I am trying to solve the following exercise:
Let $\alpha(s) = (x(s), y(s))$ be a regular curve parametrized by arc lenght.
Let $\beta(s) = \alpha(s) + \frac{1}{k(s)} \cdot n(s) $, where $k(s)$ is the curvature of $\alpha(s)$ and $n(s) = (-y(s), x(s))$ is the normal vector function.
(a) Prove that $\beta$ is differentiable if $k(s) \neq 0 \;\; \forall s$. [I've done that]
(b) Prove that, if $k(s) \neq 0 \;\; \forall s$, then $\beta$ is regular if $k'(s) \neq 0 \;\; \forall s$. [I'm stuck here].
- A curve $\alpha(s)$ is regular if $\lvert \alpha(s) \rvert \neq 0 \;\; \forall s$.
- The curvature of a curve $\alpha(s) = (x(s), y(s))$ parametized by arc length is $k(s) = \langle t'(s), n(s)\rangle$, where $t'(s) = (x''(s), y''(s))$.
What I have already done:
I have derived $\beta$ and found that
$$\beta'(s) = \alpha'(s) + \frac{k(s)}{k^2(s)} \cdot n'(s) – \frac{k'(s)}{k^2(s)} \cdot n(s).$$
I have tried proving the contrapositive, by supposing the existence of a number $s$ such that $\lvert \beta'(s) \rvert = 0$. This led me to $\beta'(s) = (0,0)$, and so
$$ \left(x' – \frac{k}{k^2}\cdot y'' + \frac{k'}{k^2}\cdot y' \;,\; y' + \frac{k}{k^2}\cdot x'' – \frac{k'}{k^2}\cdot x' \right) = (0,0). $$
Finally, I multiplied by $k^2$ to obtain two equalities:
$$ k^2 x' – k y'' + k'y' = 0 $$
$$ k^2 y' – k x'' + k'x' = 0. $$
Now I just can't seem to find a contradiction.
Any help is appreciated. Thank you in advance.
Best Answer
Hint. Try to simplify the first two terms of $$ \beta = \alpha'(s) + \frac{k(s)}{k^2(s)} n'(s) - \frac{k'(s)}{k^2(s)} n(s). $$ The curve is parametrized by arc length so $\alpha'(s)$ is equal to what vector? And use the Frenet-Serret formula to write $n'(s)$ in terms of the Frenet frame.