A particle is moving around the unit circle (the circle of radius 1 centered at the origin). At the point (.6, .8) the particle has horizontal velocity dx/dt=3. what is its vertical velocity dy/dt at that point?
AP Calculus - Reiny, Wednesday, December 15, 2010 at 8:08am
x^2 + y^2 = 1
2x dx/dt + 2y dy/dt = 0
dy/dt = (-x dx/dt)/y
= -.6(3)/.8 = - 2.25