Singular points of evolute and pedal

Question

Let $\gamma$ be a unit speed plane curve $\gamma : I \to \mathbb R^2$.

$T(s),N(s),k(s)$ are unit tangent vector, unit normal vector, curvature repectively.

Evolute and pedal are defined as:

$EV(s)=\gamma (s)+\frac{1}{k(s)}N(s),k(s) \ne 0,$

$P (s) = (\gamma (s) \cdot N(s)) N(s)$.

I want to find proofs of the following statements:

1. $s_0$ is the singular point of the evolute if and only if $k'(s_0)=0$.

2. $s_0$ is the singular point of the pedal if and only if $k(s_0)=0$.

Any help would be appreciated.

Answer 1

$\require{cancel}$First thing to notice: $\gamma$ having unit speed doesn't mean that ${\rm EV}$ and $\rm P$ also have unit speed. That being said, note that: $${\rm EV}'(s) = \cancel{{\bf T}(s)} - \frac{\kappa'(s)}{\kappa(s)^2}{\bf N}(s) + \cancel{\frac{1}{\kappa(s)}(-\kappa(s){\bf T}(s))} = -\frac{\kappa'(s)}{\kappa(s)^2}{\bf N}(s).$$With this in mind, use the definition: $s_0$ is a singular point of $\rm EV$ if and only if ${\rm EV}'(s_0) = 0$. But since ${\bf N}(s), \kappa(s) \neq 0$ for all $s \in I$ (in particular for $s_0$), by the above expression we have that ${\rm EV}'(s_0) = 0 \iff \kappa'(s_0) = 0$. The exact same approach can be done for $\rm P$, and I think you are capable of doing it, mimicking my work above.

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Since you actually tried to do it, let's do a complete solution. For the next one, we have that: $$\begin{align} {\rm P}'(s) &= (\gamma'(s)\cdot {\bf N}(s)){\bf N}(s) + (\gamma(s) \cdot {\bf N}'(s)){\bf N}(s) + (\gamma(s)\cdot {\bf N}(s)){\bf N}'(s) \\ &+ \cancelto{0}{({\bf T}(s)\cdot {\bf N}(s)){\bf N}(s)} -\kappa(s) (\gamma(s) \cdot {\bf T}(s)){\bf N}(s) -\kappa(s) (\gamma(s)\cdot {\bf N}(s)){\bf T}(s) \\ &= -\kappa(s) \left((\gamma(s) \cdot {\bf T}(s)){\bf N}(s) + (\gamma(s)\cdot {\bf N}(s)){\bf T}(s)\right)\end{align}$$We have that $\kappa(s_0) = 0 \implies {\rm P}'(s_0) = 0$, clear. Now suppose that ${\rm P}'(s_0) = 0$.

We have two possibilities:

1) $\kappa(s_0) = 0$, or;

2) $(\gamma(s_0) \cdot {\bf T}(s_0)){\bf N}(s_0) + (\gamma(s_0)\cdot {\bf N}(s_0)){\bf T}(s_0) = 0$.

Let's prove that 2) can't happen. Since $\{ {\bf T}(s_0),{\bf N}(s_0)\}$ is a orthonormal set, we get $\gamma(s_0) = 0$. In details, 2) implies that: $$\gamma(s_0)\cdot {\bf T}(s_0) = \gamma(s_0)\cdot {\bf N}(s_0) = 0,$$ and so $$\gamma(s_0) = (\gamma(s_0)\cdot {\bf T}(s_0)){\bf T}(s_0) + (\gamma(s_0)\cdot {\bf N}(s_0)){\bf N}(s_0) = 0\cdot {\bf T}(s_0)+0\cdot {\bf N}(s_0) = 0.$$ This is the extra hypothesis we need for the implication $\impliedby$ to be true ($\gamma$ not passing through the origin). You can see here, in item c). This gives the contradiction needed, hence we get $\kappa(s_0) = 0$, as wanted.