1. ## 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.

2. I believe you have to use

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

3. 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?

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

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

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

5. 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

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

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

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

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

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

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?

7. 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.