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Finding the intersection of an ellipsoid and a plane

  1. Nov 8, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the curve that is the intersection of x-y-z>-10 and x2+y2/4+z2/9=36.

    2. Relevant equations


    3. The attempt at a solution
    The best idea I have is to define x as x=y+z-10 and substitute it into the ellipsoid equation to get a function defined by y and z; the trouble is that leaves out the x.

    I need to define the curve explicitly in terms of three variables as I need to later take the gradient of the curve.

    Help appreciated. Thanks.
     
  2. jcsd
  3. Nov 8, 2015 #2

    Ray Vickson

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    You are finding the projection of the curve Cxyz onto the yz-plane (= curve Cyz). For any (y,z) in Cyz you get the point in Cxyz by putting back x = y+z+10.
     
  4. Nov 8, 2015 #3

    andrewkirk

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    What you suggested is fine. Leaving out the x doesn't matter.

    You can't take a gradient of a curve, as gradient is defined for a scalar field and a curve is not a scalar field. Do you mean that you want to calculate tangents to the curve (aka velocities)? If so then you will want your curve to be expressed in parametric form as a function ##\gamma:\mathbb{R}\to\mathbb{R}^3##.

    You can parametrise half the curve by doing the substitution you mentioned, to get an equation in y and z. Solve the equation for y in terms of z and then you can choose z to be the curve parameter for a half of the curve. The equation's solution will have a square root. Choosing the positive branch will give one half of the curve and the negative will give the other. The parametrisation of the half curve will have three equations: ##x=f(z),\ y=g(z),\ z=z##. You can turn this into a parametrisation of the full curve by introducing a parameter ##t## and using it to 'sew' the two branches together. But you may not need to go that far. It depends on what they want you to do.
     
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