Problem 199 - Calculus 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

The exact change is 0.6929, while the approximate change is 0.65.

Step 1: Calculate initial and final function values

First, determine the function values at the initial and final points. The initial point is \((x_0, y_0) = (2.00, 3.00)\) and the final point is \((x_1, y_1) = (2.05, 2.96)\). Calculate \(z_0 = f(x_0, y_0)\) and \(z_1 = f(x_1, y_1)\).\[ z_0 = f(2.00, 3.00) = (2.00)^2 + 3(2.00)(3.00) - (3.00)^2 = 4 + 18 - 9 = 13 \]\[ z_1 = f(2.05, 2.96) = (2.05)^2 + 3(2.05)(2.96) - (2.96)^2 \]Calculate the specific values:\[ z_1 = 4.2025 + 18.252 - 8.7616 = 13.6929 \]

Step 2: Find the exact change in the function

The exact change in the function \(\Delta z_{exact}\) is calculated by subtracting the initial function value from the final function value.\[ \Delta z_{exact} = z_1 - z_0 = 13.6929 - 13 = 0.6929 \]

Step 3: Compute partial derivatives

Compute the partial derivatives of the function \(z = f(x, y)\). These are necessary to estimate the change in the function:\[ \frac{\partial z}{\partial x} = 2x + 3y \]\[ \frac{\partial z}{\partial y} = 3x - 2y \]Evaluate these partial derivatives at the initial point \((2.00, 3.00)\):\[ \frac{\partial z}{\partial x}(2.00, 3.00) = 2(2.00) + 3(3.00) = 4 + 9 = 13 \]\[ \frac{\partial z}{\partial y}(2.00, 3.00) = 3(2.00) - 2(3.00) = 6 - 6 = 0 \]

Step 4: Calculate approximate change in the function

Use the differential \(dz\) to approximate the change in the function:\[ dz = \frac{\partial z}{\partial x}(2.00, 3.00) \cdot \Delta x + \frac{\partial z}{\partial y}(2.00, 3.00) \cdot \Delta y \]where \(\Delta x = 2.05 - 2.00 = 0.05\) and \(\Delta y = 2.96 - 3.00 = -0.04\):\[ dz = 13 \cdot 0.05 + 0 \cdot (-0.04) = 13 \cdot 0.05 = 0.65 \]

Step 5: Compare exact and approximate changes

Compare the exact change \(\Delta z_{exact} = 0.6929\) to the approximate change \(dz = 0.65\) to understand how closely the linear approximation represents the actual change.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental tool in multivariable calculus. They are used to examine how a function changes with respect to one of its variables while keeping the other variables constant. In the function \( z = f(x, y) = x^2 + 3xy - y^2 \), the partial derivatives will tell us how \( z \) changes if we vary \( x \) or \( y \) independently.

To find a partial derivative, you differentiate the function with respect to one variable, treating all other variables as constants. For our function:

• The partial derivative with respect to \( x \) is \( \frac{\partial z}{\partial x} = 2x + 3y \). This shows that changes in \( x \) alone will influence \( z \) through this expression.
• The partial derivative with respect to \( y \) is \( \frac{\partial z}{\partial y} = 3x - 2y \). This indicates the effect of isolated changes in \( y \) on \( z \).

Evaluating these at the point \((2.00, 3.00)\) gives specific insights:

• \( \frac{\partial z}{\partial x}(2.00, 3.00) = 13 \)
• \( \frac{\partial z}{\partial y}(2.00, 3.00) = 0 \)

This means at this specific point, \( z \) increases by 13 units for every one unit increase in \( x \) while remaining unchanged for small changes in \( y \). This understanding is crucial for predicting changes in multivariable functions.

Exact Change in Functions

The exact change in a function gives us the precise difference in function values between two points. This is calculated using the original function itself and is essential for understanding real changes over specified intervals.

For a function \( z = f(x, y) \), when changing from an initial point \((x_0, y_0)\) to a final point \((x_1, y_1)\), the exact change \( \Delta z_{exact} \) is determined by subtracting the initial function value from the final function value.

• Given \( z_0 = f(2.00, 3.00) = 13 \)
• and \( z_1 = f(2.05, 2.96) = 13.6929 \)
• the exact change is \( \Delta z_{exact} = z_1 - z_0 = 0.6929 \)

This exact change tells us the precise amount by which the function's value increases or decreases as the input variables shift from the initial to the final values. It is beneficial in contexts where precise calculations of change are necessary, such as physics or engineering scenarios where accuracy is critical.

Approximate Change in Functions

The approximate change in a function provides a simplified estimate of the change in a function's value. This is achieved through linear approximations using the partial derivatives, and it is particularly useful for small changes.

Using the differential of the function \( dz \), we approximate the change:\[ dz = \frac{\partial z}{\partial x}(2.00, 3.00) \cdot \Delta x + \frac{\partial z}{\partial y}(2.00, 3.00) \cdot \Delta y \]
Given:

• \( \Delta x = 2.05 - 2.00 = 0.05 \)
• \( \Delta y = 2.96 - 3.00 = -0.04 \)
• \( \frac{\partial z}{\partial x}(2.00, 3.00) = 13 \)
• \( \frac{\partial z}{\partial y}(2.00, 3.00) = 0 \)

The calculation becomes \[ dz = 13 \times 0.05 + 0 \times (-0.04) = 0.65 \]
The approximate change \( dz = 0.65 \) gives us a close estimate of how the function behaves near the point \((2.00, 3.00)\). While not exact, this approximation is useful for quick calculations and gaining insights when precise values are not necessary.