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

Question

Evelyn Agard · Accepted Answer

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

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

The normal vector at any point is:

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

Plane is in form:

#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:

#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:)#

Emma Aldridge · Answer

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

Christopher Agresta · Answer

undefined To find the equation for the tangent plane to the surface f(x, y) = 2 - (2/3)x - y through the point (3, -1, 1), we can follow these steps:

1. Calculate the partial derivatives of f(x, y) with respect to x and y.
2. Evaluate the partial derivatives at the given point (3, -1).
3. Use the values obtained in step 2 to find the normal vector to the tangent plane.
4. Write the equation of the tangent plane using the normal vector and the given point.

Let's go through these steps:

1. The partial derivative of f(x, y) with respect to x is -2/3, and with respect to y is -1.

2. Evaluating the partial derivatives at (3, -1):
   f_x(3, -1) = -2/3
   f_y(3, -1) = -1

3. The normal vector to the tangent plane is given by the coefficients of the partial derivatives, which are (-2/3, -1, 1).

4. Using the point (3, -1, 1) and the normal vector (-2/3, -1, 1), we can write the equation of the tangent plane as:
   -2/3(x - 3) - (y + 1) + (z - 1) = 0

Simplifying the equation, we get:
   -2/3x - y + z + 2/3 + 1 - 1 = 0
   -2/3x - y + z + 2/3 = 0

Therefore, the equation for the tangent plane to the surface f(x, y) = 2 - (2/3)x - y through the point (3, -1, 1) is -2/3x - y + z + 2/3 = 0.