Consider a particle travelling along the path given by c(t) = (−t, t, t^2). Here t denotes time. (a) Find the velocity and the speed of the particle.b) Find the unit tangent vector T(t), the principal normal vector N(t), the binormal vector B(t), the curvature κ(t) and the torsion τ (t) at time t.

(a) The velocity of the particle is the derivative of the position function with respect to time.

The position function is c(t) = (−t, t, t^2).

So, the velocity function v(t) is the derivative of c(t):

v(t) = c'(t) = (-1, 1, 2t)

This means that the velocity of the particle at time t is (-1, 1, 2t).

The speed of the particle is the magnitude of the velocity vector.

So, the speed function s(t) is the magnitude of v(t):

s(t) = |v(t)| = sqrt((-1)^2 + 1^2 + (2t)^2) = sqrt(1 + 1 + 4t^2) = sqrt(2 + 4t^2)

This means that the speed of the particle at time t is sqrt(2 + 4t^2).

(b) The unit tangent vector T(t) is the velocity vector v(t) normalized:

T(t) = v(t) / |v(t)| = (-1, 1, 2t) / sqrt(2 + 4t^2)

The principal normal vector N(t) is the derivative of the unit tangent vector T(t) normalized:

N(t) = T'(t) / |T'(t)|

The binormal vector B(t) is the cross product of the unit tangent vector T(t) and the principal normal vector N(t):

B(t) = T(t) x N(t)

The curvature κ(t) is the magnitude of the derivative of the unit tangent vector T(t):

κ(t) = |T'(t)|

The torsion τ(t) is the derivative of the binormal vector B(t) with respect to the arc length s:

τ(t) = dB(t)/ds

Note: The calculations for N(t), B(t), κ(t), and τ(t) are quite complex and require knowledge of vector calculus. The above definitions provide the general approach to find these quantities.