Review of function 2 variables ( derivatives)

 Partial derivatives

Tangent plane and linear approximation

Review the following posts:

Derivatives of a function of two variables

Calculating partial derivatives of a function of two variables without using limit

Partial derivative value using a contour map

Tangent plane to a surface at a point

Linear approximation for function of two variables

Notion of differential of a function of two variables

Absolute maxima and minima of a function of two variables

 Objective: Find the absolute maxima and minima of a function of two variables.

Absolute maxima and minima

To find the absolute maxima and minima for a function of one variable on a closed interval, we nee need the find the critical points over that interval and evaluate the function at the endpoints of the interval. In a function of two variables, the interval is replaced by a closed boundary set. A set is bounded if all of its points are found inside of a ball or a disk. In order to find the absolute extrema, we start by calculating the critical points on the boundary set and the corresponding critical values of the function f.  Then we find the maximum and minimum value of the function at the boundary set. The highest function value is the absolute maximum and the smallest function value is the absolute minimum. Before calculating these values, we need to assure that there is a possibility that they exist. The following theorem gives this assurance.

Theorem

Since we now know that a continuous function function f(x,y) defined on a closed bounded set attains its extreme values, we need to know how to find them. the following theorem allows this.

Theorem

Problem solving strategy

Finding the maximum and minimum value at the boundary set can be challenging. If the boundary set is a rectangle or a set of straight lines it is possible to parameterize the line segments and determine the maxima on each of these segments. The same approach can be used for other shapes such as circles and ellipses.

Example

Use the problem-solving strategy to find the absolute extrema of a function to determine the absolute extrema of the following function:

 

Solution

Practice

Determination of the local extremum and the saddle point of a function of two variables

 Local and global extremum in a function of two variables

The extremum of a function is either a minimum or a maximum. The purpose of determining a critical point is to determine relative maxima or minima of a function. In a function of a single variable, we determine an interval around the critical point where the value of the function for that critical point is greater or less than all the values of the function in the chosen interval. For a function of two variables we do the same consideration with the exception that the interval is a disk.

Definition 

Saddle point. Definition

Second Derivative Test. Theorem

Example

Find the critical point of the critical points of the following function. Use the second derivative test to find the local extrema.

Solution

Practice

Critical points in a function of two variables

 Objective: Use partial derivatives to locate critical points for a function of two variables

Critical points. Definition

Examples

Solution

Next, we set each of these expressions equal to zero:

Therefore x = 2 and y = -3, (2, -3) is a critical point of f.

We must also check for the possibility that the denominator of each partial derivative can be equal to zero. In this case, the partial derivative doesn't exist. Since the denominator is the same in both partial derivatives, we need to do this once.

This equation represents also an hyperbola. We should also note that the domain of consists of points satisfying the inequality:

Therefore, any points on the hyperbola are not only critical points, they are also at the boundary of the domain. Let's put the equation of the hyperbola in standard form by completing the square:

Notice that (2, -3) is the center of the hyperbola.

We make each partial derivative equal to zero which give a system of equations with x and y. 

Subtracting the second equation from the first gives: 10 y + 10 = 0  y= -1. Substiuting y in the first

equation gives 2x -2 + 4 = 0, 2x + 2 = 0, x = -1. Therefore (-1, -1) is a critical point of the given function. There are no points in R² that make either partial derivative not to exist since both of them are defined for any point (x, y).

Practice

 

Maximum directional derivative

Goal: Find a maximum directional derivative 

To find a maximum directional derivative, we use one of the properties of the gradient. This property states as follows:

Other properties of the gradient

1) 

2) 

Example

Find the direction for which the directional derivative of f(x,y) = 3x² - 4xy + 2y² at (-2, 3) is a maximum, What is the maximum value?

Solution

Practice

Find the direction for which the directional derivative of f(x,y) = 4x - xy + 2y² is a maximum at (-2,3). What is the maximum value?

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