Problem 9 - Applied Calculus

Question

Find the partial derivatives in problems. The variables are restricted to a domain on which the function is defined. \(f_{x}\) and \(f_{y}\) if \(f(x, y)=5 x^{2} y^{3}+8 x y^{2}-3 x^{2}\)

Step-by-Step Solution

Answer

\( f_x = 10xy^3 + 8y^2 - 6x \), \( f_y = 15x^2y^2 + 16xy \)

Step 1: Identify the Function

The given function is defined as \( f(x, y) = 5x^2y^3 + 8xy^2 - 3x^2 \). We need to find the partial derivatives \( f_x \) and \( f_y \).

Step 2: Differentiate with Respect to x

To find the partial derivative \( f_x \), differentiate the function \( f(x, y) = 5x^2y^3 + 8xy^2 - 3x^2 \) with respect to \( x \), treating \( y \) as a constant:- The derivative of \( 5x^2y^3 \) with respect to \( x \) is \( 10xy^3 \).- The derivative of \( 8xy^2 \) with respect to \( x \) is \( 8y^2 \).- The derivative of \( -3x^2 \) with respect to \( x \) is \( -6x \).Thus, the partial derivative \( f_x = 10xy^3 + 8y^2 - 6x \).

Step 3: Differentiate with Respect to y

To find the partial derivative \( f_y \), differentiate the function \( f(x, y) = 5x^2y^3 + 8xy^2 - 3x^2 \) with respect to \( y \), treating \( x \) as a constant:- The derivative of \( 5x^2y^3 \) with respect to \( y \) is \( 15x^2y^2 \).- The derivative of \( 8xy^2 \) with respect to \( y \) is \( 16xy \).- The derivative of \( -3x^2 \) with respect to \( y \) is \( 0 \) since it is independent of \( y \).Thus, the partial derivative \( f_y = 15x^2y^2 + 16xy \).

Key Concepts

Multivariable Calculus

Multivariable calculus is an extension of the regular calculus you might be familiar with, extending to functions of more than one variable. It essentially deals with functions that have two or more variables affecting their behavior.
Examples include functions like temperature varying in a room based on location, or pressure changes in a fluid.

• In multivariable calculus, you analyze how these variables interact and affect the overall outcome of the function.
• Understanding this interaction is crucial for predicting the behavior of complex systems in fields like physics and engineering.
• The main goal is to ascertain how changes in one variable impact the entire function.

For instance, if you have a function like \( f(x, y) = 5x^2y^3 + 8xy^2 - 3x^2 \), you employ partial derivatives to explore how alterations in either \( x \) or \( y \) will affect the function's output.
By computing the partial derivatives \( f_x \) and \( f_y \), you can determine how the function changes as each variable shifts individually while keeping the other constant.

Differentiation

Differentiation is a fundamental concept in calculus that refers to finding the rate at which a function changes at any given point. Specifically, it concerns how a function's output changes in response to changes in its inputs.
When dealing with multivariable calculus, differentiation becomes more interesting as you deal with partial derivatives.

• A partial derivative is the derivative of a function with several variables, with respect to one variable while keeping the other variables constant.
• For the function \( f(x, y) = 5x^2y^3 + 8xy^2 - 3x^2 \), finding the partial derivative with respect to \( x \), denoted as \( f_x \), involves treating \( y \) as a constant.
• Similarly, when finding the partial derivative with respect to \( y \), denoted as \( f_y \), \( x \) is treated as a constant.

The beauty of differentiation in this context lies in its ability to give insights into how each variable individually contributes to changes in the function, offering a clearer picture of the function's dynamic behavior.

Calculus Functions

Calculus functions are at the heart of both single-variable and multivariable calculus. These functions describe the relationship between inputs and outputs, representing real-world phenomena.
In multivariable contexts, calculus functions can involve several independent variables.

• The function \( f(x, y) = 5x^2y^3 + 8xy^2 - 3x^2 \) is an example, showing a relationship between two variables, \( x \) and \( y \).
• Functions can be linear, like \( ax + by + c \), or nonlinear, like the given example, involving powers and products of \( x \) and \( y \).
• Understanding these functions allows you to model and predict complex systems across various fields, from physical sciences to economics.

By applying techniques like partial differentiation, you can uncover a wealth of information about how these variables interact, enabling you to solve intricate problems more effectively. Calculus functions thus provide a framework for precisely defining and manipulating such relationships.