NetNews Usenet Archive 1992 #31/NN_1992_31.iso/spool/sci/math/17611

Internet Message Format | 1993-01-03 | 1.8 KB

Path: sparky!uunet!haven.umd.edu!news.umbc.edu!math9.math.umbc.edu!rouben
From: [e-mail] (Rouben Rostamian)
Newsgroups: sci.math
Subject: Re: need help in calculus problem.
Date: 3 Jan 1993 15:28:25 GMT
Organization: University of Maryland Baltimore Campus
Lines: 35
Message-ID: <[e-mail]>
References: <[e-mail]>
NNTP-Posting-Host: math9.math.umbc.edu

In article <[e-mail]> [e-mail] (Aye Naing) writes:
>The following equations describe a curve parameterized by 't'
>F(x,y,t)=0
>G(x,z,t)=0
>H(y,z,t)=0
>Find the equation of the tangent line.

Differentiate each equation wrt t, via the chain rule:
F_x x' + F_y y' + F_t = 0,
G_x x' + G_z y' + G_t = 0,
H_y y' + H_z z' + H_t = 0,
where F_x indicates the derivative of F wrt the x variable, etc.  (I know,
it's a bad notation, but you know what I mean.)
Solve this linear system of three equations for the three unknowns x', y', z'.
The vector (x',y',z') is the desired tangent vector.

To alleviate the concerns of another reader regarding the implicit
form of these derivatives,  let's note that no differential
equations are involved anywhere.  The solution vector is of the form:

(x'(t), y'(t), z'(t) ) = an an explicit function of x(t), y(t), z(t).  (*)

Each value of t corresponds to a point (x(t), y(t), z(t)) on the curve,
and a tangent vector, given by (*), at that point.

Of course if you are only given a specific point (x0, y0, z0) on the curve,
you first need to determine the correspoinding t value.  This amounts
to solving any one of the three original equations describing the curve.
For instance, one may solve
F(x0, y0, t0) = 0,
to compute t0 in terms of x0 and y0, and then proceed as before.  No
differential equations arise.

--
Rouben Rostamian