Directional derivative of a function of two variables (continued)

 Goals:

1. Define directional derivative as an expression of partial ,derivatives

2. Define "gradient"

Directional derivative as expression of partial derivatives

Theorem

Let z = f(x, y) a function of two variables x and y. Let's assume fₓ and fy exist and f is differentiable everywhere. Then the directional derivative of f in the direction of u = cosθi + sinθj is given by:

Dᵤf(x, y) =  fₓ(x,y) cosθ + fy(x,y)sinθ (1)

Example

Let θ = arccos(3/5). Find the directional derivative Dᵤf(x,y) of the function f(x, y) = x² - xy + 3y² in the direction of u = cos(θ)i + sin (θ)j,. What is Df(-1, 2)?

Solution

In order to apply the formula above, we must calculate the partial derivatives:

 

Let's now apply the formula. Let's notice that this example is the same as the example in the previous post where we had cosθ = 3/5 and )sinθ = 4/5.

Let's calculate  Dᵤf(-1, 2)?

Practice

Find the directional derivative Dᵤf(x,y) of 

What is Dᵤf(3,4)?

Gradient

The right hand side of equation (1) can be written as the dot product of two vectors. The first vector can be written as 

(2)

The second vector can be written as: 

 

Then the right hand side of equation (1) can be written as:

The first vector is called gradient of f. The symbol of the inversed delta is called "nabla" 

Definition

Let z = f(x,y) be a function of two variables x and y such that fx and fy exist. 

Example

Solution

a. Let's first calculate the partial derivatives in order to apply the formula of the gradient:

b. Let's do the same:

Practice