Curvature and plane curves 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
 A: Here is a sketch of how to proceed

*

*Given the curvature function $\kappa\colon[a,b]\to\mathbb R$ define the angular function $\theta\colon[a,b]\to\mathbb R$ as
$$
   \theta(t) = \int_a^t\kappa(u)\,du.
$$

*Now define the tangent function $T\colon[a,b]\to\mathbb R$ as
$$
   T(t) = (\cos\theta(t),\sin(\theta(t))
$$

*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.

*

*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 $\theta$.


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


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


*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 $\psi'\equiv0$.


*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.)
