I'm trying to solve the following problem from do Carmo's book on Differential Geometry

Let $\alpha:I\to \mathbb{R}^2$ be a regular parametrized plain curve (arbitrary parameter), define $n=n(t)$ and $k=k(t)$, where $k$ is the signed curvature. Assume that $k(t)\neq 0,t\in I$. In this situation, the curve $$\beta(t)=\alpha(t)+\frac{1}{k(t)}n(t),\quad t\in I,$$ is called the evolute of $\alpha$. Show that the tangent at $t$ of the evolute of $\alpha$ is the normal to $\alpha$ at $t$.

I found this answer that gives a proof of $\beta'(t) \cdot \alpha'(s) =0$, which then proves that the the tangent at $t$ of the evolute of $\alpha$ is orthogonal to $\alpha$ at $t$. However, it seems to me that the problem is asking for a proof that the tangent of the evolute is precisely the line that is normal to $\alpha$ at $t$, and not just that the lines are at right angles with one another. I believe (if I understand the problem statement correctly) that to conclude the proof you would need to show that $$ \beta(t) + \beta'(t)\lambda = \alpha(t) $$ for some scalar value of $\lambda$, since this would guarantee that the line $\beta(t) + \beta'(t)\lambda$ crosses $\alpha(t)$ at exactly the point where the normal line also passes through $\alpha(t)$.

I couldn't figure out a way to prove the latter condition using the $\beta'(t) \cdot \alpha'(s) =0$, and in fact, I'm not even sure if this is a good way to approach this problem. I tried explicitly substituting the expressions for $\beta(s)$ and $\beta'(s)$ found in the answer I linked at the beginning and go that the equation I want to prove is equivalent to $$ n(t) = \lambda\left(\alpha'(t)k(t)-\frac{k'(t)}{k(t)}n(t)+n'(t)\right) $$ I believe that from here it just suffices to show that both left and right sides vectors are in the same direction since then we can guarantee that the scalar $\lambda$ exists, but I didn't manage to find a way to do this.

Could anyone tell me if I'm on the right track to finishing this problem's solution? Or alternatively, does anyone know a better way in which I can show that the tangent at $t$ of the evolute of $\alpha$ is exactly the normal to $\alpha$ at $t$? Thank you!


1 Answer 1


May I suggest a simpler approach?

Hint: For a given $t\in I$, the normal line of $\alpha$ at $\alpha(t)$ is given by $$ \lambda \mapsto \alpha(t) + \lambda n(t). $$ Compare this with the definition of the evolute of $\alpha$. Which point of $\beta$ lies on the normal line of $\alpha$ through $\alpha(t)$?

  • 1
    $\begingroup$ Ohhhh, of course! For $\lambda = r = \frac{1}{k}$ the normal line hits the center of the osculating circle, and clearly $\beta(t) +(0)\beta'(t)$ is on the tangent line to the evolute, so since they're parallel and pass through the same point they're equal. Thank you! $\endgroup$
    – Robert Lee
    Oct 12, 2020 at 17:04

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