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Vector Function (Calc. 3)

  1. Sep 21, 2009 #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:

    The paraboloid z = 4x^2 + y^2
    The parabolic cylinder y = x^2


    2. Relevant equations

    z = 4x^2 + y^2
    y = x^2

    3. The attempt at a solution

    Combining the two equations:

    z = 4(sqrt(y))^2 + y^2
    z = 4y + y^2

    Choose a parameter:
    Let y = t
    z = 4t + t^2

    Therefore:

    x = sqrt(t)
    y = t

    I get the equation:

    r(t) = (sqrt(t))i + (t)j + (4t + t^2)k

    Is this correct or did I do something wrong?
    Any help appreciated, thanks!
     
  2. jcsd
  3. Sep 21, 2009 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Hi 1st1, welcome to PF!:smile:

    Careful,

    [tex]\sqrt{y^2}=|y|=\left\{ \begin{array}{lr} y & , y \geq 0 \\ -y & , y<0 \end{array} \right.[/tex]

    Instead of introducing this [itex]\pm[/itex] sign problem, try writing [itex]z[/itex] in terms of [itex]x[/itex] instead.
     
  4. Sep 21, 2009 #3
    Hey gabba, thanks for the welcome.

    I see what you mean but sqrt(y) is being squared so regardless of a positive or negative output it will be squared afterwards which would give me positive y either way. Correct me if I am overlooking something.

    Anyway, here is the work in terms of x:

    z = 4x^2 + (x^2)^2
    z = 4x^2 + x^4

    Let x = t
    z = 4t^2 + t^4
    x = t
    y = t^2

    r(t) = (t)i + (t^2)j + (4t^2 + t^4)k

    Is this correct?

    Thanks again.
     
  5. Sep 21, 2009 #4

    gabbagabbahey

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    Homework Helper
    Gold Member

    The reason that sqrt(y^2)=|y|, is that, by definition, the sqrt() function always returns a positive value.

    Other than that, your solution looks good to me!:approve:
     
  6. Sep 21, 2009 #5
    Sounds good, thank you.
     
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