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
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
1 answer