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- From: tao@fine.princeton.edu (Terry Tao)
- Subject: Re: need help in calculus problem.
- Message-ID: <1993Jan3.160653.2842@Princeton.EDU>
- Originator: news@nimaster
- Sender: news@Princeton.EDU (USENET News System)
- Nntp-Posting-Host: math.princeton.edu
- Organization: Princeton University
- References: <1i688pINNls9@usenet.INS.CWRU.Edu> <1i70mpINNcrj@news.umbc.edu>
- Date: Sun, 3 Jan 1993 16:06:53 GMT
- Lines: 32
-
- In article <1i70mpINNcrj@news.umbc.edu> rouben@math9.math.umbc.edu (Rouben Rostamian) writes:
- >In article <1i688pINNls9@usenet.INS.CWRU.Edu> cd187@cleveland.Freenet.Edu (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). (*)
- >
- >Rouben Rostamian
-
- Yes, you're right. Sorry about that.
-
- Of course, doing things the other way around (knowing the tangents first
- and then trying to work out the curve) requires a DE.. or shall I be
- corrected a second time?
-
- Terry
-