Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the total differential dz to approximate the change in \(z\) as \((x, y)\) moves from \(P\) to \(Q\). Then use a calculator to find the corresponding exact change \(\Delta z\) (to the accuracy of your calculator). See Example \(3 .\) $$ z=\ln \left(x^{2} y\right) ; P(-2,4), Q(-1.98,3.96) $$

Step-by-Step Solution

Answer

The approximate change is \( dz \approx -0.0225 \), and the exact change is \( \Delta z = -0.0225 \).

Step 1: Find the Formula for Total Differential

For a function \( z = f(x, y) \), the total differential \( dz \) is given by the formula: \( dz = \frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y}dy \). This formula allows us to approximate the change in \( z \) given small changes in \( x \) and \( y \).

Step 2: Calculate Partial Derivatives

Given \( z = \ln(x^2y) \), calculate the partial derivatives. \[ \frac{\partial z}{\partial x} = \frac{2x}{x^2y} = \frac{2}{x} \] \[ \frac{\partial z}{\partial y} = \frac{1}{x^2y} \]

Step 3: Evaluate Partial Derivatives at Point P

Substitute \( P(-2, 4) \) into the partial derivatives: \[ \frac{\partial z}{\partial x} = \frac{2}{-2} = -1 \] \[ \frac{\partial z}{\partial y} = \frac{1}{(-2)^2 \times 4} = \frac{1}{16} \]

Step 4: Calculate \(dx\) and \(dy\)

Determine the changes in \( x \) and \( y \): \( dx = -1.98 - (-2) = 0.02 \) \( dy = 3.96 - 4 = -0.04 \)

Step 5: Approximate Change in \( z \) Using Total Differential

Substitute into the total differential equation: \( dz = (-1)(0.02) + \left(\frac{1}{16}\right)(-0.04) \) \[ dz = -0.02 - 0.0025 = -0.0225 \] Thus, the approximate change is \( dz \approx -0.0225 \).

Step 6: Calculate Exact Change in \( z \)

Use a calculator to find the exact \( z \) values: \( z(P) = \ln((-2)^2 \times 4) = \ln(16) \approx 2.7726 \) \( z(Q) = \ln((-1.98)^2 \times 3.96) \approx 2.7501 \) The exact change is \( \Delta z = z(Q) - z(P) = 2.7501 - 2.7726 = -0.0225 \).

Key Concepts

Partial Derivatives

Partial derivatives are a cornerstone in multivariable calculus. They give us information on how a function changes as we tweak one variable, while keeping others constant. Imagine you have a mountain, where z is the height, and you want to understand how it changes as you move east or north. Partial derivatives help break down this problem into two simpler ones.

• For the function \(z = \ln(x^2y)\), the partial derivative with respect to \(x\), noted as \(\frac{\partial z}{\partial x}\), shows the rate at which \(z\) changes as \(x\) varies, while \(y\) is held constant.
• Similarly, \(\frac{\partial z}{\partial y}\) reveals how \(z\) varies if we only change \(y\).

Partial derivatives are calculated similarly to ordinary derivatives. For instance, in our exercise, we found:

• \(\frac{\partial z}{\partial x} = \frac{2}{x}\), indicating the effect of changing \(x\).
• \(\frac{\partial z}{\partial y} = \frac{1}{x^2y}\), which reflects the change due to \(y\).

These computations guide us in understanding how minor adjustments in \(x\) and \(y\) influence \(z\).

Total Differential

The total differential helps us approximate the change in a function due to small changes in its variables. When dealing with a function \(z = f(x, y)\), the total differential \(dz\) is calculated as a sum of the impacts from each variable's change.Given by the formula:\[ dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy \]It effectively combines the information from partial derivatives and small changes (\(dx\) and \(dy\)). In our exercise, by substituting the partial derivatives and the small changes at point \(P\), we found:

• \(dz = (-1)(0.02) + \left(\frac{1}{16}\right)(-0.04) = -0.0225\)

This result tells us how much \(z\) approximately changes as \(x\) and \(y\) shift slightly around \(P\). It makes predicting these changes manageable, especially when the exact computation is complex.

Change Approximation

Approximating change is about using mathematical insight to make quick predictions when precise calculation is cumbersome. This concept is essential when the changes in variables are minimal.In our example, use of the total differential simplified the process by offering an estimated change \(dz\) of \(-0.0225\). Actual calculation provided a nearly identical result, validating the reliability of our approximation.Why bother with approximations?

• They are often quicker and less cumbersome than computing exact changes.
• Approximate solutions serve as an excellent first check on the exactness of later detailed computations.

Such approximations thus become practical tools for anyone needing to quickly understand the behavior of a function.

Mathematical Problem Solving

Solving mathematical problems involves integrating various concepts to derive solutions effectively. In this exercise, we orchestrated knowledge of partial derivatives and total differentials to solve a real-world problem in a theoretical context.

• Recognizing when to use a total differential helps in approximating changes without solving the entire function.
• Understanding partial derivatives directs attention to the variables most affecting the function's behavior.
• Utilizing calculators complements manual calculations, offering precise values for validation.

By harmonizing analytical methods and computational tools, this exercise showcases a comprehensive approach to problem-solving. It is about balancing speed and accuracy, ensuring that our mathematical predictions align closely with precise results.