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Projection of intersection line

  1. Apr 30, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the projection of the intersection between the two surfaces

    S1: z = 4-x^2 - y^2 and
    S2: 4x^2y = 1 (x>0)

    in the xy-plane

    2. The attempt at a solution

    4-x^2 - y^2 = 4x^2y -1

    Is this all I need to do?
  2. jcsd
  3. Apr 30, 2008 #2


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    What is it that you did?
  4. Apr 30, 2008 #3
    Equalised the equations, with z = 0.
  5. Apr 30, 2008 #4


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    (You can't set two equations equal... but I think I know what you meant)
    (And you were looking for a geometric object, not an equation... but I assume you meant to say the object described by that equation)

    Does that process have anything to do with finding the intersection of the two surfaces? Or with projecting that intersection to the x-y plane? If so, then you're done. If not, then you're not.
  6. Apr 30, 2008 #5
    The problem is:

    How steep is the path C (the intersection between the curves) at the point S over the (1, 1/4)?

    My plan is to find the directional derivative in the direction of the tangent of the projection of C in the xy-plane. What do I do next then?
  7. May 1, 2008 #6


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    What you did initially was replace z in the first equation by the left side of the second equation. But there is no "z" in the second equation- in particular, it was NOT z= 4x^2y
    Since 1= 4x^2y, x^2= 1/(4y). Replacing x^2 by that in the first equation, z= 4- x^2- y^2= 4- 1/(4y)+ y^2. Now project by letting z= 0: y^2- 1/(4y)+ 4= 0.

    However, there really is no reason to project to the xy-plane. What you want is the derivative of z= 4- x^2- y^2 in the direction of the tangent to the curve x^2y= 1 in the xy-plane. What is the tangent vector to that curve? What is the derivative of z in the direction of that vector?
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