Question

What is the arclength of `f(x)=(x-2)/(x^2+3)` on `x in [-1,0]`?

Answer 1

`s ~~ 1.013` using technology.

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In cartesian,

`s = int_(a)^(b) sqrt(1 + ((dy)/(dx))^2)dx`

So, for `f(x) = (x-2)/(x^2 + 3)`:

`s = int_(-1)^(0) sqrt(1 + (d/(dx)[(x-2)/(x^2 + 3)])^2)dx`

The derivative of this function is:

`(dy)/(dx) = (x-2) cdot d/(dx)[1/(x^2 + 3)] + 1/(x^2 + 3) cdot cancel(d/(dx)[x-2])^(1)`

`= -(2x(x-2))/(x^2 + 3)^2 + 1/(x^2 + 3)`

`= (3 - 2x(x-2) + x^2)/(x^2 + 3)^2`

`= (-x^2 + 4x + 3)/(x^2 + 3)^2`

So, the square of it is:

`((dy)/(dx))^2 = ((-x^2 + 4x + 3)/(x^2 + 3)^2)^2`

Plugging it in:

`s = int_(0)^(-1) sqrt(1 + ((-x^2 + 4x + 3)/(x^2 + 3)^2)^2)dx`

... Nope, not doable with elementary functions.

So, we will just use Wolfram Alpha to obtain:

`color(blue)(s ~~ 1.013)`