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Intersection of surface and tangent plane

  1. Feb 23, 2013 #1
    1. The problem statement, all variables and given/known data

    I have a surface given by z=x^2 - y^2 and its tangent plane at the point (x,y)=(1,1) given by z = 2x-2y. I am asked to compute the intersection of the tangent plane with the surface.

    3. The attempt at a solution

    I did the obvious and set x^2-y^2 = 2x -2y to find the x,y lying on their intersection. I moved all of the terms to the RHS and used the quadratic formula to obtain x=y and x+y=2.

    If x=y, then obviously z=0. I'm a little lost on what to do with the x+y=2. I know what the two graphs look like and I know what their intersection looks like, but I am trying to justify my answer.
     
  2. jcsd
  3. Feb 23, 2013 #2

    Simon Bridge

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    You have two equations and two unknowns - what would you normally do in that situation?
     
  4. Feb 23, 2013 #3

    Dick

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    Ok, so if x=y then z=0. x=y is a plane, z=0 is another plane. So that leads you to the solution that is the intersection of those two planes, which is the line (x,x,0) where x can have any value. Do something very similar with the x+y=2 solution. You'll get the that the solution is two lines, yes?
     
  5. Feb 24, 2013 #4
    Okay, if I take x^2-y^2=(x+y)(x-y)=z and plug in 2-x=y, I get 4-4y=z. If I do the same for 2-y=x, I get 4x-4=z. Adding these two equations together gives me 2x-2y=z. So if x+y=2, then 2x-2y=z. So we have the intersection as two lines formed by the intersections of x=y, z=0 and x+y=2, 2x-2y=z.
     
  6. Feb 24, 2013 #5

    Dick

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    Well, yes, ok. So the two lines of intersection are what? Describe them both.
     
  7. Feb 25, 2013 #6
    Sorry, I'm not 100% sure what you mean. Can we not describe them as the two lines formed by the intersection of the planes x=y and z=0 and the planes x+y=2 and 2x-2y=z?

    We could describe one line by (x, x, 0) and the other by (x, 2-x, 4-4x), where x may be chosen freely for both.

    Thank you for your help!
     
  8. Feb 25, 2013 #7

    Simon Bridge

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    <puzzled>
    Did I misread? According to post #1 the intersection satisfies:
    (1) x=y
    (2) x+y=2
    (3) z=0

    ... these relations are repeated in post #4
    ... why not just sub (1) into (2)?
     
  9. Feb 25, 2013 #8

    Dick

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    Sure, that's right.
     
  10. Feb 25, 2013 #9

    Dick

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    Those relations come from factoring x^2-y^2=2x-2y. (x-y)(x+y)=2(x-y). The solutions are x-y=0 OR x+y=2. Not necessarily both at the same time.
     
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