Problem 33 - Calculus and its applications

Question

Find \(f_{x x}, f_{x y}, f_{y x},\) and \(f_{y y}\). $$f(x, y)=2 x-3 y$$

Step-by-Step Solution

Answer

The answers are \( f_{xx} = 0 \), \( f_{xy} = 0 \), \( f_{yx} = 0 \), and \( f_{yy} = 0 \).

Step 1: Differentiate with respect to x

The given function is \( f(x, y) = 2x - 3y \). To find the first partial derivative with respect to \( x \), we treat \( y \) as a constant and differentiate.\[ f_x = \frac{\partial}{\partial x}(2x - 3y) = 2 \]

Step 2: Differentiate with respect to y

Now we differentiate the function with respect to \( y \), treating \( x \) as a constant.\[ f_y = \frac{\partial}{\partial y}(2x - 3y) = -3 \]

Step 3: Find the second partial derivative with respect to x

We already have \( f_x = 2 \). Now, differentiate \( f_x = 2 \) again with respect to \( x \). Since \( f_x \) is a constant, its derivative with respect to \( x \) is 0.\[ f_{xx} = \frac{\partial}{\partial x}(2) = 0 \]

Step 4: Mix partial derivative with respect to x and y

Differentiate \( f_x = 2 \) with respect to \( y \). Since \( f_x \) is a constant, its derivative with respect to \( y \) is 0.\[ f_{xy} = \frac{\partial}{\partial y}(2) = 0 \]

Step 5: Mix partial derivative with respect to y and x

Differentiate \( f_y = -3 \) with respect to \( x \). Since \( f_y \) is a constant, its derivative with respect to \( x \) is 0.\[ f_{yx} = \frac{\partial}{\partial x}(-3) = 0 \]

Step 6: Find the second partial derivative with respect to y

Now, differentiate \( f_y = -3 \) again with respect to \( y \). Since \( f_y \) is a constant, its derivative with respect to \( y \) is 0.\[ f_{yy} = \frac{\partial}{\partial y}(-3) = 0 \]

Key Concepts

Calculus

Calculus is a branch of mathematics that helps us understand how things change. It comprises two main components: differentiation and integration. Differentiation deals with the rate at which quantities change. In simpler terms, it's about how one thing changes in relation to another at any given instant.
For functions of a single variable, differentiation gives us the slope of the tangent to the curve at any point. But, what if we have a function that depends on more than one variable? This is where multivariable calculus comes into play, extending these ideas to functions with more than one input or variable.
Here's a quick rundown of some key points:

• Calculus enables us to calculate velocities and rates of change. It's essentially the mathematics of movement and change.
• It has applications ranging from physics and engineering to economics and biology, making it a cornerstone for scientific research.
• The derivative in calculus determines how a function is changing at any point in its domain, being fundamental for optimization problems.

Multivariable Functions

When discussing multivariable functions, we're dealing with functions that have more than one input. For example, the function given in the exercise, \( f(x, y) = 2x - 3y \), depends on both \( x \) and \( y \). These types of functions require special techniques because traditional calculus methods for single-variable functions don't directly apply.
Partial derivatives come into play to help break down the complex changes occurring in multivariable functions. In partial differentiation:

• We differentiate the function with respect to one variable while keeping the others constant, just as done in Step 1 and Step 2 of the solution.
• This approach helps us see how the function changes with one variable at a time.

This special technique provides a pathway to understanding the behavior of functions influenced by multiple factors, making it highly valuable in fields such as physics, economics, and engineering.

Second Order Derivatives

Second order derivatives help us understand how the rate of change itself changes. They provide deeper insights into the behavior of functions, such as concavity and acceleration for curves.
In the exercise solution, we see second order partial derivatives with respect to each variable. Here's what that means:

• Second order derivatives, like \( f_{xx} \), tell us how the slope of the function changes as \( x \) changes. A second derivative being zero, as in our problem, indicates nothing is changing in this context.
• Mixed partial derivatives, \( f_{xy} \) and \( f_{yx} \), show how the rate of change in one direction is influenced by changes in another direction. In this case, they also evaluate to zero, confirming symmetry.

Understanding these derivatives not only predicts how functions behave but also confirms properties like the equality of mixed partials under certain conditions, known as Schwarz's theorem. This knowledge is crucial for economists, engineers, and scientists delving into systems with interdependent variables.