Answer:
a) At x=14 the slope will be given by:
[tex]\frac{dy}{dx}(14)=a\sinh \left({\frac {14-C_{1}}{a}}\right)[/tex].
b) Then, the angle between the line and the pole will be:
[tex]\phi=\pi - \theta[/tex]
where [tex]\theta[/tex] is the angle between the tangent to the catenary and the x-axis.
Explanation:
The catenary has the following general form:
[tex]y(x)==a\cosh \left({\frac {x-C_{1}}{a}}\right)+C_{2}[/tex]
a) The slope at any point will be given by the derivative of y.
[tex]\frac{dy}{dx}(x)=a\sinh \left({\frac {x-C_{1}}{a}}\right)[/tex]
At x=14:
[tex]\frac{dy}{dx}(14)=a\sinh \left({\frac {14-C_{1}}{a}}\right)[/tex].
b) The angle between the tangent to the catenary and the x-axis at a given point will be given by:
[tex]\frac{dy}{dx}(x)=tan(\theta)[/tex] ⇒ [tex]\theta=tan^{-1} (\frac{dy}{dx}(x))[/tex]
Then, the angle between the line and the pole will be:
[tex]\phi=\pi - \theta[/tex].
