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Homework Help: Parametrization homework help

  1. Jul 2, 2007 #1
    1. The problem statement, all variables and given/known data
    Find a vector function that represents the curve of the intersection of two surfaces.



    2. Relevant equations
    [tex]z^2=x^2+y^2[/tex] with plane [tex]z=1+y[/tex]


    3. The attempt at a solution
    So shouldn't it be
    [tex]r(t)=<cos(t), sin(t), 1+sin(t)>[/tex]
    since x=cos(t), y=sin(t), and z= 1+sin(t)?
    The book gives a wacky answer
     
  2. jcsd
  3. Jul 2, 2007 #2

    NateTG

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    What kind of a shape do you think
    [tex]z^2=x^2+y^2[/tex]
    is?
     
  4. Jul 2, 2007 #3
    I know [tex]z^2=x^2+y^2[/tex] is a cone.
     
  5. Jul 2, 2007 #4
    Can anyone help me, I have got probably an hour
     
  6. Jul 2, 2007 #5
    :uhh::cry::cry::cry:
     
  7. Jul 2, 2007 #6

    NateTG

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    The shape you parametrized is an elipse, but the intersection of the double-cone with side slope 1, and a plane with slope 1 is going to be a parabola.
     
  8. Jul 2, 2007 #7
    So how would I go about coming to a vector equation?
     
  9. Jul 2, 2007 #8

    NateTG

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    Since you already have:
    [tex]z=1+y[/tex]
    I'd substitute, simplify, and see what happens:

    [tex]z^2=x^2+y^2[/tex]
    [tex](1+y)^2=x^2+y^2[/tex]
    ...

    Which should work out reasonably well.
     
  10. Jul 2, 2007 #9

    jambaugh

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    There are more than one way to parameterize a curve, so your answer and the books answer needn't agree.
    [edit:]
    But I note an error. Since you've chosen [itex]x=cos(t), y=sin(t)[/itex] then [itex] x^2+y^2=1[/itex] you've implicitly added then another constraint and you cannot satisfy [itex] z^2 = x^2+y^2[/itex]. Rather try [itex] x = z\cdot\cos(t)[/itex] and [itex]y=z\cdot\sin(t)[/itex]

    [end edit:]



    With regard to vectorizing you have already done that:
    [tex]\mathbf{r}(t)=<x(t), y(t), z(t)> = x(t)\hat{\imath}+y(t)\hat{\jmath} + z(t)\hat{k}[/tex]
    These are just two ways of writing the same vector. The basis is:
    [tex]\langle 1,0,0\rangle=\hat{\imath}[/tex]
    [tex]\langle 0,1,0\rangle = \hat{\jmath}[/tex]
    [tex]\langle 0,0,1\rangle = \hat{k}[/tex]
     
    Last edited: Jul 2, 2007
  11. Jul 2, 2007 #10

    HallsofIvy

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    Since z= 1+ y, z^2= (1+y)^2= 1+ 2y+ y^2= x^2+ y^2. Cancelling the y^2 on each side, 1+ 2y= x^2 or y= (x^2- 1)/2. Taking x itself to be the parameter, we have x= t (of course, y= (t^2- 1)/2, z= 1+ y= 1+ (t^2- 1)/2
     
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