Problem 20 - APEX Calculus

Question

In Exercises \(19-22,\) use the information provided and the total differential to make the given approximation. $$ \begin{array}{l} f(-4,2)=13, f_{x}(-4,2)=2.6, f_{y}(-4,2)=5.1 . \text { Ap- } \\ \text { proximate } f(-4.12,2.07) . \end{array} $$

Step-by-Step Solution

Answer

The approximate value is 13.045.

Step 1: Identify the Known Values

From the problem, we know the following values: \( f(-4, 2) = 13 \), \( f_x(-4, 2) = 2.6 \), \( f_y(-4, 2) = 5.1 \). These represent the function value and partial derivatives at the point \((-4, 2)\).

Step 2: Determine the Changes in Variables

Calculate the changes in \(x\) and \(y\) from the given points. Let \( \Delta x = -4.12 - (-4) = -0.12 \) and \( \Delta y = 2.07 - 2 = 0.07 \).

Step 3: Use The Total Differential Formula

The total differential \(df\) is given by \( df = f_x \cdot \Delta x + f_y \cdot \Delta y \). Substitute the known values to find \( df = 2.6 \cdot (-0.12) + 5.1 \cdot 0.07 \).

Step 4: Calculate the Total Differential

Calculate each term: \( 2.6 \cdot (-0.12) = -0.312 \) and \( 5.1 \cdot 0.07 = 0.357 \). Thus, \( df = -0.312 + 0.357 = 0.045 \).

Step 5: Approximate the Function Value

The approximation of \(f(-4.12, 2.07)\) is given by \( f(-4,2) + df \). Therefore, \( f(-4.12, 2.07) \approx 13 + 0.045 = 13.045 \).

Key Concepts

Partial Derivatives

In the context of functions with multiple variables, partial derivatives are essential tools. They measure how a function changes as one particular variable changes while keeping other variables constant. This helps in understanding the behavior of a function near a specific point.

For example, if we have a function \( f(x, y) \), the partial derivative \( f_x \) is the rate of change of \( f \) with respect to \( x \), holding \( y \) constant. Similarly, \( f_y \) is the rate of change of \( f \) with respect to \( y \).

• Partial derivatives are used to form the gradient vector, which indicates the direction of the steepest ascent of a function.
• They are essential in optimization and in solving problems related to change detection.
• In the original problem, the partial derivatives \( f_x(-4,2) = 2.6 \) and \( f_y(-4,2) = 5.1 \) tell us how the function behaves as \( x \) or \( y \) changes.

Total Differential

The concept of the total differential helps approximate the change in the function when both variables change slightly. It serves as a linear approximation of how a small change in input affects the function's output.

The total differential of a function \( f(x, y) \) is given by the formula:df = f_x \cdot \Delta x + f_y \cdot \Delta y

Here, \( \Delta x \) and \( \Delta y \) are small changes in variables \( x \) and \( y \), respectively.

• The terms \( f_x \cdot \Delta x \) and \( f_y \cdot \Delta y \) represent the respective contributions of the changes in \( x \) and \( y \) to the total change in the function.
• This is particularly useful when exact computation of \( f(x, y) \) at a new point is difficult, but a good approximation is sufficient.
• In the exercise, we use the total differential to find that the approximation to the change is \( 0.045 \).

Function Approximation

Approximating a function involves finding a simpler expression or value that is close to the function's output. In many real-world scenarios, exact computation might be complex, making approximation methods very appealing.

Using the total differential, we get a linear approximation of the function's value at a nearby point. The reason this works well is that linear functions are much simpler to work with.

• Function approximation helps engineers, economists, and scientists to make predictions and decisions based on simplified models of reality.
• In our example, we find \( f(-4.12, 2.07) \approx 13.045 \) as an approximation of the function’s actual value at this point.
• The reliability of such approximations increases when the original function is smooth and continuous around the point of interest.

Change in Variables

Understanding the change in variables is crucial for calculating the total differential and using it for approximation. These small changes in variables form the building blocks of differential calculus.

When discussing changes \( \Delta x \) and \( \Delta y \), it highlights the difference between the initial and new values of the variables:

• \( \Delta x = -0.12 \) shows a decrease in \( x \), while \( \Delta y = 0.07 \) indicates an increase in \( y \).
• Quantifying these changes helps determine the respective impacts on the function’s value.
• Being able to precisely calculate these small differences is key for accurate approximations in mathematics and in practical applications.

By mastering the change in variables, one gains stronger analytical capabilities in solving real-world problems using differential calculus.