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Graph the surface represented by the vector valued function

  1. Nov 19, 2009 #1
    1. The problem statement, all variables and given/known data
    r(u,v)=2ucosv i + 2usinv j + u^4
    0<=u<=1
    0<=v<=2pi


    2. Relevant equations
    don't care about the graph as much as how the book got to the answer equation
    z=(x^2+y^2)^2/16


    3. The attempt at a solution[/b
    cosu=x/2v sinu=y/2v cosu^2+sinu2=1 (x/2u)^2+(y/2u)^2=1 x^2+y^2=(4)^2
    (x^2+y^2)=(4z)^2 (x^2+y^2)/16=z^2 z=sqrt((x^2+y^2)/16)
    Why isn't the final answer z=sqrt...
     
    Last edited: Nov 19, 2009
  2. jcsd
  3. Nov 19, 2009 #2

    lanedance

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    x=2ucosv
    y=2usinv
    z=u^4

    try substituting x & y into the equation
    z=(x^2+y^2)^2/16
    and see what you get
     
  4. Nov 19, 2009 #3

    Dick

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    What is x^2+y^2 in terms of u?
     
  5. Nov 19, 2009 #4
    I was in the middle of editing when you two responded, will you please re-look at my attempt at a solution?
     
  6. Nov 19, 2009 #5

    Dick

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    It's pretty garbled. But I'm with you through x^2+y^2=16*u^2, which is what I hope you meant to write. Then did you just put u=z? What did you do?
     
  7. Nov 19, 2009 #6
    Sorry about the garbled-ness. I thought I put more spaces in between the equations. Yes, I did substitute u=z and then I dividev both sides by 16 to get (x^2+y^2)/16=z^2. Then I took the sqrt of both sides which would give z=sqrt((x^2+y^2)/16) but the book gives an answer of z=(x^2+y^2)^2/16
     
  8. Nov 19, 2009 #7

    Dick

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    But you can't put u=z! z=u^4. What does that make u^2?
     
  9. Nov 19, 2009 #8
    u^2=sqrt(z)...got it. Brilliant. Thanks.
     
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