Problem 43 - Advanced Engineering Mathematics

Question

Find a function \(f\) such that $$ \nabla f=\left(3 x^{2}+y^{3}+y e^{x}\right) \mathbf{i}+\left(-2 y^{2}+3 x y^{2}+x e^{x}\right) \mathbf{j} $$

Step-by-Step Solution

Answer

The function is \( f(x, y) = x^3 + xy^3 + ye^x - \frac{2}{3}y^3 + C \).

Step 1: Understand the Gradient Vector

The gradient vector \( abla f \) is given as \( (3x^2 + y^3 + ye^{x})\mathbf{i} + (-2y^2 + 3xy^2 + xe^{x})\mathbf{j} \). This implies that the partial derivative of \( f \) with respect to \( x \) is \( \frac{\partial f}{\partial x} = 3x^2 + y^3 + ye^{x} \) and with respect to \( y \) is \( \frac{\partial f}{\partial y} = -2y^2 + 3xy^2 + xe^{x} \).

Step 2: Integrate with Respect to x

To find \( f \), integrate \( \frac{\partial f}{\partial x} = 3x^2 + y^3 + ye^{x} \) with respect to \( x \). This results in \( f(x, y) = x^3 + xy^3 + ye^{x} + g(y) \), where \( g(y) \) is an arbitrary function of \( y \).

Step 3: Differentiate with Respect to y

Differentiate the result of the integration with respect to \( y \): \( \frac{\partial f}{\partial y} = 3x y^2 + xe^{x} + g'(y) \).

Step 4: Equate to Given y Derivative

Set the expression from Step 3 equal to the given partial derivative with respect to \( y \): \( 3xy^2 + xe^{x} + g'(y) = -2y^2 + 3xy^2 + xe^{x} \).

Step 5: Solve for g'(y)

Simplifying the equation from Step 4, we find \( g'(y) = -2y^2 \).

Step 6: Integrate g'(y)

To find \( g(y) \), integrate \( g'(y) = -2y^2 \) with respect to \( y \), resulting in \( g(y) = -\frac{2}{3}y^3 + C \), where \( C \) is a constant.

Step 7: Formulate the Function f(x, y)

Substitute \( g(y) \) back into the expression for \( f \): \( f(x, y) = x^3 + xy^3 + ye^x - \frac{2}{3}y^3 + C \). This is the function \( f \).

Key Concepts

Partial Derivatives

In mathematics, partial derivatives help us understand how a function changes as its input variables change one at a time, while keeping the other variables constant. Consider a function of two variables, like in the exercise, where we have a function \( f(x, y) \). Here, the partial derivative with respect to \( x \), written as \( \frac{\partial f}{\partial x} \), describes how \( f(x, y) \) changes as \( x \) changes, with \( y \) held constant. Similarly, \( \frac{\partial f}{\partial y} \) tells us about changes in \( y \) while \( x \) is constant.

**Importance of Partial Derivatives**:

• They help in finding the slope of surfaces described by multi-variable functions.
• They're crucial in understanding gradient vector fields, as they define the components of these fields.
• In optimization problems, partial derivatives are used to find local maxima and minima.

In this exercise, we use partial derivatives to configure gradients through integration techniques. They outline the first step in reconstructing the function \( f(x, y) \) from its gradient vector field.

Integration

Integration is a fundamental concept in calculus used to find functions when given a rate of change, such as a derivative. It reverses the process of differentiation. To integrate a function like \( \frac{\partial f}{\partial x} = 3x^2 + y^3 + ye^x \) with respect to \( x \), we determine a function \( f(x, y) \) whose derivative with respect to \( x \) is the given function.

**Key Steps in Integration**:

• Identify the variable of integration (in this case, \( x \)). Keep other variables constant.
• Find the antiderivative for each term involving the integration variable.
• Introduce an arbitrary function of constants (in this case, \( g(y) \)) after integration. This accounts for any part of the function not dependent on the variable of integration.

In this task, after integrating, we calculated \( f(x, y) = x^3 + xy^3 + ye^x + g(y) \). The function \( g(y) \) retains terms not dependent solely on \( x \), which we further determine by analyzing the given partial derivative \( \frac{\partial f}{\partial y} \).

Vector Calculus

Vector calculus is a branch of mathematics that deals with vectors and operations applicable to them, such as dot products and cross products. One crucial concept is the gradient vector, which is like a multi-variable extension of a derivative.

**Understanding Gradient Vectors**:

• The gradient vector \( abla f \) represents the rate and direction of the fastest increase of the function \( f(x, y) \).
• This gradient comprises partial derivatives. For example, \( abla f = (3x^2 + y^3 + ye^x) \mathbf{i} + (-2y^2 + 3xy^2 + xe^x) \mathbf{j} \).
• In practical terms, the gradient vector points toward the steepest slope of the function's graph.

In exercises like this, finding the function \( f \) from its gradient involves working backwards using integration and ensuring all parts of the equation fit consistently to reconstruct \( f(x, y) \). This illustrates the interconnectedness of calculus concepts when dealing with vector fields.

Functions of Two Variables

Functions of two variables rely on both inputs to produce an output and are a natural extension from single-variable functions. In two-variable functions, such as \( f(x, y) \), both \( x \) and \( y \) influence the output value. These functions are often visualized as surfaces in a three-dimensional space.

**Features of such functions include:**

• They can be differentiated with respect to each variable, revealing how changes in each variable individually impact the output.
• These functions are crucial for modeling real-world phenomena where outcomes depend on multiple factors, such as temperature and pressure in physics.
• They provide a rich framework for optimization problems and are foundational in fields like economics and engineering.

In this problem, reconstructing \( f(x, y) \) involves recognizing that changes in the gradient correspond to changes in the output of a two-variable function. This understanding helps us derive \( f(x, y) = x^3 + xy^3 + ye^x - \frac{2}{3}y^3 + C \) from its gradient.