Question

What is the arclength of `f(x)=sqrt(x^2-1)/x` on `x in [-2,-1]`?

Answer 1

Approximately `1.48` units.

Explanation:

The arc length of a curve on `x in [a, b]` is given by

`A = int_a^b sqrt(1 + (dy/dx)^2)dx`

Taking the derivative, we get

`f'(x) = (x/sqrt(x^2 - 1)(x) - sqrt(x^2 - 1)(1))/x^2`

`f'(x) = (x^2/sqrt(x^2 - 1) - sqrt(x^2 - 1))/x^2`

`f'(x) = ((x^2 - x^2 + 1)/sqrt(x^2 - 1))/x^2`

`f'(x) = 1/(sqrt(x^2 - 1)x^2)`

So now applying the formula, we get:

`A = int_-2^-1 sqrt(1 + (1/(sqrt(x^2 - 1)x^2))^2) dx`

`A = int_-2^-1 sqrt(1 + 1/(x^6 - x^4)) dx`

`A = int_-2^-1 sqrt((x^6 - x^4 + 1)/(x^6 - x^4)) dx`

This integral doesn't have an elementary solution. According to Wolfram Alpha, the approximation for this integral would be `1.48` units.

Hopefully this helps!