Question

Find the directional derivative of f(x, y, z) 5y(x + z) = at P(4,1,-3)
in the direction from P to Q(3,2,-2).
NOTE: Enter the exact answer.

Answer 1

The directional derivative of a function f(x, y, z) at a point P in the direction of a unit vector u = <u1, u2, u3> is given by the dot product of the gradient of f at P and u:

Duf(P) = ∇f(P) · u

where ∇f(P) is the gradient of f at P.

To find the directional derivative of f(x, y, z) = 5y(x + z) at P(4, 1, -3) in the direction from P to Q(3, 2, -2), we need to first find the unit vector u in that direction:

u = Q - P / ||Q - P||

= <3 - 4, 2 - 1, -2 - (-3)> / ||<-1, 1, 1>||

= <-1/√3, 1/√3, -1/√3>

Next, we need to find the gradient of f at P:

∇f(x, y, z) = <∂f/∂x, ∂f/∂y, ∂f/∂z>

∂f/∂x = 5y, ∂f/∂y = 5(x + z), ∂f/∂z = 5y

∇f(4, 1, -3) = <5, 20, 5>

Finally, we can find the directional derivative of f at P in the direction from P to Q:

Duf(P) = ∇f(P) · u

= <5, 20, 5> · <-1/√3, 1/√3, -1/√3>

= (-5/√3) + (20/√3) + (-5/√3)

= 10/√3

Therefore, the exact value of the directional derivative of f(x, y, z) = 5y(x + z) at P(4, 1, -3) in the direction from P to Q(3, 2, -2) is 10/√3.