Vector Function of Cone & Plane Intersection Curve

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SUMMARY

The discussion focuses on finding a vector function that represents the intersection curve of the cone defined by the equation z = sqrt(x^2 + y^2) and the plane described by z = 1 + y. Two parameterizations are explored: using x as the parameter, resulting in x = t, y = (t^2 - 1)/2, and z = (t^2 + 1)/2; and using y as the parameter, yielding y = t, x = sqrt(2t + 1), and z = t + 1. It is concluded that while both parameterizations are valid, the second one only represents part of the curve, as it does not account for negative x values, which requires including x = -sqrt(2t + 1) for a complete representation.

PREREQUISITES
  • Understanding of vector functions and parameterization in multivariable calculus.
  • Familiarity with the equations of cones and planes in three-dimensional space.
  • Knowledge of solving equations involving square roots and quadratic expressions.
  • Ability to analyze and interpret parametric equations and their geometric implications.
NEXT STEPS
  • Explore the concept of parametric equations in three-dimensional geometry.
  • Study the implications of parameterization choices on the representation of curves.
  • Learn about the geometric properties of conic sections and their intersections with planes.
  • Investigate the use of implicit functions and their applications in multivariable calculus.
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Students studying multivariable calculus, mathematicians interested in geometric interpretations, and educators teaching vector functions and surface intersections.

Litcyb
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Homework Statement



Find a vector function that represents the curve of intersection of the two surfaces:
The cone z = sqrt( x^2 + y^2) and the plane z = 1+y.

Homework Equations



z = sqrt( x^2 + y^2) and the plane z = 1+y.

The Attempt at a Solution


This problem can be solved as following using x as the parameter.
x^2+y^2 = z^2 = (1+y)^2 = 1+2y+y^2. => x^2 = 1 + 2y.

x=t; y = (t^2-1)/2; z = 1+(t^2-1)/2 = (t^2+1)/2

My question is, what if we use y as the parameter,

i get ,

y=t, x=(2t+1)^(1/2) z=t+1,

is this answer also correct?
 
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Litcyb said:

Homework Statement



Find a vector function that represents the curve of intersection of the two surfaces:
The cone z = sqrt( x^2 + y^2) and the plane z = 1+y.

Homework Equations



z = sqrt( x^2 + y^2) and the plane z = 1+y.

The Attempt at a Solution


This problem can be solved as following using x as the parameter.
x^2+y^2 = z^2 = (1+y)^2 = 1+2y+y^2. => x^2 = 1 + 2y.

x=t; y = (t^2-1)/2; z = 1+(t^2-1)/2 = (t^2+1)/2

My question is, what if we use y as the parameter,

i get ,

y=t, x=(2t+1)^(1/2) z=t+1,

is this answer also correct?

Sure, with the proviso that it only represents part of the curve. For example, points with negative values of x won't appear in the second parametrization (where you also should specify t>=(-1/2)). You'd need the x=(-(2t+1)^(1/2)) solution as well to get them all.
 

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