Length of a displaced curve

Question

Particle $A$ traces a smooth closed curve $\mathcal{C}$ in $\mathbb{R}^2$. Particle $B$ is at a constant distance $d > 0$ from the particle $A$ while being orthogonal to the curve $\mathcal{C}$, tracing the curve $\mathcal{C'}$. If the length of the curve $\mathcal{C}$ is $L$, what is the length of the curve $\mathcal{C'}$?

The curves are depicted here:

The problem was stated at a physics class, so some assumptions about $\mathcal{C}$ and $\mathcal{C'}$ might be missing.

My attempt:

Let $\alpha : [0,1]\to\mathbb{R}^2$ be the parametrization of $\mathcal{C}$. Let's determine the parametrization $\beta : [0,1]\to\mathbb{R}^2$ of $\mathcal{C'}$.

The unit normal vector at the point $\alpha(t) = \begin{pmatrix} x(t) \\ y(t)\end{pmatrix} \in\mathcal{C}$ is $\vec{n} = \frac{1}{\sqrt{\dot{x}(t)^2+\dot{y}(t)^2}}\begin{pmatrix} \dot{y}(t) \\ -\dot{x}(t)\end{pmatrix}$, with $\dot{x}, \dot{y}$ being the derivatives of $x,y$. Another assumption seems to be that $\mathcal{C}$ is the inner curve, and $\mathcal{C'}$ the outer one so $\vec{n}$ has to be directed outwards. I think that is the case here, but how to check it?

Anyway, it should be $\beta(t) = \alpha(t) + d\vec{n}$:

$$\beta(t) = \left(x(t) + d\frac{\dot{y}(t)}{\sqrt{\dot{x}(t)^2+\dot{y}(t)^2}}, y(t) - d\frac{\dot{x}(t)}{\sqrt{\dot{x}(t)^2+\dot{y}(t)^2}}\right)$$

The length of $\mathcal{C'}$ is $\int_0^1 \|\beta'(t)\|\,dt$, so we need $\|\beta'\|$:

$$\beta' = \left(\dot{x} + d\frac{\dot{x}\left(\ddot{y}\dot{x}-\dot{y}\ddot{x}\right)}{\left(\dot{x}^2+\dot{y}^2\right)^\frac{3}{2}}, \dot{y} - d\frac{\dot{y}\left(\ddot{x}\dot{y}-\dot{x}\ddot{y}\right)}{\left(\dot{x}^2+\dot{y}^2\right)^\frac{3}{2}}\right)$$

$$\|\beta'\| = \dot{x}^2+\dot{y}^2+ 2d\frac{\dot{x}\left(\ddot{y}\dot{x}-\dot{y}\ddot{x}\right) - \dot{y}\left(\ddot{x}\dot{y}-\dot{x}\ddot{y}\right)}{\left(\dot{x}^2+\dot{y}^2\right)^\frac{3}{2}} + d^2\frac{\dot{x}^2\left(\ddot{y}\dot{x}-\dot{y}\ddot{x}\right)^2-\dot{y}^2\left(\ddot{x}\dot{y}-\dot{x}\ddot{y}\right)^2}{\left(\dot{x}^2+\dot{y}^2\right)^3}$$

Nothing seems to cancel here. Have I made a mistake somewhere? Will I be able to connect the result with $L$?

Answer 1

Approximate the curve as a polygon and construct the offset curve. The latter is made of line segments at distance $d$ of the original segments and of the same length, connected by circular arcs. The cumulated length of the arcs is $d$ times the angle swept by the normal, which is an integer number of revolutions (depending on the curve).

By letting the approximation be finer and finer, you converge to the true lengths and the difference in lengths is $2k\pi d$ for some $k$ ($\pm1$ for a non self-intersecting curve).

Answer 2

Assume the containing space is $\mathbb{R}^2$.

Assume the trajectory of the inner particle is a smooth, simple closed curve $C$, and let $C'$ be the trajectory of the outer particle.

Let $L,L'$ be the lengths of $C,C'$, respectively.

For the case where the region bounded by $C$ is convex, then $L'=L+2\pi d$.

Explanation:$\;$Appoximate $C$ by a convex polygon $P$ with rounded corners. By elementary geometry, the result is true for $P$, hence, taking the limit as $P$ approaches $C$, it's true for $C$ as well.

For the case where there are portions of $C$ which bend inwards, then, restricted to those portions, we get the inverse relationship $L' = L-2\pi d$.

Answer 3

The figure in the OP is somewhat misleading. I consider a random closed form in the complex plane (see the figure below, blue line), call it $z$. A curve with a uniform displacement, say $d$, normal to the curve is desired. The angle of the displacement from the curve is the tangent angle of the curve, i.e., $\alpha=\arg \dot z$ minus $\pi/2$. Therefore, the displace curve, $w$ is given by

$$w=z+de^{i(\alpha-\pi/2)}$$

The displaced curve is shown in red in the figure below. Notice that it loops in regions where the curve is concave.

Finally, the lengths of the two curves are given by

$$ s_z=\int |\dot z|~du\\ s_w=\int |\dot w|~du $$