Question

What is the arc length of `f(x)=x^2/sqrt(7-x^2)` on `x in [0,1]`?

Answer 1

The arc length of a function `f` on the interval `[a,b]` is given by:

`s=int_a^bsqrt(1+(f'(x))^2)dx`

Here, `f(x)=x^2/sqrt(7-x^2)` and I leave it to you to determine that `f'(x)=(x(14-x^2))/(7-x^2)^(3/2)`.

Then, the arc length desired is:

`s=int_0^1sqrt(1+(x^2(14-x^2)^2)/(7-x^2)^3)dx`

This has no closed form. Use a calculator to find:

`sapprox1.10458`

Answer 2

`approx 1.10458`

Explanation:

We have
`f(x)=x^2/sqrt(7-x^2)`
so
`f'(x)=(2x*sqrt(7-x^2)-x^2*1/2*(7-x^2)^(-1/2)(-2x))/(7-x^2)`
Multiplying numerator and denominator by
`sqrt(7-x^2)`
we get
`f'(x)=(2x(7-x^2)+x^3)/((7-x^2)*sqrt(7-x^2))`
so
`f'(x)=(14x-x^3)/((sqrt(7-x^2)(7-x^2))`
Using the Formula
`s=int_0^1sqrt(1+f'(x)^2)dx`
we get by a numerical method `approx 1.10458`