Re: TI-M: math class

To: ti-math@lists.ticalc.org
Subject: Re: TI-M: math class
From: Andy Selle <aselle@ticalc.org>
Date: Thu, 31 Aug 2000 01:28:41 +0200 (MET DST)
In-Reply-To: <OF61E4DFFF.A710B08A-ON8525694B.007B1785@notesnet.bgsu.edu>
Reply-To: ti-math@lists.ticalc.org

Well, since you didn't specify the details of your problem, it is
difficult to be sure that my advice is applicable, but here goes:

Probably, you forgot to take into account the multivalued inverse
functions that you are implicitely using.  For example:

Suppose you have x^2+y^2=r^2 (it's a circle with radius r centered at
x=y=0).  You solve for y and get

y = sqrt(r^2-x^2)

You graph that, and you get a semicircle on the top half of the coordinate
plane.  The reason you don't get a circle is that you haven't taken into
account that the square root has two values i.e. sqrt(4) = 2 or
sqrt(4)=-2.

So what you need to do is graph the two functions
,
| y = sqrt(r^2-x^2)
| y = - sqrt(r^2-x^2)
`

On the TI, you can do this using one line by saying

y1 = {1,-1} sqrt(r^2-x^2)

hope this helps,

Andy

--
Andy Selle <aselle@ticalc.org>
   Programming and System Administration, Survey Editor, Accounts Manager
   the ticalc.org project - http://www.ticalc.org/

On Wed, 30 Aug 2000 waltnat@bgnet.bgsu.edu wrote:

> 
> 
> hell, i can't get it to do what i want.  i have an expression, with both
> x's and y's.  i solve for y.  easy enough.  i turn the remaining expression
> into a fuction, but it doesn't behave that way. grr.
> 
> nathan
> 
> 
>

References:

Re: TI-M: math class

From: waltnat@bgnet.bgsu.edu