Tom N. answered 11/30/19
Strong proficiency in elementary and advanced mathematics
For r(t) = 3 cost j +2 sint k v(t)=-3 sint j +2 cost k and a(t) = -3 cost j -2 sint k. |v|2 = 9 sin2t + 4 cos2t and |a|2= 9 cos2t + 4 sin2t. Now d|v|2/dt =18 sint cost -8 sint cost = 10 sint cost and d|a|2/dt = 18 sint cost -8 cost sint = 10 sint cost. Using these 10 sint cost =0 or 5sin2t=0 and t= 0,π, 3π/2. So looking at |v|2 for t=0,π v2(0) and v2(π) = 4 and |v| =2 and v2(3π/2) = 9 and |v|=3. So the min is 2 and the max is 3. Doing the same for |a|2 gives a min of 2 and a max of 3 for |a|.