Osculating circle given parametric coordinates in polar form

Question

I have the following parametric equation for a curve in polar form: \begin{align} r(t) &= r_0-kt\\ \phi(t)&=h\ln(r_0-k_0t)+a \end{align} where $(r_0,h,a,k_0,k)\in\mathbb{R}$ what is the radius of the osculating circle $R(t)$? Thanks in advance.

Answer 1

Use Frenet- Serret relations with vectors $(N,T)$ in normal plane in the usual differential form after converting from polar to rectangular coordinates.

rz = 0.75; kz = 0.4; k = 0.25; h = 0.2; a = 1.25;
rho[t_] = (rz - k t); th[t_] = h Log[rz - kz t] + a ;
ParametricPlot[rho[t] {Cos[th[t]], Sin[th[t]]}, {t, -5, 2}, 
 GridLines -> Automatic, AspectRatio -> 1.25]

In the above Mathematica used function definition f[t_].

Answer 2

This is the Mathematica output, using the parameter names given.

Let me elaborate it a bit. I just started with $r'$ and $\phi'$, then rotated the vector $(r',r \phi')$ by an angle of $\phi$ to put them in XY coordinates. The norm of the derivative of the normalized resulting vector is the curvature, i.e., the inverse of the radius.