Problem 193 - Calculus Volume 3

Question

Complete each task. Show that \(f(x, y)=x^{2}+3 y\) is differentiable at every point. In other words, show that \(\Delta z=f(x+\Delta x, y+\Delta y)-f(x, y)=f_{x} \Delta x+f_{y} \Delta y+\varepsilon_{1} \Delta x+\varepsilon_{2} \Delta y\) where both \(\varepsilon_{1}\) and \(\varepsilon_{2}\) approach zero as \((\Delta x, \Delta y)\) approaches (0,0) .

Step-by-Step Solution

Answer

The function \(f(x, y) = x^2 + 3y\) is differentiable at every point.

Step 1: Expression Setup

The function given is \( f(x, y) = x^2 + 3y \). We need to show that the change in function value \( \Delta z = f(x+\Delta x, y+\Delta y) - f(x, y) \) can be expressed in the form \( f_x \Delta x + f_y \Delta y + \varepsilon_1 \Delta x + \varepsilon_2 \Delta y \).

Step 2: Expansion of Function

First, substitute \( x + \Delta x \) and \( y + \Delta y \) into the function: \[ f(x+\Delta x, y+\Delta y) = (x + \Delta x)^2 + 3(y + \Delta y). \]

Step 3: Simplification

Expand the expression: \[ (x + \Delta x)^2 = x^2 + 2x\Delta x + (\Delta x)^2 \] and \[ 3(y + \Delta y) = 3y + 3\Delta y. \] Thus, the function becomes: \[ f(x + \Delta x, y + \Delta y) = x^2 + 2x\Delta x + (\Delta x)^2 + 3y + 3\Delta y. \]

Step 4: Calculate \(\Delta z\)

Compute \(\Delta z\): \[ \Delta z = (x^2 + 2x\Delta x + (\Delta x)^2 + 3y + 3\Delta y) - (x^2 + 3y). \]This simplifies to \[ \Delta z = 2x\Delta x + (\Delta x)^2 + 3\Delta y. \]

Step 5: Identify Partial Derivatives

Find the partial derivatives: \( f_x = \frac{\partial f}{\partial x} = 2x \) and \( f_y = \frac{\partial f}{\partial y} = 3 \). Thus, we have \( f_x \Delta x + f_y \Delta y = 2x\Delta x + 3\Delta y. \)

Step 6: Express as Required Form

Rewrite the equation to compare: \[ \Delta z = 2x\Delta x + 3\Delta y + (\Delta x)^2. \]Let \( \varepsilon_1 = \Delta x \) and \( \varepsilon_2 = 0 \). The expression becomes \( \Delta z = f_x \Delta x + f_y \Delta y + \varepsilon_1 \Delta x + \varepsilon_2 \Delta y. \)

Step 7: Limit Validation

Verify that both \( \varepsilon_1 \) and \( \varepsilon_2 \) tend to zero as \( (\Delta x, \Delta y) \to (0,0) \). Since \( \varepsilon_1 = \Delta x \) and \( \varepsilon_2 = 0 \), as \( \Delta x \to 0 \) and \( \Delta y \to 0 \), both conditions hold true.

Key Concepts

Partial derivatives

Partial derivatives are a way to measure how a multivariable function changes as one of its variables is altered, while the others are held constant. In simpler terms, it describes the rate of change of a function with respect to each individual variable. For the function given, \( f(x, y) = x^2 + 3y \), we find the partial derivatives:

• With respect to \( x \): \( f_x = \frac{\partial f}{\partial x} = 2x \)
• With respect to \( y \): \( f_y = \frac{\partial f}{\partial y} = 3 \)

This tells us that for every unit increase in \( x \), \( f(x, y) \) increases by \( 2x \), and for every unit increase in \( y \), \( f(x, y) \) increases by 3. Identifying partial derivatives is crucial for understanding differentiability and is a fundamental step in solving such problems.

Limit

Limits help us handle values approaching a particular point, even if they never actually reach that point. In the context of differentiability, they allow us to explore how a function behaves as changes in the inputs become very small. For this problem, we are interested in \( \lim_{(\Delta x, \Delta y) \to (0,0)} \varepsilon_1 \Delta x + \varepsilon_2 \Delta y \). As the changes \( \Delta x \) and \( \Delta y \) reduce to zero, the terms \( \varepsilon_1 \) and \( \varepsilon_2 \) should also approach zero in a way that ensures continuity and differentiability. This aligns with understanding the continuity and smoothness of the function \( f(x, y) \). The property of limits guarantees that our function satisfies the required conditions for differentiability wherever the limit holds true.

Function expansion

Function expansion allows us to express a function in terms of its variables with slightly altered values. This method helps us comprehend how a function diverges or resembles a simpler form when each variable is incremented by a small amount, \( \Delta x \) and \( \Delta y \). In this exercise: We substitute \( x + \Delta x \) and \( y + \Delta y \) into \( f(x, y) \).

• Compute the squared term: \( (x + \Delta x)^2 = x^2 + 2x\Delta x + (\Delta x)^2 \)
• Add the linear term adjustment: \( 3(y + \Delta y) = 3y + 3\Delta y \)

The expansion yields: \( x^2 + 2x\Delta x + (\Delta x)^2 + 3y + 3\Delta y \). This expansion is critical to breaking down and evaluating how each segment contributes to the function's total change \( \Delta z \). It is particularly useful in crafting expressions that demonstrate differentiability.

Epsilon-delta definition

The epsilon-delta definition is a formal way of proving that a function is continuous (and thus, differentiable) at a certain point. This approach involves showing that for every small value \( \varepsilon \) there exists another small value \( \delta \) such that changes in the input \( \Delta x, \Delta y \) affect the change in output \( \Delta z \) in a controlled manner. In this problem, we need \( \varepsilon_1 \) and \( \varepsilon_2 \) to approach zero as \( \Delta x \) and \( \Delta y \) tend to zero.

• Consider \( \varepsilon_1 = \Delta x \) and \( \varepsilon_2 = 0 \).
• As \( \Delta x \to 0 \) and \( \Delta y \to 0 \), both \( \varepsilon_1 \Delta x \) and \( \varepsilon_2 \Delta y \) effectively become zero.

By fulfilling these conditions, we demonstrate that the change in function value \( \Delta z \) remains within acceptable bounds for every small \( \varepsilon \), making \( f(x, y) \) differentiable at any point \((x, y)\). This foundational idea is a cornerstone of calculus, ensuring that the behavior of functions is consistent and predictable.