Hyperbolic Trig

**VitaX** · February 22nd 2010, 06:43 PM

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.

**Black** · February 22nd 2010, 06:52 PM

I believe you have to use

$L=\int_{-315}^{315}\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx.$

**VitaX** · February 22nd 2010, 07:14 PM

Originally Posted by Black

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?

**VitaX** · February 22nd 2010, 09:00 PM

$y' = -127.7 \frac{dy}{dx} Sinh \left(\frac{x}{127.7}\right)\left(\frac{1}{127.7}\ right) \Longrightarrow -Sinh \left(\frac{x}{127.7}\right)$

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

Is this correct thus far? Seems like I missed something because when you substitute this into the formula, it's just a huge mess

**Black** · February 22nd 2010, 09:26 PM

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

$L=\int_{-315}^{315}\text{cosh}\left(\frac{x}{127.7}\right) dx$ ...

**VitaX** · February 22nd 2010, 11:05 PM

$L=\int_{-315}^{315} Cosh \left(\frac{x}{127.7}\right) dx$

$u=\frac{1}{127.7}x$

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

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

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?

**VitaX** · February 23rd 2010, 10:38 AM

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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