1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $$y'(x)^2 = s'(x)^2 - 1\\ y'(x) = \sqrt {s'(x)^2 - 1}\\ y(x) = \int_0^x \sqrt {s'(s)^2 - 1} ds$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.