# Calculate acceleration of particle?

#### zelscore

##### New member
The problem is as follows: "A particle moves along the curve r = 3ui + 3(u^2)j + 2(u^3)k in the direction corresponding to increasing u and with a constant speed of 6. Find the velocity and acceleration of the particle when it is at the point (3, 3, 2)"

I succesfully calculated the velocity, but I do not know how to calculate the acceleration.

I know that to find the velocity, v = dr/dt = (du/dt)*(dr/du). The latter factor is easily found to be 3i + 6uj + 6(u^2)k, while the first factor is found by realizing that
Speed = abs(velocity) = abs(du/dt)*abs(dr/du) = 6 ----> Rearranging gives the expression du/dt = 6/abs(dr/du) which is put in to the expression for v, and then I solve for (3, 3, 2) where x=3u, y=3u^2, z=2u^3.

However, to find the acceleration, I write the formula, which i'm not quite sure to be correct in the first place;
a = dv/dt = (d^2u/dt^2)*(d^2r/du^2)

The second factor is found easily by taking the derivative of r with respect to u once again, but this time I don't see any trick to find the first factor, the second derivative of u with respect to t.

#### Dr.Peterson

##### Elite Member
You started with v = dr/dt = (du/dt)*(dr/du), and you claim that dv/dt = (d^2u/dt^2)*(d^2r/du^2). But you didn't apply the product rule to differentiate the product ...