Problem 12 - APEX Calculus

Question

In Exercises \(9-12,\) a function \(2=f(x, y)\) is given. Give the indicated approximation using the total differential. $$ \begin{array}{l} f(x, y)=\ln (x-y) . \text { Approximate } f(5.1,3.98) \text { knowing } \\ f(5,4)=0 \end{array} $$

Step-by-Step Solution

Answer

The approximate value of \( f(5.1, 3.98) \) is 0.12.

Step 1: Determine the point and change in variables

We are given the function \( f(x, y) = \ln(x-y) \) and need to approximate \( f(5.1, 3.98) \). We will use the point \( (5, 4) \) as the base point and note that \(\Delta x = 5.1 - 5 = 0.1\) and \(\Delta y = 3.98 - 4 = -0.02\).

Step 2: Compute partial derivatives

To use the total differential, we need the partial derivatives of \( f(x, y) \). The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = \frac{1}{x-y} \) and with respect to \( y \) is \( \frac{\partial f}{\partial y} = -\frac{1}{x-y} \).

Step 3: Evaluate partial derivatives at the base point

Evaluate the partial derivatives at the point \( (x, y) = (5, 4) \). Thus, \( \frac{\partial f}{\partial x} (5, 4) = \frac{1}{5-4} = 1 \) and \( \frac{\partial f}{\partial y} (5, 4) = -\frac{1}{5-4} = -1 \).

Step 4: Use total differential to approximate

The total differential \( df \) is \( df = \frac{\partial f}{\partial x}\, dx + \frac{\partial f}{\partial y}\, dy \). Substitute the values: \( df = (1)(0.1) + (-1)(-0.02) = 0.1 + 0.02 = 0.12 \).

Step 5: Compute the approximate value

The approximate value of \( f(5.1, 3.98) \) is given by \( f(5, 4) + df = 0 + 0.12 = 0.12 \).

Key Concepts

Partial Derivatives

Partial derivatives are fundamental in multivariable calculus. They help us understand how a function changes in relation to one variable while keeping the others constant. If you have a function that depends on several variables, you compute the partial derivative with respect to one variable by treating all other variables as constants. In our exercise with the function \( f(x, y) = \ln(x-y) \), we calculated partial derivatives with respect to both \( x \) and \( y \):

• The partial derivative with respect to \( x \) is \( \frac{1}{x-y} \).
• The partial derivative with respect to \( y \) is \( -\frac{1}{x-y} \).

These derivatives tell us how \( f(x, y) \) changes as \( x \) or \( y \) changes individually. Evaluating these at a specific point, like \( (5, 4) \), gives us the rate of change at that point, which is crucial for approximations.

Function Approximation

Function approximation with total differentials is an effective technique when dealing with complex functions. By using the concept of a differential, we can estimate the function's value at new points near a known value. Essentially, we use information about how the function changes to make this estimate.
The total differential, in this case, is expressed as: \[ df = \frac{\partial f}{\partial x} \, dx + \frac{\partial f}{\partial y} \, dy \] Our task was to approximate \( f(5.1, 3.98) \) using a known value \( f(5, 4) = 0 \). Given the small changes \( dx = 0.1 \) and \( dy = -0.02 \), we found \( df = 0.12 \).
Thus, the function's approximate value at the new point is the sum of the base value and the total differential: \( f(5, 4) + df = 0.12 \). This provides a quick and handy method to get estimates without intense computation.

Calculus Problem Solving

Calculus problem solving often involves breaking down complex expressions into manageable parts. Techniques like computing partial derivatives and using total differentials are quintessential skills that form the backbone of solving calculus problems.
For practical problem-solving:

• Identify your function and the variables it depends on.
• Compute relevant derivatives to understand how the function behaves.
• Use those derivatives in applied problems, such as finding approximations.

In our problem, we utilized these steps methodically. By establishing base points, calculating changes, and then applying the total differential, we simplified what could otherwise be a daunting task. Practice and familiarity with these steps enhance both accuracy and efficiency in tackling calculus problems.