Line Tangent to following surface

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Homework Help Overview

The discussion revolves around finding the equation of a line tangent to the surface defined by the equation z=6-(4x^2)-(y^2) at a specific point (5,3,-103). The subject area includes concepts related to calculus and differential geometry, particularly the properties of tangent planes and lines in three-dimensional space.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to find the tangent plane to the surface and the relationship between the tangent line and the tangent plane. There are inquiries about the process of determining the normal vector and how it relates to the tangent plane.

Discussion Status

Some participants have provided guidance on the initial steps, such as finding the normal vector and understanding the concept of tangent planes. Multiple interpretations of how to approach the problem are being explored, but there is no explicit consensus on the method to proceed.

Contextual Notes

There is an emphasis on understanding the geometric implications of the problem, and participants are encouraged to share their attempts and specific points of confusion to facilitate more targeted assistance.

mopar969
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What is the solution to:
What is the equation of a line tangent to the following surface z=6-(4x^2)-(y^2) at the point (5,3,-103)
 
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hi mopar969! :wink:

show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
You ask for the equation of a line tangent to a surface. There will be a whole plane tangent to the surface and any line lying in that plane will be tangent to the surface. Do you know how to find the tangent plane to a surface?
 
I thought that maybe I need to find the normal vector first to start this problem but I am not sure please help.
 
that would certainly do it! :smile:
 
If a surface is described by f(x,y,z)= constant, then [itex]\nabla f[/itex] is perpendicular to the surface.
 
Let z = f(x,y). z is "three-dimensional" so its tangent must be "two-dimensional".

T: z -z0 = . . . you finish the rest.
 

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