Goat tied to a barnDate: Tue, 8 Nov 1994 15:10:02 -0500 (EST) Subject: Re: Dr. Math Question Hello there, Leonard! >If there is a goat tied to a rectangular barn on a 50 foot lead and the >barn is 20 feet by 20 feet (floor), what is the maximum grazing area? >Assume the goat is tied to a corner. - - Leonard Thompson - SouthWest Edgecombe H. S. - Pinetops, N. C. Great question. The way to approach this one is to divide the area into sections that you can deal with. You've probably already done that. In that case, you know where the tricky part is: on the opposite side of the barn, where the two semicircles overlap. So let's add up the area and subtract that overlap region. Picture (sort of): _________________ | | | | | | | Barn | | | | | | | _______|_______________| | | _ | \-_| This just isn't going to work. I can't do ASCII art worth beans. So I guess you'll just have to label your own picture. The straight parts of the edges of that overlap region have length 10, right? The best way I know of to find this area is to use an integral. What we need to do is find equations for the arcs of the circles which bound that region, and so that we can find the limits of integration. For simplicity, let's assume that the origin of the x-y plane is at the lower-left corner of the barn. So the arcs of the circles will have centers at the points (20, 0) and (0, 20), and they both have radius 30. So the two equations describing these arcs are (x-20)^2 + y^2 = 900 x^2 + (y-20)^2 = 900. Now to find the area of this region, we'll find the point where the two arcs intersect. We solve for y in the first equation to get y = Sqrt [900 - (x - 20)^2]. Then plug that in for y in the second equation to find x. I'll tell you something else: it's NOT easy, but you can do it with standard high school math tools, and I encourage you to press on. I get 10 - 5 Sqrt[14] as the x coordinate of the intersect point. The other root you'll get, 10 + 5 Sqrt[14], is another point of intersection that we're not interested in. Call the number 10 - 5 Sqrt[14] by the name k from here on. It'll simplify the notation. So the y coordinate of the point of intersection is the same: does that make sense from the picture? To set up the integral, we'll break up the region into two parts: to the left of the point of intersection and to the right. The area on the right is the negative of the integral from k to 0 of 20 - Sqrt[900 -x^2] dx. That's going to require a trig substitution to solve, I think. The area on the left is going to be the negative of the integral from -10 to k of Sqrt[900 - (x - 20)^2]. Also, this looks like a trig substitution. So once you evaluate these integrals and add them up, you'll have the area of the overlap region. Then you can subtract it! Enjoy, and let us know if there's anything else puzzling you, or if this doesn't make sense. -Ken "Dr." Math __________ Date: Tue, 8 Nov 1994 15:42:29 -0500 (EST) Subject: Re: Dr. Math Question Hey there! I just realized an easier way to find the point of intersection in your Barn and Goat problem. See, you can see from the picture that the two circles are going to meet on the line x = y, right? So you can just solve for the intersection of the two functions. x = y and (x - 20)^2 + y^2 = 900. Which is _SO_ much easier. You get the same answer too, which is nice. Then the rest of the problem follows the same way (the integrals part). Enjoy! -Ken "Dr." Math Date: Thu, 10 Nov 1994 09:50:25 -0500 (EST) From: Henry Oglesby Subject: goat on a wire Thank you for your help with the goat problem! I had not yet received my learning link access when the enigma landed on my desk so I used "leonard's" account! I found your second suggestion to be considerably less difficult! In summary, I would like to thank you for your help and I would hope that you will be able to assist me, or perhaps my class, on new problems! Your patient, Henry Oglesby Jr. Southwest Edgecombe High e-mail: oglesby@unctv.org Date: Mon, 14 Nov 1994 12:10:10 -0500 (EST) From: Dr. Ken Subject: Re: goat on a wire As Henry Oglesby wrote, >Doctor, > I have finished the problem and would like to compare solutions to >see if I may have made any errors. After using the arcsin substitution >for sqrt(u2-a2) I evaluated the integrals, combined my sectors, and came >up with approximately 7255 square feet of grazing space. Will you >review my file and write me a prescription if any doctoring is necessary. > >Your Patient, >Henry Oglesby asa Leonard Thompson Hey there! Glad to see you still care about the goat problem. I did the integrals, and I get (for the two tricky areas that you have to subtract) an area of 7.53197 for the left section, and an area of 83.3662 for the right section. These seem like plausible numbers, at least. So the total you subtract from 2325 Pi is going to be 90.8982, and I get a final answer of (drum roll......) 7213.3. Is that close enough to yours? Probably, if you used the approximation 3.141 for Pi. Anyway, thanks for the problem, and let us know if you need more help! -Ken "Dr." Math |
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