Problem 12 - 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 \(3 .\) $$ z=\tan ^{-1} x y ; P(-2,-0.5), Q(-2.03,-0.51) $$

Step-by-Step Solution

Answer

Approximate change in \( z \) is 0.0175; exact change is 0.0174.

Step 1: Calculate Partial Derivatives

First, we find the partial derivatives of the function \( z = \tan^{-1}(xy) \) with respect to \( x \) and \( y \). The partial derivative of \( z \) with respect to \( x \) is \( \frac{\partial z}{\partial x} = \frac{y}{1 + (xy)^2} \). The partial derivative of \( z \) with respect to \( y \) is \( \frac{\partial z}{\partial y} = \frac{x}{1 + (xy)^2} \).

Step 2: Evaluate Partial Derivatives at Point P

Substitute the coordinates of point \( P(-2, -0.5) \) into the partial derivatives: \( \frac{\partial z}{\partial x} \bigg|_P = \frac{-0.5}{1 + (-2 \times -0.5)^2} = \frac{-0.5}{1 + 1} = \frac{-0.5}{2} = -0.25 \). \( \frac{\partial z}{\partial y} \bigg|_P = \frac{-2}{1 + (-2 \times -0.5)^2} = \frac{-2}{2} = -1 \).

Step 3: Determine Changes in x and y

Find the changes in \( x \) and \( y \) from point \( P(-2, -0.5) \) to \( Q(-2.03, -0.51) \): The change in \( x \) is \( \Delta x = -2.03 - (-2) = -0.03 \). The change in \( y \) is \( \Delta y = -0.51 - (-0.5) = -0.01 \).

Step 4: Apply Total Differential Formula

Use the total differential formula to approximate the change in \( z \): \( dz = \frac{\partial z}{\partial x} \cdot \Delta x + \frac{\partial z}{\partial y} \cdot \Delta y \). Substitute the values obtained: \( dz = (-0.25)(-0.03) + (-1)(-0.01) = 0.0075 + 0.01 = 0.0175 \). So, the approximate change in \( z \) is \( 0.0175 \).

Step 5: Calculate the Exact Change in z

Calculate \( z \) at points \( P \) and \( Q \): At \( P \), \( z = \tan^{-1}(-2 \times -0.5) = \tan^{-1}(1) = \frac{\pi}{4} \approx 0.7854 \). At \( Q \), \( z = \tan^{-1}(-2.03 \times -0.51) = \tan^{-1}(1.0353) \). Use a calculator: \( z \approx 0.8028 \). The exact change \( \Delta z = 0.8028 - 0.7854 = 0.0174 \).

Key Concepts

Partial Derivatives

Partial derivatives are used in calculus to measure how a function changes as one of its variables is adjusted, while keeping all others constant. In this case, if you have a multivariable function such as our given example, \( z = \tan^{-1}(xy) \), understanding the change with respect to \( x \) and \( y \) individually is crucial. This is achieved through partial differentiation.
To find the partial derivative of \( z \) with respect to \( x \), denoted \( \frac{\partial z}{\partial x} \), you treat \( y \) as a constant and differentiate with respect to \( x \). Conversely, for the partial derivative with respect to \( y \), denoted \( \frac{\partial z}{\partial y} \), you treat \( x \) as a constant and differentiate with respect to \( y \).
The calculated partial derivatives for our function are:

• \( \frac{\partial z}{\partial x} = \frac{y}{1 + (xy)^2} \)
• \( \frac{\partial z}{\partial y} = \frac{x}{1 + (xy)^2} \)

These derivatives allow us to understand the individual effects of changing one variable at a time on the function \( z \). Knowing these effects is required for calculating the total differential, which gives a complete picture of how small changes in both variables affect the function.

Total Differential

The total differential of a function gives an approximation of how that function changes as each of its variables undergoes a small change.
For a function \( z = f(x, y) \), the total differential \( dz \) is given by the formula:
\[ dz = \frac{\partial z}{\partial x} \cdot \Delta x + \frac{\partial z}{\partial y} \cdot \Delta y \]
Here, \( \Delta x \) and \( \Delta y \) are small changes in \( x \) and \( y \), respectively.
In the problem, transitioning from point \( P(-2, -0.5) \) to \( Q(-2.03, -0.51) \), gives us \( \Delta x = -0.03 \) and \( \Delta y = -0.01 \). Applying the partial derivatives we found earlier, we use the total differential formula to find that the approximate change in \( z \) is \( 0.0175 \).
This approach is often beneficial when direct calculations are cumbersome, providing a simpler way to evaluate changes across multiple variables.

Approximate Change

An approximate change in a function represents how much we expect the function's value to change, given small adjustments in its input variables. This is especially useful in understanding how functions behave around particular points.
Total differentials give this approximation, as they factor in the influence of each small variable change on the overall function change.
Comparing the approximate change with the actual value can also point out differences that arise from assuming linearity in certain functions. In our exercise, the approximate change calculated using the total differential is \( 0.0175 \). This is remarkably close to the exact change \( \Delta z = 0.0174 \) obtained through precise computation.
This slight difference validates the utility of differentials in applications where exact computations are impractical or unnecessary. Always remember: while small errors may appear in approximation, in many scenarios they are negligible, making approximations a powerful tool.