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Curve of intersection of sufaces, find f'(1), g'(1), x, and y. (Answers included).

  1. Mar 21, 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 values of the derivatives f'(1), g'(1)?
    (b) What is the tangent line to the curve of intersection at (1,1,1) given that z = 1 + s? (Find what x and y are equal to.)

    Answers:
    f'(1) = 4
    g'(1) = 7
    x = 1 + 4s
    y = 1 + 7s

    2. Relevant equations
    At least gradients, and partial derivative taking.


    3. The attempt at a solution
    I took the gradients of each surface (the first one being f(x,y,z) and the second one being g(x,y,z) respectively). I tried to set the respective components equal to each other in an attempt to solve for something but I ended nowhere useful. I also tried to think of z = t = 1 and in order to try and find x and y. Then, I was like "Oh, wait, I have a point (1,1,1), let me just shove it into the gradients." all in order to get the correct answers and then to try and make sense of what I did but I didn't succeed :(.

    Any help in solving this problem would be greatly appreciated!
    Thanks in advance!
     
  2. jcsd
  3. Mar 27, 2012 #2

    s3a

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    Re: Curve of intersection of sufaces, find f'(1), g'(1), x, and y. (Answers included)

    Is what I wrote confusing or something? If it is, please tell me what to change that way I could get an answer.
     
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