Consider the function f(x,y)=xy+xz+yz+4 at point p=(2,-1,1)

a)find the unit vector in direction of p.
b) find the directional derivative at p in the direction of <0,-1/(sqrt2),-1/(sqrt2)>

|p| = √(4+1+1) = √6

u= p/|p| = (2/√6,-1/√6,1/√6)

Now, with u = <0,-1/√2,-1/√2)>
∇<sub<uf = ∇f•u
= (y+z)(0) + (x+z)(-1/√2) + (x+y)(-1/√2)
= (0)(0) - 1/√2 (2+1) - 1/√2 (2-1)
= -4/√2 = -2√2

Oh crap typo error, should read find the unit vector in direction of maximum increase of f in the direction of P.

To find the unit vector in the direction of point p=(2,-1,1), we need to divide each component by the magnitude of the vector. Here's how you can calculate it:

a) Find the magnitude of p:
The magnitude of a vector (a,b,c) is given by the formula sqrt(a^2 + b^2 + c^2). In this case, the magnitude of p is sqrt(2^2 + (-1)^2 + 1^2) = sqrt(4 + 1 + 1) = sqrt(6).

b) Divide each component of p by its magnitude:
The unit vector in the direction of p is given by (2/sqrt(6), -1/sqrt(6), 1/sqrt(6)).

To find the directional derivative at point p in the direction of <0, -1/(sqrt2), -1/(sqrt2)>, we'll use the formula for directional derivatives:

D_u(f) = ∇f · u

where ∇f is the gradient vector of f, and u is the unit vector in the desired direction. Let's calculate it step by step:

1) Calculate the gradient vector ∇f:
The gradient of a function f(x, y, z) is given by the vector (∂f/∂x, ∂f/∂y, ∂f/∂z).

In this case, f(x,y,z) = xy + xz + yz + 4, so:
∂f/∂x = y + z
∂f/∂y = x + z
∂f/∂z = x + y

Therefore, the gradient vector ∇f = (y + z, x + z, x + y).

2) Calculate the dot product of ∇f and u:
The dot product of two vectors (a, b, c) and (d, e, f) is given by a*d + b*e + c*f.

In this case, ∇f · u = (y + z)*(0) + (x + z)*(-1/(sqrt2)) + (x + y)*(-1/(sqrt2)).
Simplifying, we get ∇f · u = -(x + y + z)/(sqrt2).

Therefore, the directional derivative at point p in the direction of <0, -1/(sqrt2), -1/(sqrt2)> is -(x + y + z)/(sqrt2).
Substituting the values of p=(2,-1,1), we get the directional derivative as -(2 - 1 + 1)/(sqrt2) = 0/(sqrt2) = 0.