Generating curve from curvature Is there a way to plot a curve in the plane, given the curvature at $(x,y)$ is a function of $(x,y)$, called $C(x,y)$, along with initial conditions?
For example. If a curve originating from (0,0) with the slope 0 has the curvature $C(x,y)=x$ , it would "climb" up the y-axis in a shape resembling a sine curve.
Thanks.
 A: Working with curvature $k=\tfrac{1}{R}$ is like working with arc length $s$: it is very uncommon (set apart the circle...) to get tractable expressions in cartesian coordinates.
Oddly, a certain number of curves can be described as the unique solution to a relationship $k=\phi(s)$ (or sometimes $s=\psi(k))$ between $k$ and $s$. These functions $k=\phi(s)$ (or sometimes $s=\psi(k))$ are called intrinsic equations. See (https://en.wikipedia.org/wiki/Intrinsic_equation) where they are called Cesàro intrinsic equations. 
Some examples of intrinsic equations:


*

*Euler's (or Cornu's) spiral: $\tfrac{1}{r}=k=s^2$ or more generally $k=\tfrac{s^2}{a^2}$ ; see the very nice article (http://levien.com/phd/euler_hist.pdf)

*Logarithmic spiral: $\tfrac{1}{r}=k=\tfrac{a}{s}$ (a constant)

*Circle involute:  $k=\tfrac{a}{\sqrt{s}}.$

*Catenary: $k =\tfrac{a}{s^{2}+a^{2}}.$
Remarks: 
1) an intrinsic equation defines a curve up to a rotation.
2) intrinsic equations are generalizable to space curves with a third character, the torsion $\tau$ (see Serret-Frenet equations) 
A: I'm looking at this in the complex plane where it might be easier to comprehend. As pointed out elsewhere on this page, the curvature $\kappa$ is typically expressed as a function of the arc length, $s$ or the tangent angle, $\theta$. The curve can then be expressed as
$$z(s)=\int e^{i\int \kappa(s)ds}ds$$
Since the tangent angle $\theta=\int \kappa(s)ds$, we can also say
$$z(\theta)=\int \frac{e^{i\theta}}{\kappa(\theta)}d\theta$$
Now, when $z$ is expressed parametrically, say, $z=z(u)$, then it can be shown that the curvature is given by
$$\kappa=\frac{\mathfrak{Im}\{\dot z^*\ddot z\}}{|\dot z|^3}$$
I do not see a clear path to obtaining $z$ from $\kappa(z)$.
If we look at the equation in Cartesian coordinates, and allow that the curve is given by $y=f(x)$, then the curvature and signed curvature, $k$ are given by
$$
\kappa=\frac{|y''|}{(1+y'^2)^{3/2}}\\
k=\frac{y''}{(1+y'^2)^{3/2}}
$$
Again, I can't see a path to $y(\kappa)$.
References
Zwikker, C. (1968). The Advanced Geometry of Plane Curves and Their Applications, Dover Press.
Wikipedia: Curvature
