Question

Consider the parametric equation (t2+t+1,t3−3t)

Calculate the velocity vector of the associated particle as a function of time.

Answer 1

To calculate the velocity vector of the associated particle, you need to differentiate the parametric equation with respect to time. The velocity vector is the derivative of the position vector. Given the parametric equation (t^2 + t + 1, t^3 - 3t), the velocity vector is (2t + 1, 3t^2 - 3).

Step-by-step explanation:

To calculate the velocity vector of the associated particle, we need to differentiate the parametric equation with respect to time. The velocity vector is the derivative of the position vector.

Given the parametric equation (x(t), y(t)) = (t^2 + t + 1, t^3 - 3t), we can find the velocity vector by taking the derivatives of x(t) and y(t).

Derivative of x(t): dx/dt = 2t + 1

Derivative of y(t): dy/dt = 3t^2 - 3

Therefore, the velocity vector v(t) = (2t + 1, 3t^2 - 3).