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!
Best Answer
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.)