Arc length of the Inverse Function

Question

I am looking to prove that the arc length of the inverse function $f^{-1}(x)$ on the interval $[f(0),f(a)]$ is the same as the arc length of the function $f(x)$ on the interval $[0,a]$. One may see that this is true visually, as it comes from flipping the function across the line $y=x$, which does not change its length, but I want to show it analytically. Start with the arc length formula, $$\int_{f(0)}^{f(a)} \sqrt{1+\left(\frac{d}{dx}\left[f^{-1}(x)\right]\right)^2}dx.$$ Use the general form for the derivative of the inverse function and simplify to the following: $$\int_{f(0)}^{f(a)}\frac{\sqrt{1+f'(x)^2}}{f'(x)}\,dx.$$ The Substitution $y=f^{-1}(x)\,\Rightarrow dy = dx/f'(x)$ then gets very close, $$\int_0^a \sqrt{1+f'(f(y))^2}\,dy,$$ but this is not the arc length formula I am after. What I want is $$\int_0^a \sqrt{1+f'(y)^2}\,dy.$$ I have been thinking about this for longer than I care to admit, and haven't figured out what went wrong. Any help is very much appreciated.

Answer 1

It helps to use the lettering

$$L = \int_{\min\{f(0),f(a)\}}^{\max\{f(0),f(a)\}}\sqrt{1+[(f^{-1})'(y)]^2}\:dy$$

and we'll use the same substitution $y = f(x)$:

$$dy \to |f'(x)|\:dx$$

and

$$(f^{-1})'(f(x)) = \frac{1}{f'(x)} $$

gives us

$$L = \int_0^a \sqrt{1+\frac{1}{[f'(x)]^2}}\cdot |f'(x)|\:dx = \int_0^a\sqrt{[f'(x)]^2+1}\:dx$$

we take the absolute value because that is the formula for the Jacobian, which lets us blindly take the absolute value and be assured that the orientation for the bounds is always in the same direction before and after the change of variables (either increasing or decreasing for both).