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Let $\alpha(s) = (x(s)),y(s))$ be a regular plane curve that is parameterized by arclength, and let $n(s)$ be the normal vector and $k(s)$ be the curvature of $\alpha$. Consider the family of curves:

$\beta(s,r) = \alpha(s) + rn(s), -\epsilon \leq r \leq \epsilon$

I'm asked to prove that, if $s_0$ and $r_0$ are constants, then $\beta(s,r_0)$ and $\beta(s_0,r)$ are regular curves for sufficiently small $\epsilon$, that they are orthogonal, and also asked to compute the curvature of $\beta(s, r_0)$. I'm a little unsure if what I have to prove they are regular is correct (see below), and I'm having a tough time with the last two (proving they are orthogonal and compute the curvature).

$\beta'(s, r_0) = \alpha'(s) + r_0n'(s) = \alpha'(s) - r_0k(s)\alpha'(s) = \alpha'(s)(1-r_0k(s)) \Rightarrow ||\beta'(s, r_0)|| = |1 - r_0k(s)|$

So the curve above is regular for $r_0 \neq \cfrac{1}{k(s)}$.

$\beta'(s_0, r) = (\alpha(s_0))' + n(s_0) = n(s_0)$ since I'm taking the derivative with respect to $r$ and $\alpha(s_0)$ will be a constant. $\Rightarrow||\beta'(s_0, r)|| = ||n(s_0)|| = 1$ so this curve is regular for all choices of $\epsilon$.

I tried taking the dot product $\beta'(s, r_0)\cdot \beta'(s_0, r) = \alpha'(s)n(s_0)(1-r_0k(s))$ and show it to be $0$, but I didn't get anywhere. I also don't see how to get the curvature of $\beta(s,r_0)$. Is there anything I did wrong here or something I'm missing that will help me? I'd appreciate any help/nudge in the right direction.

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