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Tangents to parametric equations

  1. Dec 5, 2015 #1
    The site http://tutorial.math.lamar.edu/Classes/CalcII/TangentNormalVectors.aspx talks of "the" unit tangent vector of r(t) = f(t)*i(t)+g(t)*j(t)+h(t)*k(t) as finding "the" tangent vector r'(t) = f'(t)*i(t)+g'(t)*j(t)+h'(t)*k(t) and normalizing it, and further with finding "the" tangent line at t=t0 as r(t0) + r'(t0)*t . (If I got that right.) But if one thinks of a tangent line as a line being perpendicular to the curve at the given point, then there are an infinite number of tangent lines (and unit tangent vectors), which is, as I understand it, the reason one deals with tangent planes in respect to curves in 3-D instead of tangent lines. What am I missing here?
  2. jcsd
  3. Dec 5, 2015 #2


    Staff: Mentor

    For your curve in space r(t) = f(t)i + g(t)j + h(t)k, at a given point on the curve P(x0, y0, z0), the tangent plane contains an infinite number of lines that lie in the plane and intersect the given point. However, there is only one line in this plane that has the same direction as the curve, and that's the one they're talking about in the tutorial. The unit tangent is a unit vector in that direction.
  4. Dec 5, 2015 #3
    Ah, that makes sense. Thanks very much, Mark44
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