# Tangent of evolute and singed curvature

This is an exercise from differential geometry textbook by Do Carmo.

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 $t$.

My professor uses another book as our textbook. In that book, we only define positive curvature for curve. Hence I am a little confused about the sign. I can show the tangent of evolute is $\frac{(\frac{1}{k})^{'}}{| (\frac{1}{k})^{'} |}n(t)$, but I don't know how to cancel the absolute value.

You can restrict yourself to an arc length parametrization of the curve $\alpha$ (ask for details if it is not clear why this is!).

Then compute $$\beta'(t)=\alpha'(t)-\frac{k'(t)}{k(t)^2}n(t)+\frac{1}{k(t)}n'(t).$$ To answer your question you need to show $\langle \beta'(t),\alpha'(t)\rangle=0$. This follows directly if you use $\langle \alpha''(t),n(t)\rangle =- \langle \alpha'(t),n'(t)\rangle$ which follows from differentiating $\langle \alpha'(t),n(t)\rangle =-0$ with respect to $t$. Then use $\alpha''(t)=k(t)n(t)$.

EDIT: What do you mean by "the normal"? EDIT: Added missing sign.

EDIT: I obtain in my way $\beta'(t)=-\frac{k'}{k^2}n(t).$

• Why you say $\beta'(t)=\alpha'(t)-\frac{k'(t)}{k(t)^2}n(t)+\frac{1}{k(t)}n'(t)$? The right thing is not $\beta'(t)=\alpha'(t)+\frac{k'(t)}{k(t)^2}n(t)-\frac{1}{k(t)}n'(t)$? Nov 15, 2015 at 17:17
• I get the minus from the derivative of the term $k^{-1}$ and use the product rule otherwise. Why do you think is there a minus in front of the last term? It's just the derivative of $n$ times $k^{-1}$ and there is no minus sign.
– frog
Nov 16, 2015 at 18:37
• Why do we have to show it is perpendicular to tangent, even binormal is perpendicular to tangent, right? Sep 18, 2017 at 20:08
• The normal bundle of the curve has a 2-dimensional fibre (in our situation it Is spanned by the curvature vector and the binormal vector), hence you either show that the tangent of the evolute is a linear combination of the curvature and binormal vector, or (what is easier) you show that it is perpendicular to the tangent of the curve...
– frog
Sep 19, 2017 at 7:44

As the previous answer told, you can suppose that $$\alpha$$ is unit-speed (equivalently parametrized by arc-length), so we can change $$t$$ by $$s$$. The unit tangent vector of $$\alpha$$, $$\mathbf{t}$$, can be written as $$\alpha'(s) = \mathbf{t}(s) = (\cos \varphi(s), \sin \varphi(s)),$$ where $$\varphi(s)$$ is a smooth function. The unit normal vector $$\mathbf{n}(s)$$ is the result of rotating the unit tangent vector by $$\pi/2$$ counterclockwise, then $$\mathbf{n}(s) = (- \sin \varphi(s), \cos \varphi(s))$$ therefore, $$\frac{d\mathbf{n}}{ds} = - \varphi'(s) \mathbf{t}(s) = -k_{s}(s) \mathbf{t}(s) \qquad (1).$$ The derivative $$\varphi'(s)$$ is the same signed curvature, that is to say, $$\varphi'(s) = k_{s} (s)$$ (the subscript $$s$$ here stands for the signed curvature).

Using the equation (1) you can see that \begin{align*} \frac{d\beta}{ds} &= \mathbf{t}(s) + \left(\frac{-k'_{s}(s)}{[k_{s}(s)]^{2}}\right)\mathbf{n}(s) + \frac{1}{k_{s}(s)}\frac{d\mathbf{n}}{ds}\\ &= \mathbf{t}(s) + \left(\frac{-k_{s}'(s)}{[k_{s}(s)]^{2}}\right)\mathbf{n}(s) + \frac{1}{k_{s}(s)}(-k_{s}(s) \mathbf{t}(s)) \\& = \left(\frac{-k'_{s}(s)}{[k_{s}(s)]^{2}}\right) \mathbf{n}(s) \end{align*}

Since $$\frac{d\beta}{ds}$$ is a tangent vector to the evolute $$\beta$$, note that this vector is parallel to the normal vector of $$\alpha$$, so, the normal line of $$\alpha$$ at $$s$$ contains the point $$\beta(s)$$ and is tangent to $$\beta$$ at $$s$$.