For the position function r(t)= (2+t^2,3t), determine the approximate angle between the acceleration and velocity vectors at (3,3).

Question

a)45.7
b)56.3
c)71.7
d)44.9
e)58.2

Answer 1

2+t^2 = 3 so t = 1

dx/dt = 2t = 2 at t = 1
dy/dt = 3

d^2x/dt^2 = 2
d^2y/dt^2 = 0

so acceleration in x axis direction, theta = 0

velocity is (1,3)
tan theta = 3/1
theta = 71.6 degrees
71.6 - 0 = 71.6 degrees

Answer 2

I agree with Damon except for the fact velocity should be (2,3)
Therefore tan theta = 3/2
Which means theta = 56.3 degrees
And then subtract the velocity angle from the acceleration angle (like below)
56.3 - 0 = 56.3 degrees
So b is your answer.