66.2k views
5 votes
The graph of a function f is shown below. Find an equation of the tangent plane to the surface z=f(x,y) at the specified point. f(x,y)=26−x²−y²

User Secretive
by
7.6k points

1 Answer

6 votes

The equation for the tangent plane to the surface z=26−x²−y² at the point (2,1,23) is 4x + 2y - z = 15.

The equation for a tangent plane of an explicitly defined surface z=f(x,y) at the point (a,b,c) is:

f_x(x - a) + f_y(y - b) + (z - c) = 0

where f_x and f_y are the partial derivatives of f with respect to x and y, evaluated at the point (a,b).

In your case, the surface is defined as z=26−x²−y² and the point of interest is (2,1,23).

Therefore, we need to find f_x and f_y, evaluate them at (2,1), and plug those values into the equation above.

Here are the steps:

Find f_x and f_y:

f_x = -2x

f_y = -2y

Evaluate f_x and f_y at (2,1):

f_x(2,1) = -4

f_y(2,1) = -2

Plug the values into the equation for the tangent plane:

-4(x - 2) - 2(y - 1) + (z - 23) = 0

Simplifying the equation, we get:

-4x + 8 - 2y + 2 + z - 23 = 0

Rearranging the terms, we get the final equation for the tangent plane:

4x + 2y - z = 15

Therefore, the equation for the tangent plane to the surface z=26−x²−y² at the point (2,1,23) is 4x + 2y - z = 15.

The graph of a function f is shown below. Find an equation of the tangent plane to-example-1
User Ante
by
7.6k points