hyperbolic function

[feline]

I am struggling with this equation is cosh the same as arccos?

any help would be more than gratefully appreciated.

Q.
The sag, y, of an overhead line cable is assumed to be given by the expression:

y = Ccosh(x/C) where C is a constant and x is the distance from a pylon.

By using the exponential form of the hyperbolic function, given that C = 2.5 determine:

1) the value of y when x = 3.5
2) the value of x when y = 12.5

I have tried letting z=x/2.5 and sub into equation but Im getting nowhere, does exponential mean 1/cosh and if sp what identity is that?

Many thanks in advance.

 

[MarkFL]

The hyperbolic cosine function is not equivalent to the inverse cosine function.

The exponential form of the hyperbolic cosine function is:

\(\displaystyle \cosh(x)\equiv\dfrac{e^x+e^{-x}}{2}\)

 

[feline]

answer

Thank you for your reply
I have let z = x/2.5 and sub into the equation you gave me, I then multiplied the whole equation by 2.5 but I cant get any further when I need to find x I can find y if I have done it right.

I have 2.5 times the equation given below having e to the power of x/2.5 and -x/2.5 but I dont
know how to reverse, do i need to take logs?

I really want to solve this but am struggling, any help would be great.

 

[MarkFL]

With \(\displaystyle C=2.5=\dfrac{5}{2}\) we then have:

\(\displaystyle y(x)=\dfrac{5}{4}\left(e^{\dfrac{2x}{5}}+e^{-\dfrac{2x}{5}} \right)\)

1.) Evaluate:

\(\displaystyle y\left(\dfrac{7}{2} \right)=\dfrac{5}{4}\left(e^{\dfrac{7}{5}}+e^{-\dfrac{7}{5}} \right)\)

2.) We will need to solve for x, and that will take some algebra:

\(\displaystyle \dfrac{4}{5}y=e^{\dfrac{2x}{5}}+e^{-\dfrac{2x}{5}}\)

Square both sides:

\(\displaystyle \dfrac{16}{25}y^2=e^{\dfrac{4x}{5}}+2+e^{-\dfrac{4x}{5}}\)

Subtract 4 from both sides:

\(\displaystyle \dfrac{16}{25}y^2-4=e^{\dfrac{4x}{5}}-2+e^{-\dfrac{4x}{5}}\)

\(\displaystyle \dfrac{16y^2-100}{25}=\left(e^{\dfrac{2x}{5}}-e^{-\dfrac{2x}{5}} \right)^2\)

\(\displaystyle \dfrac{4(4y^2-25)}{25}=\left(e^{-\dfrac{2x}{5}}-e^{\dfrac{2x}{5}} \right)^2\)

Take the positive root:

\(\displaystyle \dfrac{2\sqrt{4y^2-25}}{5}=e^{-\dfrac{2x}{5}}-e^{\dfrac{2x}{5}}\)

Add \(\displaystyle 2e^{\dfrac{2x}{5}}\) to both sides:

\(\displaystyle \dfrac{2\sqrt{4y^2-25}}{5}+2e^{\dfrac{2x}{5}}=e^{-\dfrac{2x}{5}}+e^{\dfrac{2x}{5}}\)

\(\displaystyle \dfrac{2\sqrt{4y^2-25}}{5}+2e^{\dfrac{2x}{5}}=e^{\dfrac{2x}{5}}+e^{-\dfrac{2x}{5}}\)

Recall \(\displaystyle \dfrac{4}{5}y=e^{\dfrac{2x}{5}}+e^{-\dfrac{2x}{5}}\) hence:

\(\displaystyle \dfrac{2\sqrt{4y^2-25}}{5}+2e^{\dfrac{2x}{5}}=\dfrac{4}{5}y\)

\(\displaystyle 2e^{\dfrac{2x}{5}}=\dfrac{4}{5}y-\dfrac{2\sqrt{4y^2-25}}{5}\)

\(\displaystyle 2e^{\dfrac{2x}{5}}=\dfrac{4y-2\sqrt{4y^2-25}}{5}\)

\(\displaystyle e^{\dfrac{2x}{5}}=\dfrac{2y-\sqrt{4y^2-25}}{5}\)

Convert from exponential to logarithmic form:

\(\displaystyle \dfrac{2x}{5}=\ln\left(\dfrac{2y-\sqrt{4y^2-25}}{5} \right)\)

\(\displaystyle x(y)=\dfrac{5}{2}\ln\left(\dfrac{2y-\sqrt{4y^2-25}}{5} \right)\)

Now evaluate:

\(\displaystyle x\left(\dfrac{25}{2} \right)=\dfrac{5}{2}\ln\left(5-2\sqrt{6} \right)\)

 

[feline]

thank you

thank you so much for what you have provided I will look through this and hoepfully will be able to get answer out from here.
Thanks again, really appreciated.