Daedalus 329 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 0 Share this post Link to post Share on other sites

Willa 14 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... 0 Share this post Link to post Share on other sites

Daedalus 329 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 : ( 0 Share this post Link to post Share on other sites

CaptainPanic 1158 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 Share this post Link to post Share on other sites

John Cuthber 3668 Posted June 29, 2012 The area is r squared, but I haven't specified the units. 0 Share this post Link to post Share on other sites

CaptainPanic 1158 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. 0 Share this post Link to post Share on other sites

Daedalus 329 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 ; ) 0 Share this post Link to post Share on other sites

Daedalus 329 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 Share this post Link to post Share on other sites