Question

A cable hangs between a street pole and a house that are 3 m apart. The height of the cable above the ground is approximated by the function f(x) = cosh(x) + 3, with the lowest point 1 m from the house and 2 m from the pole. Use the arc length integral to compute the length of the chain. You do not have to use the Riemann sum approach to construct the integral formula; just use it.

Answer 1

Answer:

The arc length integral takes complex values.

Step-by-step explanation:

This function $f(x) = cosh(x) + 3$ has a minimun in $x=0$.

Then we can estimate that the pole is 2 m at the left of $x=0$, in $x=-2$, and the house, 1 m at the rigth: $x=1$.

The arc length we have to calculate goes from $x=-2$ to $x=1$

The arc length integral equation is:

$S=\int\limits^a_b {\sqrt{1+f'(x)} \, dx$

that is derived from the Riemann's sum

$S=\sum\limits^n_{i=1} {\sqrt{1+\Delta y_i/\Delta x_i} \, \Delta x_i$

To compute $f'(x)$ we derive $f(x) = cosh(x) + 3$ :

$f'(x)=\frac{d(cosh(x)}{dx} +\frac{d(3)}{dx}=sinh(x)$

The function $\sqrt{1+sinh(x)$ 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.