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Simplifying a arc length problem

  1. Apr 30, 2008 #1
    Simplifying an arc length problem

    I have L= Int(-2..2) sqrt(16*cosh(4*t)^2+9*sinh(4*t)^2+9)

    and can use Maple to simplify this to sqrt(25*cosh(4*t)^2)
    but I just can't see how that is done. (or how to get maple to show me the steps!)

    Can anyone help by showing the steps, including any trig substitutions/identities used?
    Last edited: Apr 30, 2008
  2. jcsd
  3. Apr 30, 2008 #2
    try the definition of cosh and sinh in terms of exponentials (or use that to derive the hyperbolic equivalent of sin^2 + cos^2 = 1)
  4. May 1, 2008 #3

    Gib Z

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    Homework Helper

    [tex] 1+ \sinh^2 x = .....[/tex] so [tex] 9 + 9 \sinh^2 (4t) = ..................[/tex]?
  5. May 1, 2008 #4
    ah! thanks - i can't believe I didn't see that - it's only one step! I often have problems seeing the stripped down identities and applying them to problems which have other factors (eg 9 and 4t etc).
    Thanks for the hint
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