To find the equation of the tangent plane, we first need to calculate the gradient vector of the function (f (x,y)) at the given point P(1,2). The gradient vector is composed of the partial derivatives of the function with respect to x and y. Thus,
V(f (x,y)) = [df/dx, df/dy]
We first differentiate f with respect to x, holding y constant (df/dx):
df/dx = 2x - 2y
And then differentiate f with respect to y, holding x constant (df/dy):
df/dy = 2y - 2x
Substitute x = 1 and y = 2 into df/dx and df/dy:
df/dx (1,2) = 2*1 - 2*2 = -2,
df/dy (1,2) = 2*2 - 2*1 = 2
So, the gradient V(f(1,2)) = [-2, 2]
Next, we define the normal vector to the plane. For a tangent plane, the normal vector is the gradient vector of the function at that point:
N = V(f(1,2)) = [-2, 2, -1]
We then form the equation of the tangent plane as (N . r) = 0, where r is the position vector [x - x0, y - y0, z - z0] and (.) denotes the dot product.
N . r = (-2)*(x - 1) + 2*(y - 2) -1*(z - 1) = 0
=> -2 x + 2 + 2y - 4 - z + 1 = 0
=> -2 x + 2y - z - 1= 0
Hence the equation of the tangent plane is -2 x + 2 y - z - 1= 0
Now, for the directional derivative, it helps us to determine the rate at which the function changes at a point P in the direction of a given vector u.
We calculate it as the dot product of the gradient vector and the direction vector u. Here, u is given as [0,1].
So, the directional derivative Duf(1,2) = Vf(1,2) . u = [-2, 2] . [0, 1] = 2
Therefore, the directional derivative of f at the point (1,2) in the direction of the vector u=(0,1) is 2.