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Tangent line to path

  1. Sep 15, 2008 #1
    Find an equation for the line tangent to the given path at the indicated value for the parameter.

    x(t)=4costi-3sintj+5tk, t=[tex]\pi[/tex]/3

    So what I did here was take x'(t) and then plugged in [tex]\pi[/tex]/3 after that to get an equation containing i, j, and k.
    x'(t)= -4sinti-3costj+5k
    x'([tex]\pi[/tex]/3)=-4[tex]\sqrt{3}[/tex]/2i-3/2j+5k
    Am I doing this right? Calc III is confusing me so much. Any help would be appreciated. This is due Thursday. Thanks!
     
  2. jcsd
  3. Sep 15, 2008 #2

    Dick

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    Looks all right so far.
     
  4. Sep 15, 2008 #3
    Alright sweet thanks a lot!
     
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