Problem 199 - Calculus Volume 3

Question

Let \(z=f(x, y)=x^{2}+3 x y-y^{2}\). Find the exact change in the function and the approximate change in the function as \(x\) changes from 2.00 to 2.05 and \(y\) changes from 3.00 to 2.96 .

Step-by-Step Solution

Answer

Exact change: 0.6689; Approximate change: 0.65.

Step 1: Evaluate the Initial Function Value

First, substitute the initial values of \(x = 2.00\) and \(y = 3.00\) into the function \(z = f(x, y) = x^2 + 3xy - y^2\). Calculate the initial value: \[\begin{align*} z_0 & = (2.00)^2 + 3 \times 2.00 \times 3.00 - (3.00)^2 \ & = 4 + 18 - 9 = 13.\end{align*}\] Thus, the initial function value \(z_0\) is 13.

Step 2: Evaluate the New Function Value

Substitute the new values of \(x = 2.05\) and \(y = 2.96\) into the function. Calculate the new value: \[\begin{align*} z_1 & = (2.05)^2 + 3 \times 2.05 \times 2.96 - (2.96)^2 \ & = 4.2025 + 18.228 - 8.7616 = 13.6689.\end{align*}\] Thus, the new function value \(z_1\) is approximately 13.6689.

Step 3: Calculate the Exact Change in the Function

The exact change in the function is found by subtracting the initial function value from the new function value: \[\Delta z = z_1 - z_0 = 13.6689 - 13 = 0.6689.\] Therefore, the exact change in the function is approximately 0.6689.

Step 4: Calculate the Partial Derivatives

Find the partial derivatives of the function with respect to \(x\) and \(y\): \(f_x(x, y) = 2x + 3y\), \(f_y(x, y) = 3x - 2y\).

Step 5: Evaluate the Partial Derivatives at Initial Values

Substitute \(x = 2.00\) and \(y = 3.00\) into the partial derivatives: \[f_x(2.00, 3.00) = 2 \times 2.00 + 3 \times 3.00 = 4 + 9 = 13\] and \[f_y(2.00, 3.00) = 3 \times 2.00 - 2 \times 3.00 = 6 - 6 = 0.\] Thus, \(f_x(2.00, 3.00) = 13\) and \(f_y(2.00, 3.00) = 0\).

Step 6: Calculate the Approximate Change in the Function

Use the linear approximation formula: \[\Delta z \approx f_x(2.00, 3.00) \cdot \Delta x + f_y(2.00, 3.00) \cdot \Delta y,\] where \(\Delta x = 2.05 - 2.00 = 0.05\) and \(\Delta y = 2.96 - 3.00 = -0.04\). Substitute in the values: \[\Delta z \approx 13 \times 0.05 + 0 \times (-0.04) = 0.65.\] Thus, the approximate change in the function is 0.65.

Key Concepts

Function Evaluation

Function evaluation is a fundamental step in calculus, especially when working with multivariable functions such as those involving multiple variables like \(x\) and \(y\). In this context, evaluating a function is the process of substituting given values of \(x\) and \(y\) into the function to find its output.

• Initially, we consider the function \(z = f(x, y) = x^2 + 3xy - y^2\). Using initial values, \(x = 2.00\) and \(y = 3.00\), we substitute and compute \(z_0 = 13\).
• For the new conditions where \(x = 2.05\) and \(y = 2.96\), we calculate the updated value \(z_1 = 13.6689\).

Function evaluation helps us understand how the function behaves at specific points. It's a crucial step for determining changes and analyzing the impact of modifications in the input values.

Linear Approximation

Linear approximation is a method used in calculus to estimate how a function changes as its variables change slightly. It's based on the function's partial derivatives and is particularly useful when exact computation is complex or unnecessary.We use the formula:\[\Delta z \approx f_x \cdot \Delta x + f_y \cdot \Delta y\]where:

• \(f_x\) and \(f_y\) are partial derivatives at initial points.
• \(\Delta x\) and \(\Delta y\) represent small changes in \(x\) and \(y\).

In this example, at \(x = 2.00\) and \(y = 3.00\), the partial derivatives are \(f_x = 13\) and \(f_y = 0\). The approximate change calculated was \(0.65\), capturing the change due to \(\Delta x = 0.05\) and \(\Delta y = -0.04\). This method gives us a quick way to gauge the function's behavior in response to small input changes.

Exact Change Calculation

Calculation of the exact change in a function involves subtracting the initial function value from its new value after changes in input.

• The exact change is computed using \(\Delta z = z_1 - z_0\).
• In our case, substituting \(z_0 = 13\) and \(z_1 = 13.6689\), results in \(\Delta z = 0.6689\).

This numerical difference represents how much the function's output has altered due to the changes in the variables. It's used for accuracy comparison against approximation methods.

Multivariable Calculus

Multivariable calculus extends the concepts of calculus to functions of several variables. It involves tackling problems where more than one input variable affects the outcome.

• Partial derivatives are crucial as they provide the rate of change with respect to one variable while keeping others constant.
• In our problem, \(f_x(x, y) = 2x + 3y\) and \(f_y(x, y) = 3x - 2y\) describe how changes in \(x\) or \(y\) independently affect \(z\).

This field is essential for modeling and solving real-world problems that involve multiple dimensions. Understanding how individual variables impact the function helps in predicting and optimizing outcomes.