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Math Help - Arc Length question

  1. #1
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    Arc Length question

    It says...

    Find the length fo the graph and compare it to the straight-line distance between the endpoints of the graph.

    f(x)=ln[sex(x)] x E [0,(pi/4)]

    how would you do this?
    please help
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  2. #2
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    ln(sex)?. That's everyone's favorite integral. What is E?.

    Do you mean arc length of ln(sec(x)), \;\ [0,\frac{\pi}{4}]


    If so, \frac{d}{dx}[ln(sec(x))=tan(x)

    \int_{0}^{\frac{\pi}{4}}\sqrt{1+tan^{2}(x)}dx=\int  _{0}^{\frac{\pi}{4}}sec(x)dx
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  3. #3
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    oops. haha sorry about that. about the E, im not sure what it means. thats what it has in my book. it doesnt exactly look like an E though, its more curved than straight. it just says f(x)= ln[sec(x)], x E [0, (pi/4)]

    but i think you're right. though when you take the derivative of sec(x) it equals sec(x)tan(x) right? i dont quite understand how you got tanx from ln[sec(x)]
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  4. #4
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    The derivative of ln(sec(x))=tan(x).

    Chain rule:

    \frac{1}{sec(x)}\cdot\sec(x)tan(x)=tan(x)

    The E may be 'element of'?. Is this it, \in?.

    If so, they just mean that's the limits of integration.

    \int_{0}^{\frac{\pi}{4}}\sqrt{1+[f'(x)]^{2}}dx

    Which we found whittles down to something easy.

    Once you find your solution to the integral, compare it to:

    \sqrt{(\sqrt{2}-1)^{2}+(\frac{\pi}{4}-0)^{2}}
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  5. #5
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    yeah! thats the E. but that makes more sense now. thank you so much!!
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