Question

the position of a particle is given by the parametric equations x(t)= ln(t^2 1) and y(t)=e^3t-1. What is the velocity vector at time t=1

Answer 1

To find the velocity vector at time t = 1, we first need to find the derivatives of x(t) and y(t) with respect to t:

x'(t) = d/dt[ln(t^2 + 1)] = (2t)/(t^2 + 1)

y'(t) = d/dt[e^(3t) - 1] = 3e^(3t)

Now, we can find the velocity vector at t = 1 by plugging in t = 1 into the derivatives:

x'(1) = (2(1))/(1^2 + 1) = 2/2 = 1
y'(1) = 3e^(3)

So, the velocity vector at t = 1 is <1, 3e^3>.