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Tangent equations

  1. Apr 27, 2005 #1
    Let C be the curve: r(t) = (t, t^2, t^3) in R3.

    Find the equation of the tangent line at P = (2,4,8).

    In this case should I calculate the unite tanget T= r'(t)/||r'(t)||?
    Or just the velocity r'(t) = 1i + 2tj + (3t^2)k? How to calculate the tangent equation at the point P?
     
  2. jcsd
  3. Apr 27, 2005 #2

    dextercioby

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    U don't need the versor (unit vector) for that drection.Just the parametrical equation for the tangent,which u already found...

    Then all u have to do is

    [tex]\vec{T}_{(2,4,8)}=\left[1\vec{i}+2t\vec{j}+\left(3t^{2}\right)\vec{k}\right]_{t=2} [/tex]

    Daniel.
     
  4. Apr 27, 2005 #3
    Why did you chose t=2 please? I don't know how to put the point (2.4.8) in the equation. Is there any formula to plug the points into r'?
     
  5. Apr 27, 2005 #4

    dextercioby

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    Yes,the point (2,4,8) is on the graph of the original curve,iff t=2...

    Daniel.
     
  6. Apr 27, 2005 #5
    A line is a 1-dimensional linear manifold: in order to specify it, you need two vectors: the first one, call it [itex]p[/itex], can be any point on the line. The second, call it [itex]q[/itex], represents the direction of the line (in technical jargon, it forms a basis for the directing space of the manifold). In this case, the line is the set

    [tex]\{ p + kq \ | \ k \in \mathbb{R} \}[/tex]

    (assuming your vector space is over [itex]\mathbb{R}[/itex]). Alternatively we just say that the line is specified by the equation

    [tex]v = p + kq[/tex]

    where [itex]k[/itex] is an arbitrary constant (in the appropriate field, here [itex]\mathbb{R}[/itex]).

    in this case, you are given that [itex]p = (2, \ 4, \ 8)[/itex] is on the tangent line. You now need to find the direction. This is what your [itex]1i + 2tj + (3t^2)k[/itex] gives you (it does not matter if the vector is normalized or not, since to make our line we scale take every scalar multiple anyway!). First you need to find the appropriate value for [itex]t[/itex]. You do this by solving [itex]r(t) = (2, \ 4, \ 8)[/itex], since you want the direction to be appropriate to that particular point. As dexter said, you find [itex]t=2[/itex], so the direction is dexter's [itex]\vec{T}_{(2, \ 4, \ 8)}=q[/itex].

    Now just put this together with what I said at the beginning to get the equation of the line.
     
  7. Apr 27, 2005 #6
    Is it enough as an answer to the tangant equation to put the velocity in t=2?
    ie: r'(t) = [1i + 2tj + (3t^2)k] at t=2. T = 1i + 4j + 12k or should I leave it in the form of dexter's?
     
  8. Apr 28, 2005 #7
    That is the direction of the tangent when [itex]t=2[/itex], not the equation of the tangent. Read my last post (carefully :smile:).
     
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