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Homework Help: Calculus III Parametric Equations

  1. Jul 8, 2008 #1
    1. The problem statement, all variables and given/known data
    I've uploaded a scan of the questions. Questions 4, 5, and 6 are given in the 3 files uploaded. They all come from the given information from the first scan of the problem.

    2. Relevant equations

    3. The attempt at a solution
    I've worked everything I could on paper; what am I missing. How should I go about finishing the problem and am I correct so far?

    Any good links for parametric equations also?

    Thanks for the help!

    Attached Files:

    • 1.jpg
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    • 2.jpg
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    • 3.jpg
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  2. jcsd
  3. Jul 8, 2008 #2


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    For part 3, you have correctly shown that the curve passes through (3, 0) for [itex]t=\sqrt{3}[/itex] and for [itex]t= -\sqrt{3}[/itex]. I confess that I don't understand the rather complicated way you do that! It is sufficient to show that [itex]y= t^3- 3t= 0[/itex] has roots 0, [itex]\sqrt{3}[/itex] and [itex]-\sqrt{3}[/itex] and that only the last two also satisfy [itex]x= t^2= 3[/itex].

    You have shown that the tangent line at [itex]t= \sqrt{3}[/itex] is [itex]y= \sqrt{3}x- 3\sqrt{3}[/itex] but you don't seem to have completed the problem for [itex]t= -\sqrt{3}[/itex].

    For (4), asking where the tangent lines are horizontal or vertical, again, you have again written a complicated calculation starting with writing [itex]x^2+ y^2[/itex] in terms of t. I can see no reason for that. Horizontal tangents occur where [itex]y'= 3t2-1= 0[/itex] and vertical tangents occur where [itex]x'= 2t= 0[/itex].
  4. Jul 8, 2008 #3
    Thanks HallsofIvy, once again, for the great help. These forums are great to get the wheels turning!
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