I am solving a problem about the relationship between the curvatures of two parallel curves. Along the way, I encountered a problem which seems intuitively correct but failed to show it rigorously. Here are some information:
Let $\alpha(s)$, $s\in [0,l]$ be a closed convex plane curve positively oriented. The curve $$\beta(s)=\alpha(s)-rn(s)$$,where $r$ is a positive constant and $n$ is the normal vector, is called a parallel curve to $\alpha$.
Now I'm trying to reparametrise $\beta$ by arclength. Define $a(s)$ by
$$a(s) = \int_0^s ||\beta'(u)|| du,$$
Let $\hat{\beta}$ be defined by $\hat{\beta}(a(s)) = \beta(s)$, so that $\hat{\beta}$ is the reparameterization of $\beta$ by arc-length.
Now, here is my question: Is it correct that $\hat{\beta}'(a) = \alpha'(s)$? It seems intuitively correct that $\hat{\beta}'(a) = \alpha'(s)$ because the tangent vectors of two parallel curves point in the same direction for each corresponding points on the two curves, but how can we show it rigorously? I got stumbled because the left hand side is a derivative with respect to $a$, while the right hand side is a derivative with respect to $s$? So how can we explain $\hat{\beta}'(a) = \alpha'(s)$ rigorously?
Note: Apology for two similar looking $a$ and $\alpha$.
Thanks for the help!
Your equation doesn't make sense as it stands. You need to be writing $\hat\beta{}'(a(s))=\alpha'(s)$. You need to write down the chain rule, relating $\hat\beta{}'(a(s))$ and $\beta'(s)$, and then you should have it.