# Function of curve length s in term of x

I am looking into a certain problem and decide to formulate it in a way that use the equation s=F(x) to describe a planar curve. Normally, we express the equation of a planar curve in the form y=f(x). We know that the arc length s at any point y is given by: $$s(x)=\int_{0}^{x}\sqrt{f'(x)^2+1}dx$$ so arc length s(x) is a function of x. I am interested in investigating different types of curves given by s=F(x) in general. What condition is necessary to ensure that the curve s=F(x) exists and is continuous? I know once we have s=F(x), we can convert it back to the form y=f(x) by the differential equation $$\left(\frac{dy}{dx}\right)^2=\left(\frac{ds}{dx}\right)^2-1$$ Is there any resource I can read on this?

Thanks

## 1 Answer

The (parameterized) planar curve is usually given by a map $$\gamma:[0,T] \rightarrow \mathbb{R}^2\\ \gamma(t)= (x(t), y(t))$$ and the arc-length is given by $$s(t) = \int_0^t \sqrt{[x'(t)]^2 + [y'(t)]^2} dt.$$

For example, any vertical line in $\mathbb{R}^2$ is impossible to be represented by $y=f(x)$.

When the curve is given by a graph of a function $f$, it is a special case of this by defining $x(t) = t, y(t) = f(t)$.

If you were given $s(x)$, and want to find $y= f(x)$, as you observed, we just need to solve the ODE $$y'(x)^2 = s'(x)^2 - 1\\ y'(x) = \sqrt {s'(x)^2 - 1}\\ y(x) = \int_0^x \sqrt {s'(s)^2 - 1} ds$$