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Math Help - Calc I: Arc length

  1. #1
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    Calc I: Arc length

    I am having some trouble with finding the length of the curve
    ln(secx), 0≤ X ≤π/4 (the last limit is pi over 4, the pi doesnt look right)

    I am just not sure how to approach taking the derivative of this problem to later incorporate into the arc length forumla:

    Any help would be great, and please include a description.
    Thanks
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  2. #2
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    Hello, euchresucks!

    Find the length of the curve: . f(x) \:=\:\ln(\sec x),\qquad 0 \leq x \leq\frac{\pi}{4}
    Exactly where is your difficulty?


    We have: . f(x) \:=\:\ln(\sec x)

    Then: . f'(x) \;=\;\frac{1}{\sec x}(\sec x\tan x) \;=\;\tan x

    And: . 1 + \left[f'(x)\right]^2 \;=\;1 + \tan^2\!x \;=\;\sec^2\!x

    Hence: . \sqrt{1 + \left[f'(x)\right]^2} \;=\;\sqrt{\sec^2\!x} \;=\;\sec x


    Therefore: . S \;=\;\int^{\frac{\pi}{4}}_0 \sec x\,dx\quad \hdots\; Got it?

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  3. #3
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    Thanks so much, i feel like such a goon but you nailed my problem right on the head. I had it in my head that the derivative of secx was sec^2(x). I don't know why but thank you so much.
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