Problem 11 - APEX Calculus

Question

In Exercises \(9-12,\) a function \(2=f(x, y)\) is given. Give the indicated approximation using the total differential. $$ \begin{array}{l} f(x, y)=x^{2} y-x y^{2} \text { . Approximate } f(2.04,3.06) \text { knowing } \\\ f(2,3)=-6 \end{array} $$

Step-by-Step Solution

Answer

\( f(2.04, 3.06) \approx -6.36 \)

Step 1: Compute First Derivatives

First, we need to find the partial derivatives of the function \( f(x,y) = x^2y - xy^2 \) with respect to \( x \) and \( y \). - The partial derivative of \( f \) with respect to \( x \) is \( f_x(x, y) = 2xy - y^2 \).- The partial derivative of \( f \) with respect to \( y \) is \( f_y(x, y) = x^2 - 2xy \).

Step 2: Evaluate Derivatives at the Given Point

Now, we need to evaluate the partial derivatives we found in Step 1 at the point \((x,y) = (2,3)\).- Evaluate \( f_x(2,3) = 2(2)(3) - 3^2 = 12 - 9 = 3 \).- Evaluate \( f_y(2,3) = 2^2 - 2(2)(3) = 4 - 12 = -8 \).

Step 3: Calculate Total Differential

The total differential \( df \) can be given by:\[df = f_x(x,y) dx + f_y(x,y) dy\]Substitute the values from Step 2:\[df = 3dx - 8dy\]

Step 4: Determine Changes in Variables

Find the changes in the \( x \) and \( y \) variables from the initial point \((2,3)\) to the new point \((2.04, 3.06)\):- \( dx = 2.04 - 2 = 0.04 \)- \( dy = 3.06 - 3 = 0.06 \)

Step 5: Approximate the Change in Function

Use the total differential to approximate the change in \( f \):\[df = 3(0.04) - 8(0.06) = 0.12 - 0.48 = -0.36\]

Step 6: Approximate the Function's Value at New Point

To find the approximate value of \( f(2.04, 3.06) \), add the change \( df \) to the original value \( f(2,3) \):\[f(2.04, 3.06) \approx f(2,3) + df = -6 - 0.36 = -6.36\]

Key Concepts

Partial Derivatives

Partial derivatives are a vital concept in multivariable calculus. When you have a function that depends on two or more variables, such as the function given by the exercise, partial derivatives tell you how the function changes as you vary one of the variables, keeping the others constant.

For the function \(f(x, y) = x^2y - xy^2\), we have two partial derivatives. The partial derivative with respect to \(x\), denoted as \(f_x(x, y)\), was calculated as \(2xy - y^2\). This derivative explains how \(f\) changes when \(x\) changes, with \(y\) held constant. Similarly, \(f_y(x, y)\) was found to be \(x^2 - 2xy\), which tells us how \(f\) changes with variations in \(y\), keeping \(x\) fixed.

• \(f_x(x, y): 2xy - y^2\)
• \(f_y(x, y): x^2 - 2xy\)

By understanding these derivatives, you gain insight into the slope of the function’s tangent planes in both the \(x\) and \(y\) directions at any given point. Evaluating these at a specific point, such as \((2,3)\), provides the exact rate of change of function values around that point.

Approximation Method

The approximation method using total differentials is a practical tool for estimating the value of functions at nearby points.

Based on linear approximation, the method builds on the idea of the tangent plane in multivariable calculus. In the given problem, the total differential \(df\) helps us approximate how the function \(f(x, y) = x^2y - xy^2\) changes as we move from the point \((2,3)\) to the point \(2.04, 3.06)\). The formula for the total differential is given by:

• \(df = f_x(x,y) dx + f_y(x,y) dy\)

This allows us to approximate changes in \(f\) based on small changes \(dx\) and \(dy\) in \(x\) and \(y\), respectively. In the solution,

• \(dx = 2.04 - 2 = 0.04\)
• \(dy = 3.06 - 3 = 0.06\)

By substituting these into the equation for \(df\), we found that the function’s value changes by approximately \(-0.36\). Such methods are commonly used in science and engineering to make quick estimates when exact values are computationally intensive to obtain.

Calculus

Calculus is the mathematical study of change and motion, employing concepts such as derivatives and integrals. In this exercise, we explore the derivative aspect of calculus through total differentials and partial derivatives.

Derivatives, whether ordinary or partial, play a crucial role in understanding how functions behave. The partial derivatives \(f_x\) and \(f_y\) are specialized derivatives that tell us the instantaneous rate of change of the function in different directions. The concept of total differential combines these derivatives to approximate how a function changes across multiple dimensions, offering a way to linearize complex functions locally.

For example,

• In a single-variable function, the derivative gives a linear approximation right on its tangent.
• In multivariable calculus, total differentials extend this idea to surfaces or higher-dimension spaces. This allows us to explore the function’s behavior not just along a line but across a plane or surface.

Calculus tools like these empower scientists and engineers to predict and model real-world situations effectively, crafting approximations and predictions that can be refined with more precise data.