A bee goes out from its hive in a spiral path given in plane polar coordinates by r = bekt q = ct where b, k, and c are positive constants. Show that the angle between the velocity vector and the acceleration vector remains constant as the bee moves outward.
r(t) = bekt q(t) = ct
3. The attempt at a solution
r'(t) = bke^(kt)
r’ ‘ (t) = bk^(2)e^(kt)
I’m lost as to what should I be doing with the information to make the jump to the proof.