The particle P moves along a space curve. At one instant it has velocity $v = (4i-2j-k)$ $m/s$. The magnitude of the acceleration is $8 m/s^2$. The angle between the acceleration and the velocity vector is $20^{\circ}$, so one can calculate that the acceleration in the direction of the velocity is $7.52$.
How can I calculate the radius of curvature from this information?
One of my attempts has been to try to imagine an infinitesimal change in velocity, $v = r\theta$. This implies $\frac{dv}{dt} = r\frac{d\theta}{dt}$. Could I perhaps know somehow what $\frac{d\theta}{dt}$ is?