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In multi variable calculus I learned the definition of curvature for a plane curve. I used the definition to find a function describing the curvature at various points. What I would like to do is to construct a plane curve given a curvature function. How would I do this?

Thanks, Bry

Bryhed
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    Parametrize your curve by arc-length $s$ and let $\theta(s)$ be the angle between the unit tangent vector and $x$-axis. i.e $\left(\frac{dx(s)}{ds},\frac{dy(s)}{ds}\right) = ( \cos\theta(s), \sin\theta(s) )$. Up to a sign, $\frac{d\theta(s)}{ds} =\kappa(s)$ where $\kappa(s)$ is the curvature. Integrate this twice gives you a horrible looking integral: $$ x(s) + y(s) i = x(0) + y(0)i + e^{i\theta(0)}\int_0^s e^{i\int_0^{s_1} \kappa(s_2) ds_2} ds_1$$ – achille hui May 16 '18 at 23:14

1 Answers1

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Here is a sketch of how to proceed

  1. Given the curvature function $\kappa\colon[a,b]\to\mathbb R$ define the angular function $\theta\colon[a,b]\to\mathbb R$ as $$ \theta(t) = \int_a^t\kappa(u)\,du. $$
  2. Now define the tangent function $T\colon[a,b]\to\mathbb R$ as $$ T(t) = (\cos\theta(t),\sin(\theta(t)) $$
  3. Finally, define the unit-speed curve $\alpha\colon[a,b]\to\mathbb R^2$ by $$ \alpha(t) = \int_a^tT(u)\,du $$ where the integral denotes the ordered pair of integrals in each of the coordinates of $T$.

Note: going the other way around, i.e., defining the angular function $\theta$ from the unit speed curve $\alpha$ is a bit trickier, so I will include this here.

  1. Start with the tangent vector $T(t)=\alpha'(t)$. Since $\|T(t)\|\equiv1$, for every $t$ we can chose a number $\theta(t)$ such that $$ T(t) = (\cos\theta(t),\sin\theta(t)). $$ Given that this definition says nothing about the continuity or regularity of the map $t\mapsto\theta(t)$, we need to find another way to define a smooth angular function $\theta$.

  2. Write $T(t) = (f(t),g(t))$. Then we know that $f$ and $g$ are smooth functions. Pick an angle $\theta_0$ such that $$ T(a)=(\cos\theta_0,\sin\theta_0). $$

  3. Define the (smooth) function $$ \theta(t) = \theta_0 + \int_a^tfg'-f'g. $$

  4. To see that $f(t)=\cos\theta(t)$ and $g(t)=\sin\theta(t)$ we need to show that the function $$ \psi = (f-\cos\theta)^2 + (g-\sin\theta)^2 $$ is zero everywhere. Since we know that $\psi(a)=0$, it is enough to show that $\psi'\equiv0$.

  5. Since $\alpha$ is unit speed, $f^2+g^2\equiv1$ and therefore $ff'+gg'\equiv0$. Compute $\psi'/2$ and using this equation along with $\theta'=fg'-f'g$ conclude that $$ \psi'/2 \equiv0 $$ (don't be intimidated by the large expression of the derivative, all terms will cancel.)

Leandro Caniglia
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