A particle P describes the curve with polar equation [itex]r = a e^{\theta \sqrt{3}} \cosh 2\theta[/itex] in such a manner that the radius vector from the origin rotates with uniform angular speed [itex]\omega[/itex]. Show that the resultant acceleration of the particle at any instant makes an angle of 30 degrees in the radius vector.
Any ideas? :yuck:
Any ideas? :yuck: