curvature as as function of tangential angle

I am reading a paper where the authors have defined the curvature $\kappa$ as \begin{align} \kappa = \dfrac{\partial \phi}{\partial r} \end{align} where $\phi(r)$ is the angle that the tangent makes with the horizontal plane and $r$ is the distance to the origin $O$, see the figure below. However, in another paper, I find a similar but different definition of curvature which is given by \begin{align} \kappa = \dfrac{1}{r}\dfrac{\partial }{\partial r} \left(r^2 \dfrac{\partial \phi}{\partial r} \right) = r\dfrac{\partial^2 \phi}{\partial r^2} + 2\dfrac{\partial \phi}{\partial r} \end{align}

Using the standard definition of curvature given by \begin{align} \kappa = \dfrac{\dfrac{d^2 y}{d x^2}}{\left[1 + \left(\dfrac{dy}{dx} \right)^2 \right]^{3/2}} \end{align} I have tried to derive the curvature as a function of the tangent angle, but so far have been unsuccessful. If anyone could help or provide a reference I would appreciate it.

• The definition in the first section should be $\kappa=\frac{\partial \phi}{\partial s}$ where $s$ is arc length not direct line distance – David Quinn Jan 31 '18 at 19:54
• @DavidQuinn I am aware of that definition for curvature, but don't think thats what the authors were intending. Is it possible to write the curvature in terms of the radial distance $r$? – Ragnar Jan 31 '18 at 19:58