Calculate the speed of a particle traveling in the xy-plane with position vector r(t)=(4-t^2,6t) at the point (0,12).
a)2(sqrt13)
b)6
c)4(sqrt13)
d)52
e)sqrt2
v(t) = dr/dt = (-2t,6)
(0,12) is on the curve, so at that point, t=2
so, that means
v(2) = (-4,6)
|v| = √(16+36) = 2√13
To calculate the speed of a particle traveling in the xy-plane, we need to find the magnitude of its velocity vector. The velocity vector is the derivative of the position vector with respect to time.
Given the position vector r(t) = (4 - t^2, 6t), we can find the velocity vector by taking the derivative with respect to time:
r'(t) = (-2t, 6)
To find the velocity vector at the point (0,12), we substitute t=0 into the velocity vector:
r'(0) = (-2(0), 6) = (0, 6)
The magnitude of the velocity vector is the speed of the particle. In this case, the magnitude is √(0^2 + 6^2).
Speed = √(0^2 + 6^2) = √(0 + 36) = √36 = 6
Therefore, the speed of the particle at the point (0,12) is 6. Therefore, the correct answer is option b) 6.