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Tangent line to curve Q

  1. May 18, 2006 #1
    Can anybody help me out with this Q?

    "A curve R in space has vector equation:

    [tex] x = (sin(\pi u), u^2 - 1, u^2 + 3u + 3) [/tex]

    u is a real number. Find a vector equation of the tangent line to R at the point (0, 0, 1)"
  2. jcsd
  3. May 18, 2006 #2


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    What are your ideas or thoughts on how to solve this problem? You need to show some work to get help.
  4. May 18, 2006 #3
    Well, I originally thought of taking the gradient of x and then plugging in the values of (0,0,1). Not sure if this is the right way to go about it though.
  5. May 18, 2006 #4


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    The derivative of a curve at a point is indeed parallel to the tangent line (if the curve has nonzero speed at the point and a tangent line). But you can't just plug in (0, 0, 1) because you only have 1 variable-u. How do you find u such that x(u) = (0, 0, 1)?
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