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

    Find $\displaystyle \lim_{\theta\rightarrow{0^{+}}}\frac{T({\theta})}{ S({\theta})}$
    Last edited by galactus; Nov 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 $\displaystyle T({\theta})$ is the area of the right triangle ABC, and $\displaystyle S({\theta})$ is the area of the segment formed by chord AB and the arc of the unit circle subtended by the central angle theta.

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

    in my opinion the limit must be zero:

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

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

    $\displaystyle
    \lim_{\theta~\rightarrow~0} \left( \frac{\frac{1}{2}(1-\cos(\theta)) \cdot \sin(\theta)}{ \frac{\theta}{360} \cdot \pi - \frac{1}{2}\sin(\theta) }\right)$ = $\displaystyle
    \lim_{\theta~\rightarrow~0} \left( \frac{180 \cdot (1-\cos(\theta)) \cdot \sin(\theta)}{ \theta \cdot \pi - 180 \cdot \sin(\theta) }\right)$ = $\displaystyle
    \lim_{\theta~\rightarrow~0} \left( \frac{180 \cdot (1-\cos(\theta))}{ \frac{\theta \cdot \pi}{\sin(\theta)} - 180 }\right)$

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

    $\displaystyle \lim_{\theta~\rightarrow~0}\left(\frac{\theta}{\si n(\theta)} \right)=1$ the numerator of the fraction will become zero and the denominator something ugly like $\displaystyle \pi - 180$ 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 $\displaystyle \frac{\theta}{360}$ maybe it should be $\displaystyle \frac{\theta}{2\pi}$.
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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 $\displaystyle \frac{\theta}{360}$ maybe it should be $\displaystyle \frac{\theta}{2\pi}$.
    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:

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

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

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

    Therefore, we have:

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

    You can use L'Hopital:

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

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

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

    L'Hopital again:

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

    L'Hopital again:

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

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