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Calculus

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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?

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

  • Calculus -

    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

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