Daedalus Posted June 23, 2012 Share Posted June 23, 2012 (edited) This can be a fun problem to solve if you like calculus. There is a cow tethered by a rope to a circular barn on the north side of a fence which extends eastward radially from the barn. The rope and the fence has a length equal to the circumference of the barn. Find the grazing area as a function of the radius of the barn. Edited June 23, 2012 by Daedalus Link to comment Share on other sites More sharing options...
Willa Posted June 25, 2012 Share Posted June 25, 2012 I haven't done calculus in a long time, so I hope I don't do anything silly... Let r = the radius of the barn. Let R = the length of the rope. Since the rope wraps around the barn exactly one time, R = 2*pi*r. The eastern piece of the area is easy, since it's just a quarter circle with area (1/4) * pi * R^2 = (1/4) * pi * (2*pi*r)^2 = pi^3 * r^2. As for the rest, notice that the rope uncurls one point at a time as you move clockwise around the barn. Therefore, integrate along the length of the rope--x ranges from 0 to R. For a particular x, the distance from the center of the barn is sqrt(x^2 + r^2). Therefore the part of this distance outside the barn is sqrt(x^2 + r^2) - r. The answer, therefore is: integral{x from 0 to 2*pi*r} (sqrt(x^2 + r^2) - r) dx. According to my integral tables, this is something awful like pi * r^2 * [sqrt(pi^2 + 1) - 2] + .5 * r^2 * ln |2*pi + sqrt(4*pi^2 + 1)| so the total area would be pi^3 * r^2 + pi * r^2 * [sqrt(pi^2 + 1) - 2] + .5 * r^2 * ln |2*pi + sqrt(4*pi^2 + 1)| I'd be very surprised if nothing were wrong here though... Link to comment Share on other sites More sharing options...
Daedalus Posted June 28, 2012 Author Share Posted June 28, 2012 I haven't done calculus in a long time, so I hope I don't do anything silly... Let r = the radius of the barn. Let R = the length of the rope. Since the rope wraps around the barn exactly one time, R = 2*pi*r. The eastern piece of the area is easy, since it's just a quarter circle with area (1/4) * pi * R^2 = (1/4) * pi * (2*pi*r)^2 = pi^3 * r^2. As for the rest, notice that the rope uncurls one point at a time as you move clockwise around the barn. Therefore, integrate along the length of the rope--x ranges from 0 to R. For a particular x, the distance from the center of the barn is sqrt(x^2 + r^2). Therefore the part of this distance outside the barn is sqrt(x^2 + r^2) - r. The answer, therefore is: integral{x from 0 to 2*pi*r} (sqrt(x^2 + r^2) - r) dx. According to my integral tables, this is something awful like pi * r^2 * [sqrt(pi^2 + 1) - 2] + .5 * r^2 * ln |2*pi + sqrt(4*pi^2 + 1)| so the total area would be pi^3 * r^2 + pi * r^2 * [sqrt(pi^2 + 1) - 2] + .5 * r^2 * ln |2*pi + sqrt(4*pi^2 + 1)| I'd be very surprised if nothing were wrong here though... Nope that's not the answer : ( Link to comment Share on other sites More sharing options...
CaptainPanic Posted June 29, 2012 Share Posted June 29, 2012 From an engineering point of view, the grazing area is about 3 fence-lengths squared, but the rope is just too short. Also, that fence was put in a really silly place. 2 Link to comment Share on other sites More sharing options...
John Cuthber Posted June 29, 2012 Share Posted June 29, 2012 The area is r squared, but I haven't specified the units. Link to comment Share on other sites More sharing options...
CaptainPanic Posted June 29, 2012 Share Posted June 29, 2012 The area is 42*C, where C is a constant, which consists of about 1/3rd of everything we know in science, and about 2/3rd of Dark Units. Link to comment Share on other sites More sharing options...
Daedalus Posted July 1, 2012 Author Share Posted July 1, 2012 I recently moved into a new apartment, and I finally have internet again. I'll give you guys another week on this one ; ) Link to comment Share on other sites More sharing options...
Daedalus Posted August 11, 2012 Author Share Posted August 11, 2012 (edited) Since no one solved this challenge (or perhaps just didn't care lol), I will go ahead and post the answer. Define the grazing area in terms of the areas [math]A_g[/math] , [math]A_t[/math] , [math]A_c[/math] , and the area of the barn: As illustrated in figure 1, the grazing area can be divided into three regions. Figure 1 - A graph of the grazing area, plus the area of the barn, divided into three distinct areas - [math]A_g[/math] , [math]A_t[/math] , and [math]A_c[/math]. [math]A_g[/math] is the area defined by integrating the curve in polar coordinates from the angle [math]\alpha[/math] to [math]\beta[/math]: [math]A_g = \frac{1}{2} \int_{\alpha}^{\beta} \left | \text{R} \right |^2 \, d\theta \ \ \ \text{where R is a vector valued function.}[/math] [math]A_t[/math] is the area of the triangle excluded by integrating the curve in polar coordinates : [math]A_t = \frac{1}{2}\, r \left (2 \pi r\right) = \pi r^2[/math] [math]A_c[/math] is one-fourth the area of a circle with the radius being the full length of the rope: [math]A_c = \frac{1}{4}\, \pi \left(2 \pi r\right)^2 = \pi^3 r^2[/math] The grazing area is the sum of [math]A_g[/math] , [math]A_t[/math], and [math]A_c[/math] minus the area of the barn: [math]A® = A_g + A_t + A_c - \pi r^2[/math] Substituting the values of [math]A_g[/math] , [math]A_t[/math], and [math]A_c[/math] into the equation for the grazing area we get: [math]A® = A_g + A_t + A_c - \pi r^2 = \frac{1}{2} \int_{\alpha}^{\beta} \left | \text{R} \right |^2 \, d\theta + \pi^3 r^2[/math] (Note: [math]A_t[/math] and the area of the barn cancel out) Define the perimeter of [math]A_g[/math] as a function of ɸ: As illustrated in figure 2, the length of the rope wrapped around the barn is [math]r \phi[/math], and the length of the rope not touching the barn is [math]L \left(\phi\right) = 2 \pi r - r \phi = r \left(2 \pi - \phi \right)[/math]: Figure 2 - A graph illustrating the rope as it wraps around the barn. Since we know that the length of the rope wrapped around the barn is [math]r \phi[/math], we can define the position where the rope leaves the surface of the barn as a vector valued function of [math]\phi[/math]: [math]\text{P}\left( \phi \right) = \left \langle \, r \, \text{cos} \, \phi, \ r \, \text{sin} \, \phi \, \right \rangle[/math] The angle of the rope not touching the barn is equal to [math]\phi + \pi / 2[/math]. This allows us to define a vector valued function encapsulating the length and angle for this section of the rope as: [math]\text{V}\left( \phi \right) = \left \langle \, L\left(\phi \right) \, \text{cos} \left(\phi + \pi / 2\right), \ L\left(\phi \right) \, \text{sin} \left(\phi + \pi / 2\right) \, \right \rangle[/math] Now that we have defined [math]\text{P}\left( \phi \right)[/math] and [math]\text{V}\left( \phi \right)[/math] , we can define the perimeter of [math]A_g[/math] as follows: [math]\text{R}\left( \phi \right) = \text{P}\left( \phi \right) + \text{V}\left( \phi \right)[/math] Solve for [math]A_g[/math] and complete the equation for the grazing area: Transform [math]\text{R}\left( \phi \right)[/math] into polar coordinates so that we can integrate it as a polar function: [math]A_g = \frac{1}{2} \int_{\alpha}^{\beta} \left | \text{R} \right |^2 \, d\theta[/math] The square of the magnitude of [math]\text{R}\left( \phi \right)[/math] is the equal to the square of the length of the radius from the center of the barn to the perimeter of [math]A_g[/math]: [math]\left | \text{R} \right |^2 = r^2 \left(1+\left(2 \pi - \phi\right)^2\right)[/math] The angle of this radius with respect to the [math]x[/math] axis is the arctangent of the [math]y[/math] component of [math]\text{R}\left( \phi \right)[/math] divided by the [math]x[/math] component: [math]\theta = \text{tan}^{-1} \left( \frac{\left(2 \pi - \phi\right) \text{sin}\left(\phi + \pi / 2\right) + \text{sin} \, \phi}{\left(2 \pi - \phi\right) \text{cos}\left(\phi + \pi / 2\right) + \text{cos} \, \phi} \right)[/math] Solve for [math]d\theta[/math]: [math]d\theta = \frac{\left(2 \pi - \phi \right)^2}{1+\left(2 \pi - \phi\right)^2} \, d\phi[/math] Substitute [math]\left | \text{R} \right |^2[/math] and [math]d\theta[/math] into the equation for [math]A_g[/math] (note: since we changed the variable to [math]\phi[/math] , we must integrate from [math]0[/math] to [math]2 \pi[/math]: [math]A_g = \frac{1}{2} \int_{0}^{2 \pi} r^2 \left(1+\left(2 \pi - \phi\right)^2\right) \frac{\left(2 \pi - \phi \right)^2}{1+\left(2 \pi - \phi \right)^2}\, d\phi = \frac{1}{2} r^2 \int_{0}^{2 \pi} \left(2 \pi - \phi \right)^2 \, d\phi = \frac{4 \pi^3 r^2}{3}[/math] Substitute [math]A_g[/math] into the equation for the grazing area and simplify: [math]A® = A_g + \pi^3 r^2 = \frac{4 \pi^3 r^2}{3} + \frac{3 \pi^3 r^2}{3} = \frac{7 \pi^3 r^2}{3}[/math] Edited August 11, 2012 by Daedalus 1 Link to comment Share on other sites More sharing options...
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