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Math Help - [SOLVED] Another Arc Lengths problem

  1. #1
    Senior Member mollymcf2009's Avatar
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    [SOLVED] Another Arc Lengths problem

    Find the arc length:

    y = 3 + \frac{1}{3} cosh(3x) for 0 \leq x \leq 3

    I got for my integral:

    \int\limits^{3}_{0} \sqrt{1 + sinh^2(3x)} dx

    = \int\limits^{3}_{0}1 + sinh(3x) dx


    What do I do from here? Should I substitute w/ double angle?
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  2. #2
    Senior Member mollymcf2009's Avatar
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    Ok, I figured out that I can just integrate this as is:

    \int\limits^{3}_{0} 1 + sinh(3x) dx

    = x + \frac{1}{3} cosh (3x) |^{3}_{0} **BUT THIS IS THE ORIGINAL PROBLEM?!?!?!?

    I got:

    \frac{8 + cosh(9)}{3}<br />
    Which is wrong. Any help is GREATLY appreciated!!
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  3. #3
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    mr fantastic's Avatar
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    Quote Originally Posted by mollymcf2009 View Post
    Find the arc length:

    y = 3 + \frac{1}{3} cosh(3x) for 0 \leq x \leq 3

    I got for my integral:

    \int\limits^{3}_{0} \sqrt{1 + sinh^2(3x)} dx

    = \int\limits^{3}_{0}1 + sinh(3x) dx


    What do I do from here? Should I substitute w/ double angle?
    Doesn't \cosh^2 A - \sinh^2 A = 1 ?

    In which case, won't 1 + \sinh^2 (3x) = \cosh^2 (3x) ?

    In which case won't your integral become \int \limits^{3}_{0} \cosh (3x) \, dx ?
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