Answer:
The arc length integral takes complex values.
Step-by-step explanation:
This function [tex]f(x) = cosh(x) + 3[/tex] has a minimun in [tex]x=0[/tex].
Then we can estimate that the pole is 2 m at the left of [tex]x=0[/tex], in [tex]x=-2[/tex], and the house, 1 m at the rigth: [tex]x=1[/tex].
The arc length we have to calculate goes from [tex]x=-2[/tex] to [tex]x=1[/tex]
The arc length integral equation is:
[tex]S=\int\limits^a_b {\sqrt{1+f'(x)} \, dx[/tex]
that is derived from the Riemann's sum
[tex]S=\sum\limits^n_{i=1} {\sqrt{1+\Delta y_i/\Delta x_i} \, \Delta x_i[/tex]
To compute [tex]f'(x)[/tex] we derive [tex]f(x) = cosh(x) + 3[/tex] :
[tex]f'(x)=\frac{d(cosh(x)}{dx} +\frac{d(3)}{dx}=sinh(x)[/tex]
The function [tex]\sqrt{1+sinh(x)[/tex] in the range of x between x=-1 to x=0 takes complex values that prevent calculating the sum or the integral within the scope of the real values.
