Question

How do you find the equations for the tangent plane to the surface `f(x,y)=2-2/3x-y` through (3,-1,1)?

Answer 1

`2 x +3 y + 3z = 6`

Explanation:

First write as a level surface `phi(bbx)`, with `z = f(x,y):`

• `phi(bbx) = z-2+2/3x+y = 0`

The normal vector at any point is:

`bbn = nabla phi = (: phi_x, phi_y, phi_z :) = (: 2/3,1,1:)`

Plane is in form:

• `(bbr - bbr_o)*bbn = 0`

`implies (:x-3 ,y+1, z-1 :) * (:2/3,1,1:) = 0`

`2/3 x - 2 + y + 1 + z - 1 = 0`

`:. 2 x +3 y + 3z = 6`

NB in case you are not familiar with the directional derivative , another way to find the normal vector is to take partial derivatives of `f(x,y)`

So `f_x = - 2/3 qquad f_y = - 1`

Then take the vector product of these tangent vectors:

• `{((:1,0,- 2/3:), (larr f_x)),((:0, 1,-1:),(larr f_y)):}`

`bbn = det [(hat x, haty, hatz),( 1,0,- 2/3),(0, 1,-1)]`

`= hat x(2/3) - hat y (- 1) + hat z (1) = (: 2/3,1,1:)`

Answer 2

The tangent plane to the plane `z = 2 - (2/3)x - y` is itself.