Problem 32 - Calculus Early Transcendentals

Question

If \(z=x^{2}-x y+3 y^{2}\) and \((x, y)\) changes from \((3,-1)\) to \((2.96,-0.95),\) compare the values of \(\Delta z\) and \(d z\)

Step-by-Step Solution

Answer

\(\Delta z = -0.7189\) and \(dz = -0.73\); they are close in value.

Step 1: Calculate Initial z and Final z

First, we need to compute the initial value of \( z \) when \( (x, y) = (3, -1) \) and the final value of \( z \) when \( (x, y) = (2.96, -0.95) \). - For the initial point \((3, -1)\):\[ z = 3^2 - 3(-1) + 3(-1)^2 = 9 + 3 + 3 = 15 \]- For the final point \((2.96, -0.95)\):\[ z = (2.96)^2 - 2.96(-0.95) + 3(-0.95)^2 \]Calculating each term:\( (2.96)^2 = 8.7616 \), \( 2.96(-0.95) = -2.812 \), and \( 3(-0.95)^2 = 2.7075 \).Thus, the final \( z = 8.7616 + 2.812 + 2.7075 = 14.2811 \).Here, the initial \( z \) is 15, and the final \( z \) is 14.2811.

Step 2: Calculate \(\Delta z\)

Now, we find \( \Delta z \), which is the change in \( z \):\[ \Delta z = z_{\text{final}} - z_{\text{initial}} = 14.2811 - 15 = -0.7189 \]

Step 3: Calculate Partial Derivatives

We need the partial derivatives of \( z \) with respect to \( x \) and \( y \):- \( \frac{\partial z}{\partial x} = 2x - y \).- \( \frac{\partial z}{\partial y} = -x + 6y \).At \( (x, y) = (3, -1) \):- \( \frac{\partial z}{\partial x} = 2(3) - (-1) = 6 + 1 = 7 \).- \( \frac{\partial z}{\partial y} = -(3) + 6(-1) = -3 - 6 = -9 \).

Step 4: Calculate Differential \(dz\)

The differential \( dz \) is given by:\[ dz = \frac{\partial z}{\partial x} \cdot dx + \frac{\partial z}{\partial y} \cdot dy \]Where \( dx = 2.96 - 3 = -0.04 \) and \( dy = -0.95 + 1 = 0.05 \).Substituting the values:\[ dz = 7(-0.04) + (-9)(0.05) = -0.28 - 0.45 = -0.73 \]

Step 5: Compare \(\Delta z\) and \(dz\)

Finally, we compare \( \Delta z \) and \( dz \):- \( \Delta z = -0.7189 \)- \( dz = -0.73 \)The values of \( \Delta z \) and \( dz \) are very close to each other, indicating a small error when using differentials to estimate the change in \( z \).

Key Concepts

Partial Derivatives

Partial derivatives play a fundamental role in differential calculus, especially for functions with more than one variable. In our exercise, the function is defined as \(z = x^2 - xy + 3y^2\). Partial derivatives are used to understand how the function changes with respect to changes in one variable, while keeping others constant. They help identify the rate of change along different axes or directions of the function's input space, allowing us to explore and analyze more complex surfaces.

• Partial derivative with respect to \(x\): This tells us how \(z\) changes if \(x\) changes by a small amount, while \(y\) remains unchanged. In our problem, it's given by \(\frac{\partial z}{\partial x} = 2x - y\).
• Partial derivative with respect to \(y\): This describes the change in \(z\) due to a small change in \(y\), with \(x\) held constant. The formula provided is \(\frac{\partial z}{\partial y} = -x + 6y\).

When we calculated these at the point \((3, -1)\), we obtained \(\frac{\partial z}{\partial x} = 7\) and \(\frac{\partial z}{\partial y} = -9\). These values help us measure how much \(z\) is expected to change when \(x\) and \(y\) change slightly.

Differentials

Differentials provide an approximation of how a small change in input variables affects the output of a multivariable function. In this exercise, we use them to predict the change in \(z\) given small changes in \(x\) and \(y\). The differential \(dz\) is calculated using the formula:\[ dz = \frac{\partial z}{\partial x} \cdot dx + \frac{\partial z}{\partial y} \cdot dy \] Here, \(dx\) and \(dy\) are the small changes in \(x\) and \(y\), respectively.

To find \(dz\), we first determine \(dx\) and \(dy\):

• \(dx = 2.96 - 3 = -0.04\)
• \(dy = -0.95 - (-1) = 0.05\)

Using the partial derivatives calculated before, we substitute these values:\[ dz = 7(-0.04) + (-9)(0.05) = -0.28 - 0.45 = -0.73 \]
Differentials give us a linear approximation of \(\Delta z\), which makes them very useful for smaller changes.

Change in Functions

The change in functions over intervals is a core aspect of understanding calculus. In this problem, we identify two ways to quantify the change in \(z\) as \(x\) and \(y\) change: actual change \(\Delta z\) and approximate change using the differential \(dz\).

• \(\Delta z\): This is computed by finding the difference between \(z\) at the final and initial points. In our case, \(\Delta z = 14.2811 - 15 = -0.7189\). This value reflects the actual change in function value given the exact changes in variables.
• Use of \(dz\): As an approximation, differentials help approximate \(\Delta z\) when the changes in variables are very small. For small adjustments, \(dz\) provides a close estimate for \(\Delta z\), as seen here where \(dz = -0.73\) is quite close to \(\Delta z = -0.7189\).

Understanding both forms of changes aids in grasping how accurate differentials can be for predictions and highlights their limits when changes are not infinitesimally small.