:24:24 From: Sadia Subject: Calculus and integration 1) A goat is tied to the corner of a 5-by-4-meter shed by a 10-meter piece of rope. What is the area grazed by the goat? 2) The shed is now a circle with radius r, and the rope is 2r. What is the area grazed now? I first tried using the formula for the area of a sector, but this created further questions. My college friends say the answer lies in the formula for the equation of a circle, but I have not studied calculus and the books I have don't even seem to be in English. The second part is about integration, but again I have no previous knowledge of calculus at all. I am not after an answer but guidance, as I can easily get the answer from a friend, but I want to know how it's done for my own benefit. Date: 11/28/2001 at 15:36:38 From: Doctor Rob Subject: Re: Calculus and integration Thanks for writing to Ask Dr. Math, Sadia. No calculus is required, just some analytic geometry. Draw the following diagram: __,..+...__ ,-' `-. ,-' `-. ,' `. ,' `. ,' `. / \ / \ | | | 5 B 5 A | E o---------o---------o + | /| | | \ 5/ |4 |4 | \ / C| 5 | / \ /_.-o-----__--oD / `o'---'''''' | ,' G \ 6 | ,' \ |6 ,' '._ | ,-' `-._ | _,-' ``--o--'' F ABCD is the rectangular shed, and the rope is tied at A. The area the goat can graze consists of the 3/4 of the circle of radius 10 with center A whose boundary is EBADFE, plus a sector of a circle of radius 5 and center B whose boundary is EBGE, plus a sector of a circle of radius 6 and center D whose boundary is FDGF, plus the two triangles BCG and DCG. The tricky part is to determine the length of the segment CG. Let d be the length of CG. By using the Law of Cosines on the two triangles, one gets (d^2+4^2-5^2)/(2*4*d) = cos(<BCG), (d^2+5^2-6^2)/(2*5*d) = cos(<DCG) = cos(3*Pi/2-<BCG) = -sin(<BCG). Now square both expressions and add then, and, since sin^2(<BCG) + cos^2(<BCG) = 1, you get (d^2+4^2-5^2)^2/(2*4*d)^2 + (d^2+5^2-6^2)^2/(2*5*d)^2 = 1. When fractions are cleared and the expressions are simplified, this becomes 41*d^4 - 2402*d^2 + 3961 = 0. Solving this, and rejecting extraneous roots, one finds that d = sqrt([1201-800*sqrt(2)]/41), or approximately 1.303177637 meters. Using Heron's Formula for the area of a triangle when you know all three sides gives the areas of the two triangles, which are (160*sqrt[2]-150)/41 and (250*sqrt[2]-260)/41 square meters. Using the Law of Cosines to compute <GDC and <GBC gives <GDF = Pi/2 - <GDC, = arcsin([65+40*sqrt(2)]/123) = 1.418084117... radians, <GBE = Pi/2 - <GBC, = arcsin([12+20*sqrt(2)]/41) = 1.383671627... radians. Of course the angle <EAF at the center of the 3/4 circle with radius 10 is <EAF = 3*Pi/2 radians. Now the sector area formula K = (1/2)*r^2*theta, where r is the radius and theta the central angle, will give you the areas of the three circle sectors. Add these to the areas of the two triangles, and you'll have the total graze area. Filling in the details and the rest I'll leave to you. A problem rather simple to state ended up using some fairly complicated math, and the answer turned out to be quite messy! Feel free to write again if I can help further. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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