Question

Find the directional derivative of f(x,y,z)=yx+z^4 at the point (3,2,1) in the direction of a vector making an angle of 2π/3 with ∇f(3,2,1).

Answer 1

To find the directional derivative of the function f(x,y,z)=yx+z^4 at the point (3,2,1) in the direction of a vector making an angle of 2π/3 with ∇f(3,2,1), we can use the formula: Directional derivative = ∇f(3,2,1) · u, where ∇f(3,2,1) is the gradient of the function at the given point and u is the unit vector in the given direction.

Step-by-step explanation:

To find the directional derivative of the function f(x,y,z)=yx+z^4 at the point (3,2,1) in the direction of a vector making an angle of 2π/3 with ∇f(3,2,1), we can use the formula:

Directional derivative = ∇f(3,2,1) · u

where ∇f(3,2,1) is the gradient of the function at the given point and u is the unit vector in the given direction.

First, we need to find ∇f(3,2,1), which is the gradient of the function at the point (3,2,1). The gradient is a vector that consists of the partial derivatives of the function with respect to each variable. In this case, the gradient is given by:

∇f(3,2,1) = (∂f/∂x, ∂f/∂y, ∂f/∂z)

= (y, x, 4z^3)

Substituting the given point (3,2,1) into the gradient, we have:

∇f(3,2,1) = (2, 3, 4)

Next, we need to find the unit vector u in the direction of a vector making an angle of 2π/3 with ∇f(3,2,1). To do this, we can use the formula for the unit vector:

u = (cos(2π/3), sin(2π/3))

= (-1/2, √3/2)

Finally, we can calculate the directional derivative:

Directional derivative = ∇f(3,2,1) · u

= (2, 3, 4) · (-1/2, √3/2)

= 2(-1/2) + 3(√3/2)

= -1 + 3√3/2

Therefore, the directional derivative of f(x,y,z)=yx+z^4 at the point (3,2,1) in the direction of a vector making an angle of 2π/3 with ∇f(3,2,1) is -1 + 3√3/2.

Answer 2

Step-by-step explanation:

You don't need to know u . Since your function is differentiable everywhere.

We get that..

(5,1,−4)=∇(5,1,−4)⋅||||Duf(5,1,−4)=∇f(5,1,−4)⋅u||u||

But you also know that..

12=cos3=⋅∇(5,1,−4)||||||∇(5,1,−4)||⟹12=cos⁡π3=u⋅∇f(5,1,−4)||u||||∇f(5,1,−4)||⟹∇(5,1,−4)⋅||||=12||∇(5,1,−4)||=…∇f(5,1,−4)⋅u||u||=12||∇f(5,1,−4)||=…

And there you go!
I hope this helps! UwU