Question

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

Answer 1

Arc length of `f(x)` is `sqrt2` unit.

Explanation:

`f(x) = (x^2-2x)/(2-x) ; x in [-2,-1]`

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

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

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

`f'(x) = ((2-x)(2x-2-x))/(2-x)^2` or

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

`:. [f'(x)]^2=1` . Total length of the arc from x = a to x = b is

`int_a^b sqrt(1+[f'(x)]^2)dx`

`L=int_(-2)^(-1) sqrt(1+1)dx` or

`L=int_(-2)^(-1) sqrt(2) dx` or

`L=sqrt2 [x]_-2^-1or L= sqrt2[-1-(-2)]` or

`L= sqrt2[-1+2] =sqrt2`

Arc length of `f(x)` is `sqrt2` unit. [Ans]