A differentiable function f(x,y) has the property that f(2,2)=2 and fx(2,2)=1 and fy(2,2)=2. Find the equation of the tangent plane at the point on the surface z=f(x,y) where x=2, y=2.

for f(x,y,z) = 0,

the plane is

fx(2,2)(x-2) + fy(2,2)(y-2) + fz(2,2)(z-2) = 0

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

To find the equation of the tangent plane at a point on the surface z = f(x, y), we need the values of the function f(x, y) and its partial derivatives at the specific point.

Given information:
f(2, 2) = 2
f_x(2, 2) = 1
f_y(2, 2) = 2

The equation of the tangent plane at the point (2, 2, f(2, 2)) will be of the form:

z - z_0 = A(x - x_0) + B(y - y_0)

where (x_0, y_0, z_0) represents the point on the surface.

Using the given values, we have:
x_0 = 2
y_0 = 2
z_0 = f(2, 2) = 2

Substituting these values, the equation becomes:
z - 2 = A(x - 2) + B(y - 2)

Now, we need to determine the values of A and B. For this, we'll use the partial derivatives:

f_x(2, 2) = 1
f_y(2, 2) = 2

The values of A and B are actually the partial derivatives of z with respect to x and y, evaluated at the given point.

Therefore:
A = f_x(2, 2) = 1
B = f_y(2, 2) = 2

Now, we can substitute these values into the equation of the tangent plane:

z - 2 = 1(x - 2) + 2(y - 2)

Simplifying further:
z - 2 = x - 2 + 2y - 4
z - 2 = x + 2y - 6
z = x + 2y - 4

So, the equation of the tangent plane at the point (2, 2) on the surface z = f(x, y) is z = x + 2y - 4.

To find the equation of the tangent plane at a point on a surface, we need to use the concept of the gradient vector. The gradient vector (∇f) of a function f(x, y) is a vector that points in the direction of the steepest ascent of the function at a given point. It is given by (∇f) = (fₓ, fᵧ), where fₓ represents the partial derivative of f with respect to x, and fᵧ represents the partial derivative of f with respect to y.

In this case, we are given that f(2, 2) = 2, fₓ(2, 2) = 1, and fᵧ(2, 2) = 2. So, we have the values of the function and its partial derivatives at the point (2, 2).

The equation of the tangent plane at the point (2, 2, f(2, 2)) on the surface z = f(x, y) can be written as:

z - f(2, 2) = fₓ(2, 2)(x - 2) + fᵧ(2, 2)(y - 2)

Substituting the given values, we have:

z - 2 = 1(x - 2) + 2(y - 2)

Simplifying further, we get:

z - 2 = x - 2 + 2y - 4

Combining like terms, we have:

z = x + 2y - 4

Therefore, the equation of the tangent plane at the point on the surface z=f(x,y) where x=2, y=2 is z = x + 2y - 4.