Thread: particle/curvature question

Hi guys can anyone help me out with this?

A moving particle at time t ∈ [0, 10] (seconds) has position vector in metres from the origin (0, 0, 0) given by the vector function r(t) = (10 − t)i + (t^2
− 10t)j + sin tk.
i. Describe the path of the particle, as seen from above (the positive k-direction), and also describe it in three dimensions.
ii. Find the curvature of the path, at t = 2π ≈ 6.28 seconds.
iii. Find the angle between the path (at start and end-points) and the k-direction.

Thanks

Originally Posted by katie

Hi guys can anyone help me out with this?

A moving particle at time t ∈ [0, 10] (seconds) has position vector in metres from the origin (0, 0, 0) given by the vector function r(t) = (10 − t)i + (t^2
− 10t)j + sin tk.
i. Describe the path of the particle, as seen from above (the positive k-direction), and also describe it in three dimensions.

Mr F says:

x = 10 - t => t = 10 - x .... (1)

y = t^2 - 10t .... (2)

Substitute (1) into (2).

ii. Find the curvature of the path, at t = 2π ≈ 6.28 seconds.

Mr F says: Substitute into the formula.

iii. Find the angle between the path (at start and end-points) and the k-direction.

Mr F says: Use the dot product.

Thanks

..

when you say substitute into the formula is that just into the r(t) formula?

and the dot product, i am not sure what the other vector is to be using the dot product with?

thanks

Originally Posted by jimmy

when you say substitute into the formula is that just into the r(t) formula?

and the dot product, i am not sure what the other vector is to be using the dot product with?

thanks

The curvature formula.

Direction of motion given by dr/dt. Evaluate at t = 0 and t = 10. Consider the dot product of each with k ....