# Show that the tangent line of the evolute is the normal line to curve.

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!

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 $$\alpha(t)$$?
• 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! Oct 12, 2020 at 17:04