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Math Help - triangle limit

  1. #1
    Eater of Worlds
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    triangle limit

    Here's a cool little limit problem you may wish to tackle.

    If T({\theta}) is the area of the right triangle ABC, and S({\theta}) is the area of the segment formed by chord AB and the arc of the unit circle subtended by the central angle theta.

    Find \lim_{\theta\rightarrow{0^{+}}}\frac{T({\theta})}{  S({\theta})}
    Last edited by galactus; November 24th 2008 at 05:39 AM.
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  2. #2
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    Quote Originally Posted by galactus View Post
    Here's a cool little limit problem you may wish to tackle.

    If T({\theta}) is the area of the right triangle ABC, and S({\theta}) is the area of the segment formed by chord AB and the arc of the unit circle subtended by the central angle theta.

    Find \lim_{\theta\rightarrow{0^{+}}}\frac{T({\theta})}{  S({\theta})}
    Hello,

    in my opinion the limit must be zero:

    Area of triangle ABC: a_{ABC} = \frac{1}{2}(1-\cos(\theta)) \cdot \sin(\theta)

    Area of the segment: a_{seg} = \frac{\theta}{360} \cdot \pi \cdot 1^2 - \frac{1}{2}\sin(\theta) \cdot 1^2

    <br />
\lim_{\theta~\rightarrow~0} \left( \frac{\frac{1}{2}(1-\cos(\theta)) \cdot \sin(\theta)}{ \frac{\theta}{360} \cdot \pi  - \frac{1}{2}\sin(\theta) }\right) = <br />
\lim_{\theta~\rightarrow~0} \left( \frac{180 \cdot (1-\cos(\theta)) \cdot \sin(\theta)}{ \theta \cdot \pi  - 180 \cdot \sin(\theta) }\right) = <br />
\lim_{\theta~\rightarrow~0} \left( \frac{180 \cdot (1-\cos(\theta))}{ \frac{\theta \cdot \pi}{\sin(\theta)}  - 180 }\right)

    Since \lim_{\theta~\rightarrow~0}(1-\cos(\theta) = 0 and

    \lim_{\theta~\rightarrow~0}\left(\frac{\theta}{\si  n(\theta)} \right)=1 the numerator of the fraction will become zero and the denominator something ugly like \pi - 180 but it will not be zero.

    Thus the complete fraction must be zero.
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  3. #3
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    Earboth, are you sure you used radians. You wrote \frac{\theta}{360} maybe it should be \frac{\theta}{2\pi}.
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    Earboth, are you sure you used radians. You wrote \frac{\theta}{360} maybe it should be \frac{\theta}{2\pi}.
    Hello,

    no, I used degree. In my opinion it doesn't matter whether you use radians or degree.
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  5. #5
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    Area of triangle:

    T({\theta})=\frac{1}{2}sin{\theta}(1-cos{\theta})

    Area of circular segment is given by \frac{1}{2}r^{2}({\theta}-sin{\theta}). Our radius = 1.

    S({\theta})=\frac{1}{2}({\theta}-sin{\theta})

    Therefore, we have:

    \lim_{\theta\rightarrow{0}^{+}}\frac{T({\theta})}{  S({\theta})}=\lim_{\theta\rightarrow{0}^{+}}\frac{  sin{\theta}-sin{\theta}cos{\theta}}{{\theta}-sin{\theta}}

    You can use L'Hopital:

    \lim_{\theta\rightarrow{0}^{+}}\frac{cos{\theta}+s  in^{2}{\theta}-cos^{2}{\theta}}{1-cos{\theta}}

    identity: sin^{2}{\theta}-cos^{2}{\theta}=-cos(2{\theta}):

    =\lim_{\theta\rightarrow{0}^{+}}\frac{cos{\theta}-cos({2\theta})}{1-cos{\theta}}

    L'Hopital again:

    =\lim_{\theta\rightarrow{0}^{+}}\frac{2sin(2{\thet  a})-sin{\theta}}{sin{\theta}}

    L'Hopital again:

    =\lim_{\theta\rightarrow{0}^{+}}\frac{4cos(2{\thet  a})-cos{\theta}}{cos{\theta}}=\boxed{3}

    Of course, you can manipulate it without L'Hopital. Whatever you want to do.
    Last edited by galactus; November 24th 2008 at 05:39 AM.
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