Finding parametric equations of a tangent line

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Homework Help Overview

The discussion revolves around finding the parametric equations of a tangent line at a specific point on the curve defined by the intersection of a surface and a plane. The subject area includes multivariable calculus and vector analysis.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the use of gradients to determine normal directions for the surfaces involved and question how to find a tangent vector that is perpendicular to these normals. There is uncertainty about the correct application of the gradient and the cross product in this context.

Discussion Status

The discussion is ongoing, with participants attempting to clarify their understanding of gradients and normal vectors. Some guidance has been offered regarding the relationship between normals and tangents, but there is no explicit consensus on the approach to take.

Contextual Notes

Participants express confusion about the definitions and applications of gradients and vectors, indicating a need for further clarification on these concepts.

grog
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Homework Statement



Find parametric equations of the tangent line at the point (-2,2,4) to the curve of intersection of the surface z=2x2-y2 and the plane z=4

Homework Equations



Not sure


The Attempt at a Solution



Not sure quite how to approach this. take the gradient of 2x^2-y^2 and just plug for x=rcos\Theta and y=r sin\Theta ?

That seems too simple..
 
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The gradient gives you a normal direction for each surface. Then the tangent is perpendicular to both normals. How can you find a vector perpendicular to two other vectors?
 
so I would end up with 4x-2y and zero for the two normal vectors? and then take the cross product of the two? I think I may be confusing some concepts here.
 
grog said:
so I would end up with 4x-2y and zero for the two normal vectors? and then take the cross product of the two? I think I may be confusing some concepts here.

Probably. The gradient is a vector. Those don't look like vectors. Better check the definition of 'gradient'.
 

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