I run to the following problem which says if you have a smooth curve that is evolving over time (say finite length at the beginning) then

$$\frac{d}{dt}(curve \; length \; at \; time \; t)=-\int_{curve} k\cdot v \; ds,$$

where $k$ is curvature of the curve and $v$ is velocity of point on curve. $ds$ represents integration by parametrization by arclength.

I have tried proving this but I can not get out without re parametrazing curves. Is there some neater way to do this?

  • $\begingroup$ How can that be true? The curvature is a scalar function, how would you take the dot product with the velocity? (And if it is a scalar product instead of the dot product, it wouldn't work either, since the result would have to be a scalar, not a vector.) $\endgroup$ – Lukas Geyer Oct 25 '12 at 6:10
  • $\begingroup$ Take the dot to just be multiplication, not dot product. The result is a scalar so I don't understand your parenthetical objection. $\endgroup$ – Tyler Brabham Oct 25 '12 at 6:21
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    $\begingroup$ doesn't curvature point in the normal direction? $\kappa \vec{N}$ $\endgroup$ – cactus314 Oct 25 '12 at 20:27
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    $\begingroup$ This is only true if the endpoints are fixed, or at least do not move along the locally tangential direction. For a counterexample, consider a straight line segment that changes its length but not its orientation. $\endgroup$ – Rahul Oct 26 '12 at 4:50
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    $\begingroup$ The problem is false as stated. Hopefully he clarifies what happens at the endpoints. Maybe they are fixed or something where the endpoint contributions vanish. Maybe the curve is closed. $\endgroup$ – cactus314 Oct 26 '12 at 9:14

Here's a nice crib-sheet for differential geometry of curves in space.

Let $\gamma: [a,b]\times [0,1] \to \mathbb{R}^3$ be a family of curves. The rate of change of arc-length is: $$ \frac{d}{dt}(\mathrm{arc length})=\frac{d}{dt}\int_{a}^{b} ||\gamma'(s,t)|| ds = \int_{a}^{b} \frac{d}{dt} ||\gamma'(s,t)|| ds $$ I will ignore the last term since there's no dependence on endpoints in your question. The size of the tangent vector is inner product of $\gamma' = \frac{d\gamma}{ds}$ with itself (square-root). $$ ||\gamma'(s,t)|| = ( \gamma'\cdot \gamma')^{1/2}$$ Differentiate both sides with respect to time and use the chain rule from calculus: $$ \frac{d}{dt} ||\gamma'(s,t)|| = \frac{\frac{d\gamma'}{dt}\cdot \gamma'}{\sqrt{ \gamma'\cdot \gamma'}} = \frac{dv}{ds} \cdot T $$

where $T$ is the unit tangent vector. $$ T = \frac{\gamma'}{ ( \gamma'\cdot \gamma')^{1/2}} $$ Velocity is relative to the deformation parameter $s$ not the curve paramter $t$. Since partial derivatives commute, we can relate $\gamma', v$: $$ \frac{d \gamma'}{dt} = \frac{\partial^2 \gamma}{\partial s \partial t} = \frac{\partial}{\partial s} \frac{\partial \gamma}{\partial t} = \frac{dv}{ds}$$

Then... integration by parts: $$\frac{d}{dt}(\mathrm{arc length})= \int \frac{dv}{ds} \cdot T ds = -\int v \cdot \frac{dT}{ds}ds= - \int v \cdot (\kappa N) ds $$ Curvature is the derivative of the unit tangent vector $\fbox{$\frac{dT}{ds} = \kappa N$}$. It was missing in your problem that curvature points in the direction of the normal.

For small pieces of arc, the curvature is constant and the curve can be approximate by the arc of a circle. The arc-length is just $ds = r d\theta$.

The maximum potential for growth is when the arc moves outward in the normal direction. If each curve moves tangentially, the arc-length doesn't change. In general, the arc grows according to the normal component of the velocity, $\Delta (ds) = (v \Delta t \cdot N)d\theta $. enter image description here

Globally we can imagine the curve swept out by $\gamma(\cdot,t)$. Our change in arc-length is $\frac{d^2A}{dt^2}$. This area should grow the most if it expands out in the normal direction, so it should be proportional to $v\cdot N$. Sharper turns should grow faster and we just showed it should be proportional to curvature $k$. So we get $\frac{d}{dt}(\mathrm{arclength}) =-\int v\cdot (kN) ds$.

enter image description here

  • $\begingroup$ Can you explain a bit why $\frac {d \gamma \prime}{dt}$ is equivalent to $\frac {dv}{ds}$? $\endgroup$ – Tyler Brabham Oct 25 '12 at 23:59
  • $\begingroup$ For the integration by parts, I think you need to state the assumption that $v\cdot T=0$ at the endpoints. $\endgroup$ – Rahul Oct 26 '12 at 4:52
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    $\begingroup$ His problem statement needs clarification about endpoint effects. What about straight lines? Let $\gamma(s,t)=(st,0,0)$ for $s,t \in [0,1]$. Then $\mathrm{arclength}=t$ and $\frac{d}{dt}(\mathrm{arclength})=1$, but curvature of line is 0 and so RHS $\int k\cdot v = 0$. $\endgroup$ – cactus314 Oct 26 '12 at 9:11

This is not as comprehensive an answer as John Mangual's. Consider it, rather, supplemental information to provide intuition, drawn from the textbook, Discrete and Computational Geometry: Eq.5.1


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