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Vector function intersection

  1. Sep 7, 2014 #1
    1. The problem statement, all variables and given/known data

    "Find a vector function that represents the curve of intersection of the two surfaces."

    2. Relevant equations

    Cone: [tex]z = \sqrt{x^2 + y^2}[/tex] Plane: [tex]z = 1+y[/tex]

    3. The attempt at a solution

    I began by setting [tex]x=cos t[/tex], so that [tex]y = sin t[/tex] and [tex]z = 1+sin t[/tex]. At this point, however, I am stuck. I think I need to set something equal to something else, but I am not sure what.
     
  2. jcsd
  3. Sep 7, 2014 #2
    [strike]You have 2 equations in 3 variables and you want to get 3 equations in 4 variables. This means you can only introduce one new equation where your new variable will appear. Otherwise you will have 4 equations in 4 variables and you no longer have the one degree of freedom required to get a curve. Here you have introduced two new arbitrary equations. Try just setting x = t and then try to express y and z in terms of just t.[/strike] Sorry, I thought you arbitrarily set y = sint, but it seems that you just made a calculation mistake (because you write "so that y = sint", which is wrong). It will be much easier if you just set x = t and then you solve y and z in terms of x (=t).
     
    Last edited: Sep 7, 2014
  4. Sep 7, 2014 #3

    HallsofIvy

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    Do you have any particular reason for setting [itex]x= cos(t)[/itex] and [itex]y= sin(t)[/itex]?
    Surely, not from the first equation?

    If you set [itex]x= cos(t)[/itex] and [itex]y= sin(t)[/itex] then you are saying that [tex]z= \sqrt{sin^2(t)+ cos^2(t)}= 1[/tex] for all t- and that is NOT true.

    The obvious thing to do, since [itex]z= \sqrt{x^2+ y^2}[/itex] and [itex]z= 1+ y[/itex], is to set [itex]x+ y= \sqrt{x^2+ y^2}[/itex]. Then, after squaring both sides, [itex]1+ 2y+ y^2= x^2+ y^2[/itex] so that [itex]2y= x^2- 1[/itex] and [itex]y= (1/2)x^2- (1/2)[/itex].

    Now, let x= t as the parameter.
     
  5. Sep 7, 2014 #4
    Awesome! Thank-you.
     
  6. Sep 7, 2014 #5

    LCKurtz

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    Nevermind my previous post (deleted).
     
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