Problem 30 - APEX Calculus

Question

Form a function \(z=f(x, y)\) such that \(f_{x}\) and \(f_{y}\) match those given. $$ f_{x}=\frac{2 x}{x^{2}+y^{2}}, \quad f_{y}=\frac{2 y}{x^{2}+y^{2}} $$

Step-by-Step Solution

Answer

Integrate \( f_x \) w.r.t. \( x\) and verify with \( f_y\). Function is \( z = \ln(x^2 + y^2)\).

Step 1: Integrate with Respect to x

To find the function \( z=f(x, y) \), start by integrating \( f_{x} = \frac{2x}{x^2 + y^2} \) with respect to \( x \). This results in:\[\int \frac{2x}{x^2 + y^2} \, dx = \ln(x^2 + y^2) + g(y)\]where \( g(y) \) is an arbitrary function of \( y \), which acts as a constant with respect to \( x \).

Step 2: Differentiate the Result with Respect to y

Differentiate \( z = \ln(x^2 + y^2) + g(y) \) with respect to \( y \) to ensure it matches \( f_{y} \).\[\frac{\partial z}{\partial y} = \frac{2y}{x^2 + y^2} + g'(y)\]

Key Concepts

Partial Derivatives

Understanding partial derivatives is essential when dealing with multivariable functions, like those of two variables, say \(x\) and \(y\). A partial derivative represents how a function changes as one of the variables changes, while the other variables are held constant.
This is a key component in multivariable calculus.In this context, the given expressions \(f_{x}\) and \(f_{y}\) are partial derivatives of a function \(f(x, y)\). The notation \(f_{x}\) represents the partial derivative of \(f\) with respect to \(x\), which is how much \(f\) changes if \(x\) changes slightly, while \(y\) remains fixed. Similarly, \(f_{y}\) is the partial derivative with respect to \(y\).

• Not dependent on each other: Each partial derivative only manipulates one variable at a time.
• The Goal: By knowing \(f_{x}\) and \(f_{y}\), we reconstruct the original function \(z=f(x, y)\).

This involves procedures like integration and differentiation, which are topics that go hand-in-hand in exploring multivariable calculus.

Function Integration

Function integration, in the multivariable context, involves finding the original function when given its partial derivatives. Beginning with the partial derivative \(f_{x} = \frac{2x}{x^2 + y^2}\), integration with respect to \(x\) helps retrieve part of the function \(z = f(x, y)\). This process is called \'integrating with respect to \(x\)\'.
During integration, variables that are not being directly integrated are treated as constants. In this case, \(y\) acts as a constant, which means a result of integration includes an arbitrary function \(g(y)\).

• Finding the Integral: When integrating \(\frac{2x}{x^2 + y^2}\), it simplifies to \(\ln(x^2 + y^2) + g(y)\).
• Role of \(g(y)\): Since \(g(y)\) depends only on \(y\), it represents the integration constant for the \(x\)-integration, still holding a potential derivative in terms of \(y\).

Ensuring the function representation is valid for both inputs \(x\) and \(y\), we proceed with differentiating the integrated result by \(y\) to discover \(g'(y)\).

Arbitrary Function

The concept of an arbitrary function is pivotal when performing integration with partial derivatives in multivariable calculus. An arbitrary function, like \(g(y)\) in our exercise, embodies the constant of integration but depends on the ignored variable, \(y\), during the initial integration.
This function acts as a placeholder for changes that might occur due to that ignored variable. It ensures the new expression continues to match other partial derivatives.

• Flexibility and Functionality: By allowing \(g(y)\), the function can account for variation in \(y\) that wasn’t addressed during the integration by \(x\).
• Subsequent Steps: After finding \(z = \ln(x^2 + y^2) + g(y)\), differentiating with respect to \(y\) integrates this arbitrary function \(g(y)\). Comparing this result with the known \(\frac{2y}{x^2 + y^2}\), the equation \(g'(y) = 0\) reveals that \(g(y)\) is constant.

Thus, \(g(y)\) ensures the reconstructed function \(z = f(x, y)\) aligns with both partial derivatives provided, demonstrating its importance in accurate integration within multivariable calculus.