[TI-M] Re: Math problem help


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[TI-M] Re: Math problem help



Yep, use (x - h)^2 + (y - k)^2 = r^2, and differentiate with respect to t.  
You should end up with an equation with h, k, x, y, dx/dt, and dy/dt.  You 
know 5 of these 6 values, so you should be able to solve for dy/dt.

(x - h)^2 + (y - k)^2 = r^2
2*(x - h)*dx/dt + 2*(y - k)*dy/dt = 0
dy/dt = (h - x)*(dx/dt)/(y - k) = (2 - 1)*(.25)/(-1 - (-3)) = .125 
inches/second

Someone else please verify this solution; I'm pretty sure it's correct, but 
you never know...

- JayEll

In a message dated 5/7/01 5:22:25 PM Mountain Daylight Time, 
daracerz@crosswinds.net writes:


> I just got a question from school and was woundering if any of you are able
> to help me out with this a little.
> 
> Heres the question:
> If a pt is moving on a circle with a center of (2,-3), and the rate of
> change of the x is .25 inches/sec. then find the rate the y changes when x=1
> and y=-1.  Radius is Rad 5.
> 
> Can someone please state the formula i need to use and the derivative of
> that formula so I can plug in the rest of the values.  Thx.  I believe the
> formula to use is (x-h)^2+(y-k)^2=r^2, but I'm not sure.  Any help
> appreciated.  Thx
> 




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