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.


2 Answers 2


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 $\beta'(t)=-\frac{k'}{k^2}n(t).$

  • $\begingroup$ 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)$? $\endgroup$ Nov 15, 2015 at 17:17
  • 1
    $\begingroup$ 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. $\endgroup$
    – frog
    Nov 16, 2015 at 18:37
  • $\begingroup$ Why do we have to show it is perpendicular to tangent, even binormal is perpendicular to tangent, right? $\endgroup$
    – john doe
    Sep 18, 2017 at 20:08
  • $\begingroup$ 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... $\endgroup$
    – 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 $s$.


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