About the curvature of a curve

Question

Let $\alpha : I \rightarrow R^2$ be a smooth curve. For each $t \in I$ consider $N(t)$ the normal unit vector at the point $\alpha(t)$. Fix $\lambda > 0 $ a constant and define the parallel curve $\gamma(t) = \alpha (t) + N(t) \lambda$. Someone can give me a help to calculate the curvature of $\gamma$ with respect to the curvature of $\alpha$? My intuition says that will be something like:

$$K_{\gamma}(t) = \frac{K_{\alpha} (t)}{ 1 + \lambda K_{\alpha(t)}}$$

where $K_{\gamma}$ and $K_{\alpha}$ are the curvatures of $\gamma$ and $\alpha$ respectively. Someone can give me a help ? I am not good with these things ...

Thanks in advance

Answer 1

You are right. Directly compare radii of parallels in normal direction using constant normal distance difference $ \lambda$ from reciprocal of curvatures.

$$1/K_{\gamma}(t) = 1/K_{\alpha} (t) + \lambda $$

which is what you gave put into radius form.

Answer 2

Suppose t is the arc-length parameter of $\alpha$.
$\dot\gamma=\dot\alpha+\lambda\dot N=\dot\alpha - k\lambda\dot\alpha=(1-k\lambda)\dot\alpha$
$\ddot{\gamma}=-\lambda\dot k\dot\alpha+(1-k\lambda)kN$
So the curvature is $$\frac{\det(\dot\gamma, \ddot\gamma)}{|\dot\gamma|^3}=\frac{k(1-k\lambda)^2}{(1-k\lambda)^3}=\frac{k}{1-k\lambda}$$ The difference of the sign comes from different direction setting of N. So yes you're right.

Answer 3

Suppose that $t$ is the arc length of $\alpha$, then $$\dot\gamma=\dot\alpha+\lambda\dot N \\=T(t)-\lambda K_\alpha T(t)\\=(1-\lambda K_\alpha) T(t)\\||\dot\gamma||=(1-\lambda K_\alpha)$$Thus the arc length of $\gamma$ satisfies $$\frac{ds}{dt}=(1-\lambda K_\alpha)$$ and the unit tangent vector of $\gamma$ is $$U(s)=\frac{\dot\gamma}{||\dot\gamma||}=T(t)$$Then $$\frac{dU}{ds}=\frac{dT}{ds}\\=\frac{dT}{dt}*\frac{dt}{ds}\\\frac{dU}{ds}=\frac{dT}{dt}*\frac{1}{1-\lambda K_\alpha}$$ and so $$K_\gamma= K_\alpha*\frac{1}{1-\lambda K_\alpha}$$

Answer 4

The unit speed condition is not essential, the only required condition is $\alpha$ to be regular. If $\alpha$ is unit speed computations are simpler. Anyway, if $\alpha$ is regular the curve $\gamma(t)$ is regular whenever $\lambda k_{s} \neq 1$, where $k_{s}$ is the signed curvature of $\alpha$. To see this, let $s$ be the arc-length parameter of $\alpha$ and $\mathbf{t}$ be the unit tangent vector of $\alpha$, then $$ \frac{d \gamma}{dt} = \frac{d \alpha}{dt} + \lambda \frac{d\mathbf{n}}{dt} = \frac{ds}{dt}\mathbf{t} + \lambda \frac{d\mathbf{n}}{ds} \frac{ds}{dt} , $$ but it is known that $\frac{d\mathbf{n}}{ds} = -k_{s} \mathbf{t}$, so $$ \frac{d\gamma}{dt} = \frac{ds}{dt} \mathbf{t} - \lambda k_{s} \frac{ds}{dt} \mathbf{t} = (1 - \lambda k_{s}) \frac{ds}{dt} \mathbf{t}. \qquad (1) $$ Therefore, $\gamma$ is regular whenever $1 - \lambda k_{s} \neq 0$, as desired. An additional fact is that the signed curvature of $\gamma$ is $$ k_{s}^{\lambda} = \frac{k_{s}}{\lvert 1 - \lambda k_{s} \rvert }. $$ To show this, let $s^{\lambda}$ be the arc-length parameter of $\gamma$, then by equation (1) we have $$ \frac{d s^{\lambda}}{dt} = \lvert 1 - \lambda k_{s} \rvert \frac{ds}{dt}. $$ Let the $\mathbf{t}^{\lambda}$ be the unit tangent vector of $\gamma$, then $$ \mathbf{t}^{\lambda} = \frac{d \gamma}{d s^{\lambda }} = \frac{d\gamma}{dt}\frac{dt}{ds^{\lambda}} = \frac{d\gamma}{dt}/\frac{ds^{\lambda}}{dt} = (1- \lambda k_{s}) \frac{ds}{dt} \mathbf{t}\left/\lvert 1 - \lambda k_{s} \rvert \frac{ds}{dt}\right. = \pm \mathbf{t} = \epsilon \mathbf{t}. $$ Then, the unit normal vector of $\gamma$ is ${n}^{\lambda} = \epsilon \mathbf{n}$. By definition, the signed curvature $k_{s}^{\lambda} $of $\gamma$ is given by $$ k_{s}^{\lambda} \mathbf{n}^{\lambda} = \frac{d \mathbf{t}^{\lambda}}{d s^{\lambda}} = \frac{d \mathbf{t}^{\lambda}}{dt}\left/ \frac{ds^{\lambda}}{dt}\right. = \epsilon \frac{d \mathbf{t}}{dt}\left/ \lvert 1 - \lambda k_{s} \rvert\frac{ds}{dt}\right. $$ that is to say,

$$ k_{s}^{\lambda} \mathbf{n}^{\lambda} = \epsilon \lvert 1 - \lambda k_{s} \rvert ^{-1} \frac{dt}{ds} \frac{d \mathbf{t}}{dt} = \epsilon \lvert 1 - \lambda k_{s} \rvert ^{-1} \frac{d \mathbf{t}}{ds} = \epsilon \lvert 1- \lambda k_{s} \rvert ^{-1} k _{s} \mathbf{n} =k_{s}\lvert 1 - \lambda k_{s} \rvert^{-1} (\epsilon \mathbf{n}) = k_{s} \lvert 1 - \lambda k_{s} \rvert^{-1} \mathbf{n}^{\lambda}, $$ which implies $$ k_{s}^{\lambda} = \frac{k_{s}}{ \lvert 1 - \lambda k_{s}\lvert.} $$

The curve $\gamma$ is parallel to $\alpha$ in the sense that the distance bewteen $\gamma$ and $\alpha $ is $\lVert \gamma(t) - \alpha (t) \rVert = \lambda$, for all $t$, and $(\gamma(t) - \alpha(t))\cdot \mathbf{t}(t) = 0 $, for all $t$; however this is not the case for $\lambda$ too large, as is shown in the next example

where $\alpha(t) = (2 \cos t, \sin (t))$ and $\gamma = \alpha + \lambda \mathbf{n}$.