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Tangent line with slope = 0.

  1. Feb 13, 2014 #1
    1. The problem statement, all variables and given/known data

    I was given parametric equations.

    x(t) = a(t)
    y(t) = b(t)
    z(t) = c(t)

    where a, b, and c are functions that depend on t.

    I was supposed to find equation of the tangent line at t = f given:

    x(f)= m
    y(f) = n
    z (f) = o

    where m,n,o are some constant numbers

    and given
    x'(f)=y'(f)=z'(f)= 0


    2. Relevant equations

    I think this is the relevant equation though its not given.
    r(t) = <Px,Py,Pz>+tv

    3. The attempt at a solution

    I'm super confused here. Given that the derivative of x(t),y(t),z(t) are all 0 at t=f. Then There is no slope whatsoever. That means its basically 3 intersecting planes where x=m, y=n, and z=o. Which isn't a line but a point.

    If I use the equation for a line r(t) = <Px,Py,Pz>+tv then,

    <Px,Py,Pz> = <m,n,o>
    and v = <0,0,0>

    so r(t) = <m,n,o>+t<0,0,0> = <m,n,o>

    But <m,n,o> is a vector with direction, while I already know that the tangent is only a point (m,n,o) given by the intersection of planes x=m, y=n, z=o.

    Is <m,n,o> vector really the tangent line, I feel like there is no 'line' as its a point.
     
  2. jcsd
  3. Feb 13, 2014 #2

    Simon Bridge

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    I think you need to infer that the surface is not discontinuous at t=f - or the problem does not make sense.
    To help you understand the situation - have a play around with the conditions:
    i.e. x=t-f+m, y=(t-f)^2+n, z=(t-f)^3+o ... what's the tangent for t=f?

    If you have a 3D plotting program, pick some values for f,m,n,o and see what the surface looks like.
     
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