non linear function derivative

[AmySaunders]

A particle moves in a circular orbit described by the equation x^2 +y^2=25. As it passes through (4.3), its y coordinate is decreasing at a rate of 3 units/sec. What is the rate of change of the x coordinate the instant the particle is passing through the point (4,3)?


I figured I would dreivate the position equation, x^2 + y^2=25, to arrive at the velocity equation, 2x dx +2y dy, and set that equal to -3, the velocity in the y direction at that point. Of course it doesn't come out right, though, as I am not sure how to account for the velocity in the x direction at that point. How do you separate velocity into x and y components?

Thank you for your help!

Amy

[nona.m.nona]

A particle moves in a circular orbit described by the equation x^2 +y^2=25. As it passes through (4.3), its y coordinate is decreasing at a rate of 3 units/sec. What is the rate of change of the x coordinate the instant the particle is passing through the point (4,3)?

Since you are given information on rates of change with respect to time, you might want to consider differentiating with respect to time, rather than with respect to x or y. Also, it is not the derivative of x² + y² which is given as changing at a rate of -3 when passing through the point (4, 3), but the derivative of x (with respect to t).

[AmySaunders]

Thank you so much for your help. I derivated with respect to time. So x dx/dt + y dy/dt =0. Solve algebraically and get dx/dt= -ydy/xdt and plug in the given values. The velocity, -3, is dy/dt, and x and y are 4 and 3, respectively. The answer is 9/4.

Thanks again!