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 \(3 .\) \(z=\ln \left(x^{2} y\right) ; P(-2,4), Q(-1.98,3.96)\)

Short answer

The approximated change is \(-0.03\) and the exact change is \(-0.0232\).

Step-by-step solution

Calculate the Partial Derivatives

To use the total differential to approximate the change in function value, start by calculating the partial derivatives of the function with respect to each variable. For the function \( z = \ln(x^2 y) \), find \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \). The partial derivative with respect to \( x \) is found by using the chain rule:\[ \frac{\partial z}{\partial x} = \frac{1}{x^2 y} \cdot 2xy = \frac{2}{x} \] The partial derivative with respect to \( y \) is:\[ \frac{\partial z}{\partial y} = \frac{1}{x^2 y} \cdot x^2 = \frac{1}{y} \]

Compute the Total Differential

The formula for the total differential \( dz \) is given by:\[ dz = \frac{\partial z}{\partial x} \cdot dx + \frac{\partial z}{\partial y} \cdot dy \]Using the partial derivatives found in Step 1, substitute these values and the given points. Let \( dx = -1.98 + 2 = 0.02 \) and \( dy = 3.96 - 4 = -0.04 \).Thus, at point \((-2, 4)\):\[ dz = \frac{2}{-2} \cdot 0.02 + \frac{1}{4} \cdot (-0.04) = -0.02 - 0.01 = -0.03 \]

Find Exact Change for \( \, \Delta z \)

Calculate the exact change in \( z \) when moving from \( P \) to \( Q \) by directly evaluating the function at both points. At \( P(-2, 4) \):\[ z = \ln((-2)^2 \times 4) = \ln(16) \approx 2.7726 \]At \( Q(-1.98, 3.96) \):\[ z = \ln((-1.98)^2 \times 3.96) \approx \ln(15.6024) \approx 2.7494 \]The exact change \( \Delta z \) is:\[ \Delta z = 2.7494 - 2.7726 = -0.0232 \]

Compare Approximated and Exact Changes

Compare the approximate change found using the total differential with the exact change computed using the function directly.The approximation using the differential is \(-0.03\) and the exact value is \(-0.0232\). This indicates that the total differential is a reasonable approximation for small changes, with the differential slightly overestimating the change.

Key concepts

Partial Derivatives

Partial derivatives help us understand how a function changes as we vary one of its variables, while keeping the others constant. They are the building blocks for understanding the behavior of multivariable functions. For the function in the exercise, which is given by \( z = \ln(x^2 y) \), we need to find out how changes in \( x \) and \( y \) independently affect \( z \).

• To find the partial derivative of \( z \) with respect to \( x \), you treat \( y \) as a constant and apply derivative rules to \( x^2 y \).
• Similarly, finding the partial derivative with respect to \( y \) involves treating \( x \) as a constant.

The results \( \frac{\partial z}{\partial x} = \frac{2}{x} \) and \( \frac{\partial z}{\partial y} = \frac{1}{y} \) tell us the rate of change of \( z \) along the axes of \( x \) and \( y \) respectively. These derivatives are essential for using approximation techniques in calculus.

Total Differential

The total differential, often denoted as \( dz \), provides an approximation of how a small change in one or more variables affects the function's output. It uses the partial derivatives calculated before to provide a linear approximation of the function.
The formula for the total differential is:

• \( dz = \frac{\partial z}{\partial x} \cdot dx + \frac{\partial z}{\partial y} \cdot dy \)

Plugging in the values for partial derivatives and the changes \( dx \) and \( dy \), you can approximate how \( z \) changes when you slightly adjust \( x \) and \( y \).
This is ideal for small changes and provides a straightforward method for estimating the change in multivariable functions without having to calculate them explicitly every time.
In the exercise, using the total differential at the point \((-2, 4)\), the change \( dz \) was calculated to be approximately \(-0.03\).

Approximation Methods

Approximation methods in calculus serve to estimate values that might be challenging to compute exactly. The total differential is an excellent technique for estimating changes in multivariable contexts.
By using partial derivatives, we create a linear approximation of a function, which is practical for small adjustments or changes.
This is particularly useful when we want to quickly understand the behavior of complex functions with respect to small changes in their inputs.

• Beyond the total differential, approximation methods often include Taylor series and linearization, which expand the function into manageable terms.

These methods help in simplifying calculations and understanding how a function behaves locally, i.e., near a particular point.

Exact Change in Calculus

Exact change refers to evaluating the precise difference in the function's output when transitioning between two points. In our problem, this was done by computing \( \Delta z \), which is the change from point \( P \) to \( Q \) for the function \( z = \ln(x^2 y) \).
Unlike approximation, exact change involves evaluating the function at specific coordinates and finding their difference.

• For instance, at \( P(-2, 4) \), the function evaluates to \( \ln(16) \), and at \( Q(-1.98, 3.96) \), it evaluates to \( \ln(15.6024) \).
• Finding the difference gives \( \Delta z = -0.0232 \), which is the exact change.

By comparing the results of exact and approximated changes (using the total differential), we gain insights into the accuracy of our approximations, demonstrating how effective they are for small adjustments in variables.