If a particle moves in the xy-plane so that at time t>0 its position vector is (e^(t²), e^(-t³)), then its velocity vector at time t=3 is...?
Thanks so much!
(e^(t^2))'=e^(t^2)2t
(e^(-t^3))'=e(-t^3)(-3t^2)
velocity vector at time t=3 is (6e^9,-27e^(-27))
x = e^(t^2) y = e^(-t^3)
Vx = dx/dt = 2t*e^(t^2)
Vy = dy/dt = -3t^2*e^(-t^3)
Vx(3) = 6*8103 = 48,619
Vy(3) = -27*1.88*10^-12 = -5*10^-11
Motion along the y direction becomes negligible for t>1.
To find the velocity vector at time t=3, we need to first find the derivative of the position vector with respect to time.
The position vector of the particle is given by (e^(t²), e^(-t³)).
To find the velocity vector, we need to differentiate each component of the position vector with respect to time.
So, the derivative of the vector (e^(t²), e^(-t³)) with respect to time is given by:
(d/dt) (e^(t²), e^(-t³))
= (d/dt) (e^(t²))i + (d/dt) (e^(-t³))j
To differentiate e^(t²), we use the chain rule. The derivative of e^(t²) with respect to t is 2te^(t²).
Similarly, the derivative of e^(-t³) with respect to t is -3t^2e^(-t³).
Therefore, the velocity vector at time t=3 is:
2te^(t²) = 2(3)e^(3²) = 18e^9
-3t^2e^(-t³) = -3(3)^2e^(-(3³)) = -27e^(-27)
So, the velocity vector at time t=3 is approximately (18e^9, -27e^(-27)).
Please note that this is an approximation since we are not given the actual values of e^(9) and e^(-27).
To find the velocity vector of the particle at time t=3, we need to differentiate the position vector with respect to time.
Given position vector, r(t) = (e^(t²), e^(-t³))
To differentiate the position vector, we'll differentiate each component separately.
Differentiating the first component:
dr/dt = d(e^(t²))/dt = 2t * e^(t²)
Differentiating the second component:
dr/dt = d(e^(-t³))/dt = -3t² * e^(-t³)
Therefore, the velocity vector at time t=3 is:
v(3) = (2t * e^(t²), -3t² * e^(-t³))
Substituting t=3 into the velocity vector:
v(3) = (2*3 * e^(3²), -3*3² * e^(3³))
v(3) = (6 * e^9, -27 * e^(27))
Therefore, the velocity vector at time t=3 is (6 * e^9, -27 * e^(27)).