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Thread: grazing goat

  1. #1
    Eater of Worlds
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    grazing goat

    This problem may be a little cliche, but I will post it anyway for those who have not seen it.

    A goat is tied to a silo of radius 12 feet by a tether 36 feet long. Calculate the area the goat can graze.

    Here is a hackneyed diagram. It kind of resembles a limacon.
    Last edited by galactus; Nov 24th 2008 at 05:38 AM.
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  2. #2
    Behold, the power of SARDINES!
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    If I remember correctly it is a cardioid.

    I think Stuart called it
    "Clara the calculus cow"

    This problem made me crazy for about a week.

    A cool more advanced way to solve this is to use Greene's thorem.
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  3. #3
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    The length of the tether from where the goat is tied to the end of its rope is $\displaystyle 36-12{\theta}$.

    That's because the rope wraps around the silo before it becomes tangent to the silo at that point.

    The total angle the goat grazes is 3 because $\displaystyle 12{\theta}=36$

    There is a small piece it can not reach because the semi-diamter of the silo is slightly longer than 36 feet. Therefore, we don't have to integrate to Pi, but slightly less than that.

    Limits of integration are $\displaystyle {\theta}=[0,3]$.

    The semicircular region is easy $\displaystyle \frac{36^{2}\pi}{2}=648{\pi}$

    The region we are interested in is the second and fourth quadrants where the goat goes around the silo.

    $\displaystyle 2\int_{0}^{3}\frac{1}{2}(36-2{\theta})^{2}d{\theta}=1296$

    So, the total area is $\displaystyle 648{\pi}=1296=3331.75 \;\ ft^{2}$

    I'd like to see the Green's theorem method. I will look into it.
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  4. #4
    Behold, the power of SARDINES!
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    I will work a variation of the problem.

    Suppose that a cow is tied to a silo of radius r with a rope of length $\displaystyle \pi r$

    See the diagram below

    grazing goat-silo.jpg

    using trig and the fact that the rope is always at a right angle to the silo we can find $\displaystyle d_1,d_2,d_3,d_4$ and parametric equations of the curve.

    Note the length of the hypotenuse of the small triangle is $\displaystyle r\theta$
    becuase it have the same length as the arc of the circle subtended by the angle theta.


    $\displaystyle d_1=r\cos(\theta)$
    $\displaystyle d_1=r\theta\sin(\theta)$

    so x as a function of theta is $\displaystyle x=d_1+d_2$ so

    $\displaystyle x(\theta)=r(\cos(\theta+\theta\sin(\theta))$

    using a similar argument for y we get

    $\displaystyle
    y(\theta)=r(\sin(\theta)-\theta\cos(\theta))
    $

    if we graph this from $\displaystyle -\pi \mbox{ to } \pi$ we get
    Note for the graph I will use r=12
    grazing goat-graph1.jpg

    Now for the other half of the equation we want the half circle
    centered at (-r,0) with a radius of $\displaystyle \pi r$

    $\displaystyle x(\theta)=-r +\pi r \cos(theta)$
    $\displaystyle y(\theta)=\pi r \sin(theta)$

    graphed in green theta $\displaystyle \frac{\pi}{2}\mbox{ to } \frac{3\pi}{2}$

    grazing goat-graph2.jpg

    These two curves enclose the total area for the cow, but this is too much area we need to remove the silo.

    grazing goat-graph3.jpg

    lets add one more line
    grazing goat-graph4.jpg

    let $\displaystyle c_1$ be the blue curve
    let $\displaystyle c_2$ be the purple curve
    let $\displaystyle c_3$ be the green curve

    Now for Green's theorem

    $\displaystyle \iint_A1dA=-\int_{\partial A}ydx$

    $\displaystyle -\int_{c_1 \cup c_2}ydx=-\int_{c_1}ydx-\int_{c_2}ydx $

    but dx=0 on the purple curve so we get...

    $\displaystyle -\int_{-pi}^{pi}r(\sin(\theta)-\theta\cos(\theta))r\theta\cos(theta)d\theta=\frac {1}{3}r^2\pi^3+r^2\pi$

    The area of the other half circle is $\displaystyle \frac{1}{2}r^2\pi^3$

    Then we subtact off the area of the silo to get

    $\displaystyle A=\frac{5}{6}\pi^3r^2$

    whew that was longer than I remember.
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