Vector Calculus A particle of mass m move along the path σ(t)={t^2,sin t, cos t} Calculate the force acting on the particle t=0.

It seems like σ(t) is the positive function of the particle. Compute its acceleration by differentiating this function twice.

σ(t) = {t ², sin(t ), cos(t )}

σ'(t) = {2t, cos(t ), -sin(t )}

σ''(t) = {2, -sin(t ), -cos(t )}

When t = 0, the acceleration on the particle is

σ'' (0) = {2, -sin(0), -cos(0)} = {2, 0, -1}

Then the force acting on the particle at time t = 0 is

F = {2m, 0, -m}

by Newton's second law.

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