Finding the intersection of an ellipsoid and a plane

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SUMMARY

The discussion focuses on finding the intersection of the ellipsoid defined by the equation x² + y²/4 + z²/9 = 36 and the plane x - y - z > -10. The proposed method involves substituting x with y + z - 10 to express the curve in terms of y and z. It is clarified that the gradient cannot be taken of a curve, as gradients apply to scalar fields, and instead, tangents or velocities should be calculated. The solution suggests parametrizing the curve using a parameter t to combine both branches derived from the square root solution of the equation.

PREREQUISITES
  • Understanding of ellipsoids and their equations
  • Knowledge of parametric equations in three-dimensional space
  • Familiarity with calculus concepts such as gradients and tangents
  • Ability to manipulate algebraic equations and substitutions
NEXT STEPS
  • Learn about parametrizing curves in three-dimensional space
  • Study the properties of ellipsoids and their intersections with planes
  • Explore the concept of gradients and tangents in vector calculus
  • Investigate the use of square roots in solving parametric equations
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Conservation
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Homework Statement


Find the curve that is the intersection of x-y-z>-10 and x2+y2/4+z2/9=36.

Homework Equations

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.
 
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Conservation said:

Homework Statement


Find the curve that is the intersection of x-y-z>-10 and x2+y2/4+z2/9=36.

Homework Equations

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.

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

Conservation said:
I need to later take the gradient of the curve
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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