Question

Electrical wires suspended between two towers from a catenary as shown in the figure below are modeled by the equation

=10(ex/20+e−x/20),−20≤x≤20

Where x and y are measured in meters. Find the length of the suspended cable if the towers are 40 meters apart

Answer 1

Can you post the full question, please?

Answer 2

Where x and y are measured in meters. Find the length of the suspended cable if the towers are 40 meters apart\[y = 10(e \frac{ x }{ 20 } + e -\frac{ x }{ 20 } ) , -20\le x \le 20\]

Answer 3

i posted the rest

Answer 4

Judging by "catenary" I suspect you mean
\[y=10\left(e^{x/20}+e^{-x/20}\right)\]
right?

Answer 5

yes

Answer 6

Okay, so this is a run-of-the-mill arc length problem. I presume you have some background in integral calculus. The arc length \(L\) is given by
\[L=\int_a^b dS\]
where \(dS=\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}~dx\).

So,
\[L=\int_{-20}^{20}\sqrt{1+(y')^2} dx\]
Does this look familiar?

Answer 7

yes this looks familiar

Answer 8

Okay, so you have
\[y'=10\left(\frac{1}{20}e^{x/20}-\frac{1}{20}e^{-x/20}\right)=\frac{1}{2}\left(e^{x/20}-e^{-x/20}\right)\]
Squaring that gives
\[(y')^2=\frac{1}{4}\left(e^{x/20}-e^{-x/20}\right)^2=\frac{1}{4}\left(e^{x/10}-2+e^{-x/10}\right)\]
So the arc length is given by
\[\begin{align*}L&=\int_{-20}^{20}\sqrt{1+\frac{1}{4}\left(e^{x/10}-2+e^{-x/10}\right)} dx\\\\
&=\int_{-20}^{20}\sqrt{\frac{1}{2}+\frac{1}{4}\left(e^{x/10}+e^{-x/10}\right)} dx
\end{align*}\]
Looks ... fun.

Answer 9

Due to symmetry, you can write
\[L=2\int_{0}^{20}\sqrt{\frac{1}{2}+\frac{1}{4}\left(e^{x/10}+e^{-x/10}\right)}~dx\]

Answer 10

Maybe if you write the exponential portion as a hyperbolic trig function:
\[\frac{1}{4}\left(e^{x/10}+e^{-x/10}\right)=\frac{1}{2}\cosh\frac{x}{10}\]
Then
\[\begin{align*}L&=2\int_{0}^{20}\sqrt{\frac{1}{2}+\frac{1}{2}\cosh\frac{x}{10}}~dx\\
&=\sqrt2\int_{0}^{20}\sqrt{1+\cosh\frac{x}{10}}~dx
\end{align*}\]
and let's use a substitution to clean up the integrand a bit: \(u=\dfrac{x}{10}\), then \(10~du=dx\):
\[L=10\sqrt2\int_{0}^{2}\sqrt{1+\cosh u}~du\]

Answer 11

thank you soo much the first way is very helpful!

Answer 12

You're welcome. This integral's a bit tricky... Do you know your hyperbolic trig identities?

Answer 13

Oh hold on, it looks like there's a slightly easier way if you introduce the trig a bit earlier:
\[y'=\frac{1}{2}\left(e^{x/20}-e^{-x/20}\right)=\sinh\frac{x}{20}\]
Then
\[(y')^2=\sinh^2\frac{x}{20}\]
So the arc length is given by
\[\begin{align*}L&=2\int_{0}^{20}\sqrt{1+\sinh^2\frac{x}{20}}~dx\\\\
&=2\int_{0}^{20}\sqrt{\cosh^2\frac{x}{20}}~dx\\\\
&=2\int_{0}^{20}\cosh\frac{x}{20}~dx
\end{align*}\]
I hope I didn't make a careless mistake anywhere.

Answer 14

yes the first and third way is a lot similar to the way we would solve arc length in class

Answer 15

All that's left is computing the integral. I'll leave it to you, but I'll be around if you need help.

Answer 16

yes i think i got it from here so to finish would it be \[[2(10) \sinh \frac{ x }{ 20 }]_{0}^{20} = 20\sin (1) =\]

Answer 17

for the third way

Answer 18

Almost, you should have a factor of 20 where you have 10.
\[\int \cosh\frac{x}{a}~dx=a\sinh\frac{x}{a}+C\]
So in fact it would be \(40\sinh(1)\). Be sure you use \(\sinh\) and not \(\sin\) when computing.

Answer 19

Thank you! how would you complete the first way having some trouble , just trying to figure out both way of doing the problem