# Path of a particle with Brownian heading

This is more of a sanity check. Let $$\Gamma(t)$$ be the random trajectory of a 2D particle at time $$t$$ that has a prescribed velocity vector $$\dot\Gamma(t)=[v(t)\cos(B_t), v(t)\sin(B_t)]$$, where the speed $$v(t)$$ is deterministic and the heading $$B_t$$ is a standard Brownian motion (Wiener process). Is the following assessment true?

Almost surely, for every sample path $$\omega$$ of $$B_t$$, $$\Gamma(t, \omega)\in \text{C}^1$$ but not in $$\text{C}^2$$.

My reasoning is simple. A sample path of $$B_t$$ is almost surely continuous so $$\Gamma(t)=\left[\int_0^{t}v(s)\cos(B_s(\omega))\text{d}s, \int_0^{t}v(s)\sin(B_s(\omega))\text{d}s\right],$$ is well-defined as a Riemann or Lebesgue integral, with derivative $$\dot\Gamma(t)$$ as above. But since $$B_t$$ is almost surely not differentiable $$\dot\Gamma(t, \omega)$$ is not differentiable.

Finally, does anyone know a reference that deals with this type of processes where Brownian motion directly appear in the orientation?

Your formulation has a small problem. Maybe the integral should read $$\int_0^t v(s)\cos(B_s(\omega))ds$$?
Suppose $$v$$ is nice, say bounded. The map $$t\mapsto \int_0^t v(s)\cos(B_s(\omega))ds$$ is $$P$$-almost surely Lipschitz. Therefore, this map has a derivative Lebesgue-almost everywhere $$P$$-a.s., which equals $$v(t)\cos(B_t(\omega))$$.