Arc Length of electrical cables

Question

A cable is hung between two towers that are 40 meters apart. The cable takes the shape of a catenary whose equation is $$y=10\left(e^{\frac{x}{20}}+e^\frac{-x}{20}\right)$$ $$-20\leq x\leq20$$

Find the arc length of the cable.

Using the equation below we first need to find $f'(x)$ $$\int_a^b \sqrt{1+\left(f'(x)\right)^2} dx$$

So finding $f'(x)$ results in $$f'(x)=\frac{1}{2}e^{\frac{x}{20}}-\frac{1}{2}e^{\frac{-x}{20}}$$

Plugging that into our equation and squaring it gives $$\int_{-20}^{20}\sqrt{1+\frac{{{e^\frac{x}{10}+e^\frac{-x}{10}}}}{4}}$$

This is where I am stuck because how am I meant to put all these terms into a squared form in order to cancel out the square root?

Answer 1

$$\sqrt{1+(f'(x))^2}=\sqrt{1+\frac{1}{4}\left(e^{x/10}-2+e^{-x/10}\right)}=\sqrt{\frac{1}{4}\left(e^{x/10}+2+e^{-x/10}\right)}=\frac{1}{2}\sqrt{(e^{x/20}+e^{-x/20})^2}=\frac{1}{2}\left(e^{x/20}+e^{-x/20}\right)$$

Answer 2

If $f(x)=10(e^{x/20}+e^{-x/20})$ then $f'(x)=\tfrac12(e^{x/20}-e^{-x/20})$ and then $$(f')^2(x)=\tfrac14 (e^{x/10}-2+e^{-x/10})=\tfrac14 (e^{x/10}+e^{-x/10})-\tfrac12$$