Finding eqn of tangent plane without eqn of surface

In summary, to find the tangent plane of a surface at a given point without the equation of the surface, you can use the two given curves on the surface to find two vectors that lie on the surface. By taking the cross product of these two vectors, you can find the normal vector to the tangent plane at the given point. This normal vector can then be used to define the equation of the tangent plane.
  • #1

Homework Statement


I need to find the tangent plane of a surface S at a point P without being given the eqn of the surface. I am also given that two curves lie on this surface


Homework Equations


Point P: (2,1,3)
Curve 1: <2+3t, 1 - (t^2), 3 - 4t + (t^2)>
Curve 2: <1+ (u^2), 2(u^3) - 1, 2u+1>


The Attempt at a Solution

 
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  • #2
Here's an outline.

- To find the equation of the plane you need to find the direction of the normal vector.

- Use the two given surface curves to find two vectors that lie in (are parallel to) the surface.

- Given the above two vectors it should be easy to find a vector perpendicular to both.
 
  • #3
So after I've found direction of tangent line to the two curves I cross them to find a normal vector to plane which i use to define the plane. But this final plane I define, is it the equation of the tangent plane through P already or is it only the equation of the surface?
 
  • #4
dispiriton said:
So after I've found direction of tangent line to the two curves I cross them to find a normal vector to plane which i use to define the plane. But this final plane I define, is it the equation of the tangent plane through P already or is it only the equation of the surface?
It's the equation of the tangent plane.
 
  • #5
Assuming both those curves pass through the point of interest (easily checked) then you simply want to find the tangent vector of both curves at said point. Both these vectors will lie in the tangent plane.

From there simply take the cross product of the two vectors to get the normal vector and you'll have your plane!
 

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