Problem 28 - APEX Calculus

Question

Form a function \(z=f(x, y)\) such that \(f_{x}\) and \(f_{y}\) match those given. $$ f_{x}=x+y, \quad f_{y}=x+y $$

Step-by-Step Solution

Answer

The function is \( f(x, y) = \frac{x^2}{2} + xy + \frac{y^2}{2} + K \).

Step 1: Integrate with respect to x

To form the function \( f(x, y) \), we start by integrating \( f_x = x + y \) with respect to \( x \). The integral of \( x + y \) with respect to \( x \) is \( \frac{x^2}{2} + xy + C(y) \), where \( C(y) \) is an arbitrary function of \( y \).

Step 2: Differentiate with respect to y

Differentiate the function \( f(x, y) = \frac{x^2}{2} + xy + C(y) \) with respect to \( y \) to find \( f_y \). This result should match the given \( f_y = x + y \). Differentiating gives \( f_y = x + C'(y) \).

Step 3: Solve for C(y)

Set \( f_y = x + C'(y) = x + y \). Equating, we find \( C'(y) = y \). Integrate \( C'(y) \) with respect to \( y \) to find \( C(y) = \frac{y^2}{2} + K \), where \( K \) is a constant.

Step 4: Form the complete function f(x, y)

Substitute \( C(y) = \frac{y^2}{2} + K \) back into the function. We get:\[ f(x, y) = \frac{x^2}{2} + xy + \frac{y^2}{2} + K \]. This is the required function \( z = f(x, y) \).

Key Concepts

Partial Derivatives

Partial derivatives are essential in multivariable calculus as they help us understand how a function changes with respect to one variable while keeping others constant. In the context of the given exercise, we look at a function of two variables, say \( f(x, y) \). The partial derivative of the function concerning \( x \), represented as \( f_x \), tells us how the function \( f \) changes as \( x \) changes while keeping \( y \) constant.
This is similar to the role of \( f_y \), which shows how \( f \) changes as \( y \) changes.
The exercise presents \( f_x = x + y \) and \( f_y = x + y \), meaning each variable influences the function in the same linear manner.

• This expression indicates that for any small increase in \( x \) or \( y \), the function \( f(x, y) \) grows linearly.
• Partial derivatives provide a powerful way to approach optimizing and understanding multivariable functions, particularly when designing surfaces or models that change based on multiple factors.

Integration with Respect to a Variable

Integration plays a crucial role when reconstructing a function from its derivatives, as showcased in the exercise. Here, we begin with the partial derivative \( f_x = x + y \), indicating how the function \( f(x, y) \) changes as \( x \) changes. To find the original function \( f(x, y) \), we integrate \( x + y \) with respect to \( x \).
Performing this integration yields:

• When integrating \( x \), we obtain \( \frac{x^2}{2} \).
• Similarly, integrating a constant \( y \) with respect to \( x \) gives us \( xy \).

Thus, the integral expression becomes \( \frac{x^2}{2} + xy + C(y) \), where \( C(y) \) is some function of \( y \) that appears as an integration constant.
This method enables us to form a tentative function that matches the derivative with respect to \( x \) and is the initial step to rediscover the full function \( f(x, y) \).

Differentiation with Respect to a Variable

After integrating with respect to one variable, we differentiate the resultant expression with respect to another variable. In this problem, we have differentiated the expression obtained from integrating \( f_x \) with respect to \( y \), to ensure it aligns with the provided \( f_y = x + y \).
To do this:

• We differentiate \( \frac{x^2}{2} + xy + C(y) \) with respect to \( y \).
• The differentiation of \( xy \) with respect to \( y \) gives \( x \).
• For \( C(y) \), differentiation turns it into \( C'(y) \).

This gives us \( f_y = x + C'(y) \). We compare this with the given \( f_y = x + y \) and solve for \( C'(y) = y \).
Finally, we integrate \( C'(y) = y \) back with respect to \( y \) to identify \( C(y) = \frac{y^2}{2} + K \), completing the function as \( f(x, y) = \frac{x^2}{2} + xy + \frac{y^2}{2} + K \).
This step ensures our function model is accurate and satisfies the provided differential relationships.