Daedalus' Fifth Challenge

[Daedalus]

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

[Willa]

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...

 

 

[Daedalus]

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 : (

[CaptainPanic]

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.

[John Cuthber]

The area is r squared, but I haven't specified the units.

[CaptainPanic]

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.

[Daedalus]

I recently moved into a new apartment, and I finally have internet again. I'll give you guys another week on this one ; )

 

 

[Daedalus]

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