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Math Help - Improper integrals/ techniques, urgent help needed.

  1. #1
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    Improper integrals/ techniques, urgent help needed.



    In the figure, A is at (1,0) and the segment BD is the height of OAB. Let R be the ratio of the area of DAB to that of the shaded region formed by deleting OAB from the circular sector subtended by angle theta. Find lim R theta-> 0+

    Ok so I know that this basically means find the derivative of R but I have no idea how to set up the problem and as soon as I know how to set it up I can solve it. Help please?
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  2. #2
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    Quote Originally Posted by wmm808 View Post


    In the figure, A is at (1,0) and the segment BD is the height of OAB. Let R be the ratio of the area of DAB to that of the shaded region formed by deleting OAB from the circular sector subtended by angle theta. Find lim R theta-> 0+

    Ok so I know that this basically means find the derivative of R but I have no idea how to set up the problem and as soon as I know how to set it up I can solve it. Help please?
    area of the sector = \frac{r^2 \theta}{2} = \frac{\theta}{2}

    area of triangle OBA = \frac{\sin{\theta}}{2}

    area of triangle OBD = \frac{\sin{\theta}\cos{\theta}}{2}


    R = \frac{\frac{\theta}{2} - \frac{\sin{\theta}}{2}}{\frac{\sin{\theta}}{2} - \frac{\sin{\theta}\cos{\theta}}{2}}

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

    I got \lim_{\theta \to 0^+} R(\theta) = \frac{1}{3}

    I had to use L'Hopital to get it ... maybe someone else can work it another way.
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    Quote Originally Posted by skeeter View Post
    area of the sector = \frac{r^2 \theta}{2} = \frac{\theta}{2}

    area of triangle OBA = \frac{\sin{\theta}}{2}

    area of triangle OBD = \frac{\sin{\theta}\cos{\theta}}{2}


    R = \frac{\frac{\theta}{2} - \frac{\sin{\theta}}{2}}{\frac{\sin{\theta}}{2} - \frac{\sin{\theta}\cos{\theta}}{2}}

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

    I got \lim_{\theta \to 0^+} R(\theta) = \frac{1}{3}

    I had to use L'Hopital to get it ... maybe someone else can work it another way.
    i think you got the ratio flipped. it should be the shaded region over the triangle while you did the triangle over the shaded region. the limit therefore should be 3 instead of 1/3.
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    By Taylor's method..

    \lim_{\theta\rightarrow{0}}\ \frac{\theta-Sin\theta}{Sin\theta-Sin\theta\ Cos\theta}=\lim_{\theta\rightarrow{0}}\ \frac{f(\theta)}{g(\theta)} = \frac{\frac{d}{dx}f(\theta)}{\frac{d}{dx}g(\theta)  }\ at\ \theta=0,\ =\frac{f'(0)}{g'(0)}

    \frac{f'(\theta)}{g'(\theta)}=\frac{1-Cos\theta}{Cos\theta(1-Cos\theta)+Sin\theta\ Sin\theta} =\frac{1-Cos\theta}{Cos\theta\ (1-Cos\theta)+1-Cos^2\theta}

    =\frac{1-Cos\theta}{Cos\theta(1-Cos\theta)+(1-Cos\theta)(1+Cos\theta)} =\frac{1}{Cos\theta+(1+Cos\theta)}\ evaluated\ at\ \theta=0

    giving\ \frac{1}{3}= the ratio of the shaded region to the triangle DBA
    Last edited by Archie Meade; January 31st 2010 at 07:58 AM.
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  5. #5
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    No,
    skeeter did it correctly.
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  6. #6
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    Quote Originally Posted by Archie Meade View Post
    No,
    skeeter did it correctly.
    Let R be the ratio of the area of DAB to that of the shaded region
    oblixps is correct, I flipped it.

    oh well ... fractional dyslexia strikes again.
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  7. #7
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    Eye strain!!
    i worked out the shaded to the triangle independently
    and saw you posted skeeter!
    so i didnt write it upside down but still got caught out!

    i thought you were being ticked off for implying the triangle
    was one third of the area of the shaded piece at the limit...
    Last edited by Archie Meade; January 31st 2010 at 08:11 AM.
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