What is the velocity vector for a moving particle with a position vector r(t)=(2/t, ln(t))?
a) (2,e^t)
b) (2ln(t), tln(t)) '/>
c) (-2/t^2, 1/t)
d) (ln(t), 2/t)
My answer is "c" but is negative. Can you tell me if my answer is correct?
x = 2/t
y = ln t
Vx = dx/dt = -2/t^2
Vy = dy/dt = 1/t
so C is correct
To find the velocity vector for a moving particle, we need to differentiate the position vector with respect to time (t). Let's find the derivative of the position vector r(t):
r'(t) = (d/dt)(2/t, ln(t))
To find the derivative, we differentiate each component of the position vector separately.
For the first component, we use the chain rule:
(d/dt)(2/t) = 2(d/dt)(1/t) = 2(-1/t^2) = -2/t^2
For the second component, we use the derivative of the natural logarithm:
(d/dt)(ln(t)) = (1/t)
Therefore, the velocity vector is:
r'(t) = (-2/t^2, 1/t)
So, the correct answer is option c).