Let $\alpha:I\to R^2$ be a regular parametrised plane curve (arbitrary parameter), and define $n=n(s)$ and $k=k(t)$ to be the normal and curvature respectively. Assume $k(t)\neq0$, $t\in I$. In this situation, the curve
$$\beta(t)=\alpha(t)+\frac{1}{k(t)}n(t),\text{ }t\in I$$
is called the evolute of $\alpha$.
From Wikipedia, an evolute is the envelope of the normals to a curve.
And now I am trying to show that statement above.
So, consider the normal lines of $\alpha$ at two neighboring points $t_1, t_2, t_1\neq t_2$. Let $t_1$ approach $t_2$ and then we need to show that the intersection points of the normals converge to a point on the trace of the evolute of $\alpha$.
My approach is by using the traditional equation of line in $R^2$, but it seems that that approach is inadequate and cannot relate to the trace of the evolute of $\alpha$.
So I tried writing the equations of the normal lines:
$y=\alpha''(t_1)x+\alpha(t_1)$
$y=\alpha''(t_2)x+\alpha(t_2)$
Then $\alpha''(t_1)x+\alpha(t_1)=\alpha''(t_2)x+\alpha(t_2)$ and
$x(\alpha''(t_1)-\alpha''(t_2))=\alpha(t_2)-\alpha(t_1)$
If $t_1$ approaches $t_2$, then $\alpha(t_1)$ approaches $\alpha(t_2)$. And the possible way to continue is by substituting $x=\frac{\alpha(t_2)-\alpha(t_1)}{(\alpha''(t_1)-\alpha''(t_2))}$ to the equation to get $y$ and show that the points $(x,y)$ converges to a point on the trace of $\beta$. But I have the feeling that it will be very lengthy and I doubt if my approach is correct.
My questions:
1. Any help on how to continue or use different approach?
2. Do we have to parametrise the curve by arc length first before proceeding?
Many thanks in advance!
Things simplify a lot if we assume that the primary curve $\alpha$ is given with its arc length as parameter: $$\alpha:\quad s\mapsto z(s):=\bigl(x(s),y(s)\bigr)\ .$$ Its tangent vector $t(s):=\bigl(\dot x(s),\dot y(s)\bigr)$ as well as its normal vector $n(s):=\bigl(-\dot y(s),\dot x(s)\bigr)$ are then automatically unit vectors. I omit the "$(s)$" in the sequel. Let $\theta:={\rm arg\,}t$ be the polar angle of $t$. The curvature $\kappa$ along $\alpha$ is then given by $$\kappa:=\dot\theta=\dot x\ddot y-\dot y\ddot x\ .$$ As $\dot x^2+\dot y^2\equiv1$ one has $\dot x\ddot x+\dot y\ddot y=0$, and this implies $$\kappa\dot x=(\dot x\ddot y-\dot y\ddot x)\dot x=(\dot x^2+\dot y^2)\ddot y=\ddot y,\qquad \kappa\dot y=-\ddot x\ .\tag{1}$$ Consider now the evolute $\beta$ of $\alpha$. For its construction let $\rho(s):={1\over\kappa(s)}$ be the curvature radius of $\alpha$ at $z(s)$. A parametric representation of $\beta$ is then given by $$\beta:\quad s\mapsto e(s):=z(s)+\rho(s)\>n(s)\tag{2}$$ (of course $s$ is not the arc length parameter on $\beta$). We now look at the tangent vectors to $\beta$. From $(2)$ we get $$\dot e=(\dot x,\dot y)+\rho(-\ddot y,\ddot x)+\dot\rho\>n\ .\tag{3}$$ From $(1)$ and $\rho\kappa=1$ it then follows that the first two terms in $(3)$ cancel, so that we are left with $$\dot e=\dot\rho\>n\ .$$ From this we can draw the following conclusion: The evolute point $e(s)$ lies on the normal (a line $\nu=\nu(s)$) to the curve $\alpha$ at $z(s)$, and the tangent to $\beta$ at $e(s)$ is parallel to this line $\nu$. It follows that the family ${\cal N}$ of normals $\nu(s)$ is the family of tangents to $\beta$, so that $\beta$ is indeed the envelope of the family ${\cal N}$.
(I have omitted some technical assumptions about nonvanishing of certain quantities.)