What is the velocity vector for a moving particle with a position vector r of t equals the components 2 over t, the natural log of t ? (10 points)
A) components 2, e to the t power
B) components 0, t times the natural log of t
C) components negative 2 times t, 1 over t
D) components 0, 1 over t
r(t) = <2/t, lnt>
v(t) = dr/dt = <-2/t^2, 1/t>
To find the velocity vector for a moving particle with a position vector of r(t) given by the components 2/t and ln(t), we need to take the derivative of the position vector with respect to time.
The position vector r(t) has two components: x(t) = 2/t and y(t) = ln(t).
Step 1: Find the derivative of x(t) with respect to t.
To find the derivative of 2/t with respect to t, we use the power rule of differentiation.
dx(t)/dt = d(2/t)/dt = (-2/t^2).
Step 2: Find the derivative of y(t) with respect to t.
To find the derivative of ln(t) with respect to t, we use the chain rule of differentiation.
dy(t)/dt = d(ln(t))/dt = 1/t.
Thus, the velocity vector v(t) = (dx(t)/dt, dy(t)/dt) = (-2/t^2, 1/t).
Therefore, the correct answer is C) components -2t, 1/t.