non linear function derivative  TOPIC_SOLVED

Limits, differentiation, related rates, integration, trig integrals, etc.

non linear function derivative  TOPIC_SOLVED

Postby AmySaunders on Fri Mar 18, 2011 7:08 pm

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
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Re: non linear function derivative

Postby nona.m.nona on Fri Mar 18, 2011 7:23 pm

AmySaunders wrote: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).
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Re: non linear function derivative

Postby AmySaunders on Wed Mar 23, 2011 6:52 pm

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!
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