SOLUTION TO PROBLEM 8

To simplify our considerations, consider the boundary of this region. Part is easy to determine - when the cow is pulling away from the silo, the boundary is a semi-circle of radius $\pi r$. This area is $\pi\cdot (\pi r)^2/2=r^2\cdot \pi^3/2$.

Let us turn our attention to the other part of the boundary. The region captured within this boundary may be broken up into two halves. Let us focus on the upper half. In this case, when the cow is at the end of its tether, part of the rope (length $r(\pi-\theta)$) will be against the silo while the other part of the rope (length $r\theta$) will be pulled taught. Note that the part of the rope that is pulled taut creates a line that is tangent to the circular boundary of the silo.

Remark on Using Parametric Equations:        Suppose that x(t), y(t) are the coordinates of a graph lying above the x-axis, then if the graph begins at t=a and ends at t=b, the integral (from substitution)

$$\int_{a}^b y(t)\cdot x'(t)\,dt$$

gives the area trapped below the curve (and above the x-axis).

To create equations parameterizing the graze-able region, begin with the rope wrapped tightly around the silo and the cow pressed up against the far side. Note that as the cow begins to back up and move away from the silo, an angle is formed between the point where the rope last contacts the silo, the center of the silo and where the cow started. Call this angle $\theta$. If the cow is pulling at the rope to keep it taut, the cow will have coordinates

$$x(\theta )=r(\cos (\theta)+\theta\cdot \sin (\theta)),\,\,\,y(t)=r(\sin (\theta)-\theta\cdot\cos(\theta)).$$

Therefore, including the area that is covered up by half of the silo, the upper-right region traced out by the cow has area

$$r^2\int^0_\pi (\sin (\theta)-\theta\cdot \cos(\theta))\cdot\theta\cdot\cos(\theta)\,d\theta.$$ (1)

To evaluate this integral, we need to show that

$$\begin{eqnarray*}\int x\cos(x)\sin(x)\, dx & = & \frac{1}{2}x\sin^2(x)-\frac{1}{... ...c{1}{4}x^2\sin(2x)- \frac{1}{8}\sin(2x)+\frac{1}{4}x\cos(2x)+C. \end{eqnarray*}$$

These can be done using standard techniques: double angle formulae and integration by parts. As a result, integral (1) has value $r^2(\frac{1}{2}\pi+\frac{1}{6}\pi^3)$. To get the area that is available for grazing, double this value, remove the area covered by the silo (namely $\pi r^2$) and add in the semi-circular region. The area available for grazing is therefore

$$\frac{5\pi^3r^2}{6}.$$

Questions and/or comments should be directed to Judy Downey or Griff Elder
 

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Last modified:   Wed Feb 28 19:43:01 CST 2001