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Math Help - Hyperbolic Trig

  1. #1
    Member VitaX's Avatar
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    Hyperbolic Trig

    The Gateway Arch to the West in St. Louis has the shape of an inverted catenary. Rising 630 ft at its center and stertching 630 ft across its base, the shape of the arch can be approximated by

    y=-127.7 cosh\left(\frac{x}{127.7}\right) + 757.7 for -315\le x \le 315

    Approximate the total length of the arch.

    Is this a problem involving derivatives of integrals because I'm kind of lost at this word problem. I'm thinking take the derivative with respect to x.
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  2. #2
    Member Black's Avatar
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    I believe you have to use

    L=\int_{-315}^{315}\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx.
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  3. #3
    Member VitaX's Avatar
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    Quote Originally Posted by Black View Post
    I believe you have to use

    L=\int_{-315}^{315}\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx.
    Ok I'll give it a go and see what I get. How did you come to the conclusion though that it's that formula I have to use?
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  4. #4
    Member VitaX's Avatar
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    y' = -127.7 \frac{dy}{dx} Sinh \left(\frac{x}{127.7}\right)\left(\frac{1}{127.7}\  right) \Longrightarrow -Sinh \left(\frac{x}{127.7}\right)

    \left(\frac{dy}{dx}\right)^2 = \left(-Sinh\left(\frac{x}{127.7}\right)\right)^2 \Longrightarrow Sinh^2 \left(\frac{x}{127.7}\right)

    Is this correct thus far? Seems like I missed something because when you substitute this into the formula, it's just a huge mess
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  5. #5
    Member Black's Avatar
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    Yes.

    When you put it in the equation (which is the arc length equation, by the way), you get

    L=\int_{-315}^{315}\sqrt{1+\text{sinh}^2\left(\frac{x}{127.  7}\right)}\,dx.

    Use the identity \text{cosh}^2(t)-\text{sinh}^2(t)=1 to get

    L=\int_{-315}^{315}\text{cosh}\left(\frac{x}{127.7}\right) dx ...
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  6. #6
    Member VitaX's Avatar
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    L=\int_{-315}^{315} Cosh \left(\frac{x}{127.7}\right) dx

    u=\frac{1}{127.7}x

    du = \frac{1}{127.7}dx \rightarrow 127.7du=dx

    L=127.7 \int Cosh(u) du \rightarrow L=127.7[Sinh(u)]

    From here is it just as simple as substituting in Sinh(x)=\frac{e^x-e^{-x}}{2}

    New limits being u = 2.4667 and u=-2.4667

    L=127.7\left[\frac{e^{2.4667}-e^{-2.4667}}{2}\right]

    L = 747 ft What I got. Do you sub Sinh(x)=\frac{e^x-e^{-x}}{2} before or after you take the integral?
    Last edited by VitaX; February 23rd 2010 at 10:36 AM.
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  7. #7
    Member VitaX's Avatar
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    Had some spare time so thought I'd try out what else I had in mind for solving this as I'm not sure entirely which is right. L=\int_{-315}^{315} Cosh\left(\frac{x}{127.7}\right)dx

    L=\int_{-315}^{315} \frac{e^{\frac{x}{127.7}}+e^{-\frac{x}{127.7}}}{2}dx

    L=\frac{1}{2} \int_{-315}^{315} e^{\frac{x}{127.7}} + e^{-\frac{x}{127.7}} dx

    L=\frac{1}{2}\left[127.7 e^{\frac{x}{127.7}} - 127.7e^{-\frac{x}{127.7}}\right]

    L=63.85 \left[\left(e^{2.4667} - e^{-2.4667}\right) - \left(e^{-2.4667}-e^{2.4667}\right)\right]

    L=1494 ft

    I'll wait for clarity from someone on here if this is the right approach for solving an integral with a hyperbolic function.
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