Problem 193 - Calculus 3

Question

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 is differentiable since \(\varepsilon_1 = \Delta x\) and \(\varepsilon_2 = 0\) both approach zero as \((\Delta x, \Delta y) \rightarrow (0,0)\).

Step 1: Understand Differentiability Criteria

For the function \(f(x, y) = x^2 + 3y\) to be differentiable at every point, the change in \(z\) when \((x, y)\) changes by small amounts \((\Delta x, \Delta y)\) must be expressible as \(f_x \Delta x + f_y \Delta y + \varepsilon_1 \Delta x + \varepsilon_2 \Delta y\), where both \(\varepsilon_1\) and \(\varepsilon_2\) approach zero as \((\Delta x, \Delta y)\) tends to \((0, 0)\).

Step 2: Find Partial Derivatives

Compute the partial derivatives \(f_x\) and \(f_y\) for the function. For \(f(x, y) = x^2 + 3y\):- The partial derivative with respect to \(x\) is \(f_x = 2x\).- The partial derivative with respect to \(y\) is \(f_y = 3\).

Step 3: Calculate Expression for \(\Delta z\)

Using the function \(f(x, y)\), calculate \(f(x + \Delta x, y + \Delta y)\):\[\begin{align*}f(x + \Delta x, y + \Delta y) &= (x + \Delta x)^2 + 3(y + \Delta y) \&= x^2 + 2x \Delta x + (\Delta x)^2 + 3y + 3 \Delta y.\end{align*}\]Then, calculate \(\Delta z\) as follows:\[\Delta z = f(x + \Delta x, y + \Delta y) - f(x, y) = 2x \Delta x + (\Delta x)^2 + 3 \Delta y.\]

Step 4: Match with Differentiability Formula

Write the expression for \(\Delta z\) in terms of \(f_x\) and \(f_y\):- \(\Delta z = 2x \Delta x + 3 \Delta y + (\Delta x)^2\).Identify the remaining terms:- According to the criteria, the term \((\Delta x)^2\) can be split as: - \((\Delta x)^2 = \varepsilon_1 \Delta x + \varepsilon_2 \Delta y\), where: - \(\varepsilon_1 = \Delta x\) and \(\varepsilon_2 = 0\).Make sure both \(\varepsilon_1\) and \(\varepsilon_2\) approach zero as \((\Delta x, \Delta y) \rightarrow (0,0)\).

Step 5: Verify \(\varepsilon_1\) and \(\varepsilon_2\) Tend Toward Zero

Examine \(\varepsilon_1 = \Delta x\) and \(\varepsilon_2 = 0\):- \(\varepsilon_1 = \Delta x\) obviously approaches zero as \(\Delta x\) tends towards zero.- \(\varepsilon_2 = 0\) is already zero and remains unaffected by \((\Delta x, \Delta y)\).Thus, both are confirmed to approach zero as required.

Key Concepts

Partial Derivatives

Partial derivatives help us understand how a function changes with respect to each variable individually. For a function of two variables, like our example function, the partial derivative with respect to one variable measures how the function changes when only that variable changes, while keeping the other one constant.

For the function \(f(x, y) = x^2 + 3y\), the partial derivatives are found as follows:

• The partial derivative with respect to \(x\), denoted as \(f_x\), is \(2x\). This tells us that the rate of change of the function in the \(x\) direction is \(2x\).
• The partial derivative with respect to \(y\), denoted as \(f_y\), is \(3\). This implies that the function changes at a constant rate of 3 as \(y\) changes.

By calculating these derivatives, you can determine how changes in each variable \(x\) or \(y\) affect the function \(f(x, y)\), which is a key concept in multivariable calculus.

Multivariable Calculus

Multivariable calculus extends the techniques of single-variable calculus to functions of more than one variable. This area of mathematics is crucial for understanding complex systems where various factors interact simultaneously.

A function of two variables, like \(f(x, y) = x^2 + 3y\), depends on both \(x\) and \(y\). The behavior of such a function is not just about the paths that variables \(x\) or \(y\) can individually take, but about the surface described by these variables together.

Concepts like gradients, directional derivatives, and partial derivatives are important tools in multivariable calculus that provide us with comprehensive insights into these functions. The partial derivatives we calculated in the previous section are one such tool, providing a snapshot of the function's behavior with changes in individual dimensions. This captures the essence of multivariable calculus, centering around its ability to analyze and predict the behavior of functions across multidimensional spaces.

Epsilon-Delta Definition

In calculus, the epsilon-delta definition is a formal framework that defines what it means for a function to be continuous or differentiable. In the context of multivariable functions, we say a function is differentiable at a point \((x, y)\) if small changes in the input result in changes in the output that can be closely approximated by a linear function.

For the function \(f(x, y) = x^2 + 3y\), the definition of differentiability involves showing that the difference \( \Delta z = f(x + \Delta x, y + \Delta y) - f(x, y) \) can be expressed in terms of its partial derivatives and error terms. The error terms, denoted as \(\varepsilon_1\) and \(\varepsilon_2\), must approach zero as \((\Delta x, \Delta y)\) approaches \((0, 0)\).

We derived that \(\Delta z = 2x \Delta x + 3 \Delta y + (\Delta x)^2 \), and identified \((\Delta x)^2\) as an error term. By letting \(\varepsilon_1 = \Delta x\) and \(\varepsilon_2 = 0\), both of which clearly approach zero as \(\Delta x\) and \(\Delta y\) approach zero, we confirm the differentiability of the function across all points. This epsilon-delta definition helps in rigorously proving how closely a tangent plane at a point approximates the function's behavior near that point.