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Curve of intersection of surfaces problem (Answer included).

  1. Feb 22, 2012 #1

    s3a

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    1. The problem statement, all variables and given/known data
    "Given that near (1,1,1) the curve of intersection of the surfaces
    x^4 + y^2 + z^6 - 3xyz = 0
    and
    xy + yz + zx - 3z^8 = 0

    has the parametric equations x = f(t), y = g(t), z = t with f, g, differentiable.

    (a) What are the derivatives f'(1), g'(1)?

    (b) What is the tangent line to the curve of intersection (1,1,1) in the forms
    x = 1 + ___ s, y = 1 + ___ s, z = 1+s"


    2. Relevant equations
    Formula for gradient.


    3. The attempt at a solution
    For the last part, I think I see that t = 1 + s but I'm not sure.

    For the first part, I computed the gradients and tried (and failed) to equate them since the surfaces are intersecting.

    gradF(x,y,z) = (4x^3 + 3yz) i + (2y - 3xz) j + (6z^5 - 3xy) k
    gradG(x,y,z) = (y+z) i + (x+z) j + (y+x - 24z^7) k

    The correct answers are: f'(1) = 4, g'(1) = 7, x = 1 + 4s, y = 1 + 7s.

    Any help in solving this problem would be greatly appreciated!
    Thanks in advance!
     
  2. jcsd
  3. Feb 22, 2012 #2

    Dick

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    You don't want to set the gradients equal. If you take their cross product you'll get a tangent vector to the intersection curve. Any idea why? And how can you use that to solve your problem?
     
  4. Feb 22, 2012 #3

    s3a

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    I know that taking the cross product will give a vector perpendicular to each gradient vector and I can picture it in my mind geometrically. As for how that will help me solve my problem, I'm guessing I should take the x,y, and z components and make them functions of t and then plug in t = 1 but I don't know how to proceed after the cross product step (algebraically speaking).

    Here is the vector obtained from the cross product:
    [(2y - 3xz)(y+x - 24z^7) - (x+z)(y+x - 24z^7)]i - [(4x^3 + 3yz)(y+x - 24z^7) - (6z^5 - 3xy)(y+z)]j + [(4x^3 + 3yz)(x+z) - (2y - 3xz)(y+z)]k
     
  5. Feb 22, 2012 #4

    Dick

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    Put x=1, y=1 and z=1 in before you take the cross product. That will make your life much easier!
     
  6. Feb 22, 2012 #5

    s3a

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    Oh, right. Lol!

    so it's: 16 i + 160 j + 16 k ?

    z = t, so t = 16 = x, so y = 10t. But these aren't correct. What am I doing wrong?
     
    Last edited: Feb 22, 2012
  7. Feb 22, 2012 #6

    Dick

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    That's not what I get. What are the two gradients at (1,1,1)? I've got to confess I didn't check your gradients. And I think there are some problems there. Like shouldn't the i component of grad(F) be 4x^3 - 3yz?
     
  8. Feb 22, 2012 #7

    s3a

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    You're right but do I have another mistake?

    I now get the cross product as:
    16 i + 28j + 4k
     
  9. Feb 22, 2012 #8

    Dick

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    Ok, that's what I get. So stop worrying about that part. Any thoughts on where to go from there? That's a tangent to the intersection curve.
     
  10. Feb 22, 2012 #9

    s3a

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    I had an idea but it seems wrong.

    My idea was:
    I have the tangent vector to the curve. I'm assuming the curve has one variable when (being t) so having the tangent is having the derivative or something like that such that x = f'(t) = 16 but that's not the case.
     
  11. Feb 22, 2012 #10

    Dick

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    If the curve is parametrized by (f(t),g(t),t) then the tangent vector is the derivative of that at t=1. That has to be parallel to (16,28,4), it doesn't have to equal it.
     
  12. Feb 22, 2012 #11

    s3a

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    I have the same problem with different equations and I am getting it wrong and I checked twice and I am wondering if I got it right by fluke before or not.

    Curves:
    x^8 + y^4 + z^5 - 3xyz = 0
    and
    xy + yz + zx - 3z^4 = 0

    where everything else is the same.

    I got:

    T = -14 i + 54 j + 8k but that's not the answer. It's parallel like you said if I'm correct but I can't divide each component by 8.
     
  13. Feb 22, 2012 #12

    Dick

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    You can't change the whole problem and expect to get the same answer. You have (f'(1),g'(1),1) is parallel to (16,28,4). So (f'(1),g'(1),1)=k*(16,28,4). What is k?
     
  14. Feb 22, 2012 #13

    s3a

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    k = 4
     
  15. Feb 22, 2012 #14

    Dick

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    I don't think so. 4*4 isn't 1. It's close though.
     
    Last edited: Feb 22, 2012
  16. Feb 22, 2012 #15

    s3a

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    Oh, I saw the k multiplying the other vector.

    k = 1/4 ?
     
  17. Feb 22, 2012 #16

    Dick

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    Yes, k=1/4. So f'(1) and g'(1) are?
     
  18. Feb 22, 2012 #17

    s3a

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    4 and 7. I had successfully gotten that though ( thanks to you so thank you :) ) but for the same problem with different equations, I can't shrink the x,y and z components such that the z component is 1.
     
  19. Feb 22, 2012 #18

    Dick

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    Different equations are different problems and have different answers. I don't know what you are asking now.
     
  20. Feb 22, 2012 #19

    s3a

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    I now solved it. I took the gradients and found the tangent vector from the cross product to be:

    T = -14 i + 54 j + 8k

    and

    then I should of divided each component by 8 but being in the sleepy state I am in, I thought I had to have each component be an integer which is obviously not the case.

    Thanks for all your help!
     
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