[TI-M] Re: Math problem help

To: <ti-math@lists.ticalc.org>
Subject: [TI-M] Re: Math problem help
From: "Mark F." <daracerz@crosswinds.net>
Date: Mon, 7 May 2001 19:47:40 -0400
References: <3c.b520a77.28288a72@aol.com>
Reply-To: ti-math@lists.ticalc.org


Ok Thanks very much.  I followed your work there and does turn out corrrect
for me too.  Thx again.

----- Original Message -----
From: <JayEll64@aol.com>
To: <ti-math@lists.ticalc.org>
Sent: Monday, May 07, 2001 7:32 PM
Subject: [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
> >
>
>
>

References:

[TI-M] Re: Math problem help

From: JayEll64@aol.com