[TI-M] Re: Math problem help

To: ti-math@lists.ticalc.org
Subject: [TI-M] Re: Math problem help
From: JayEll64@aol.com
Date: Tue, 8 May 2001 18:13:45 EDT
Reply-To: ti-math@lists.ticalc.org

Hmmm, good point, but the radius actually doesn't affect the derivative at 
all except that it defines the circle.

Also, within the context of the problem, I think the rate of change of y is 
not dy/dx but dy/dt; i.e., the rate of change of y with respect to time 
(since the rate of change of x was given with respect to time).  So you would 
implicitly differentiate with respect to t rather than x...

At least that's how I interpretted the problem...

- JayEll

In a message dated 5/8/01 5:09:13 AM Mountain Daylight Time, 
noveck@pluto.njcc.com writes:

> Here goes:
> 
> Start with:
> (x-h)^2+(y-k)^2=r^2
> (x-2)^2+(y+3)^2=25
> 
> Take the derivative with respect to x to find dy/dx:
> 2*(x-2)+2*(y+3)*dy/dx=0
> 2(x-2)=-2(y+3)*dy/dx
> dy/dx=(2(x-2))/-2(y+3)
> dy/dx=-(x-2)/(y+3)
> 
> Plug in your given point:
> dy/dx=-(1-2)/(-1+3)
> dy/dx=-(-1)/(2)
> dy/dx=1/2
> 
> But there's another substantial issue with this problem: (1,-1) is _not_ on
> a circle with center of (2,-3) and a radius of 5.  If you neglect the given
> radius and instead assume that the radius is equal to whatever value would
> actually cross through that point, then this answer should be what you want.
>