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?

Question

I don't know what equation to use or anything for this problem. Please help :) It is much appreciated!

Answer 1

the equation must be
x^2 + y^2 = 1
(since .6^2 + .8^2 = 1)

2x dx/dt + 2y dy/dt = 0

divide by 2 and substitute
.6(3) + .8 dy/dt = 0
dy/dt = -1.8/.8 = - 9/4 or - 2.25