sqrt(25 + 9) = sqrt(34)
tan theta = -3/5 in quadrant IV
theta = -31 deg or 329 deg CC from x axis
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4-t^2 = 0 so t = 2
6 t = 12 so we knew t = 2
dx/dt = -2t
dy/dt = 6
at t = 2
dx/dt = Vx = -4
dy/dt = Vy = 6
v = sqrt(16+36) = sqrt (52) = 7.21
tan theta = 6/-4 in quadrant II
theta = -56.3 or 180-56.3=123.7 deg
1. What are the magnitude and direction angle for the vector u=<5,-3>
2. Calculate the speed of a particle in the xy-plane with positive vector r(t)=(4-t^2, 6t) at the point (0,12).
1 answer