A vector parallel to the tangent to the curve x=3t^(4/3) y=2t^3 -1 z= 2/(t^2) at the point (3,-3,2) on the curve is:
the answer is <-2,3,2>
how do you get this??
the tangent vector is ∇v = (dx/dt,dy/dt,dz/dt)
∇v = <4∛t,6t,-4/t^3>
v(-1) = <3,-3,2>, so
∇v = <-4,6,4>
Note that <-2,3,2> is parallel to ∇v(-1)