If a particle moves in the xy-plane so that at time t>0 its position vector is (e^(t²), e^(-t³)), then its velocity vector at time t=3 is...?

Question

Thanks so much!

Answer 1

(e^(t^2))'=e^(t^2)2t
(e^(-t^3))'=e(-t^3)(-3t^2)
velocity vector at time t=3 is (6e^9,-27e^(-27))

Answer 2

x = e^(t^2) y = e^(-t^3)
Vx = dx/dt = 2t*e^(t^2)
Vy = dy/dt = -3t^2*e^(-t^3)

Vx(3) = 6*8103 = 48,619
Vy(3) = -27*1.88*10^-12 = -5*10^-11

Motion along the y direction becomes negligible for t>1.