How do you find the equations for the tangent plane to the surface #f(x,y)=2-2/3x-y# through (3,-1,1)?
2 Answers
Jul 6, 2018
Explanation:
First write as a level surface
#phi(bbx) = z-2+2/3x+y = 0#
The normal vector at any point is:
Plane is in form:
#(bbr - bbr_o)*bbn = 0 #
NB in case you are not familiar with the directional derivative , another way to find the normal vector is to take partial derivatives of
So
Then take the vector product of these tangent vectors:
#{((:1,0,- 2/3:), (larr f_x)),((:0, 1,-1:),(larr f_y)):}#
Jul 8, 2018
The tangent plane to the plane