Problem 31 - Calculus Early Transcendentals

Question

If \(z=5 x^{2}+y^{2}\) and \((x, y)\) changes from \((1,2)\) to \((1.05,2.1)\) compare the values of \(\Delta z\) and \(d z\)

Step-by-Step Solution

Answer

\(\Delta z = 0.9225\) and \(dz = 0.9\); \(dz\) approximates \(\Delta z\) closely.

Step 1: Calculate Initial Value of z

First, calculate the initial value of \( z \) using the formula \( z = 5x^2 + y^2 \) at \((x, y) = (1, 2)\).\[ z = 5(1)^2 + (2)^2 = 5 \times 1 + 4 = 9 \]

Step 2: Calculate New Value of z

Next, calculate the new value of \( z \) when \( (x, y) \) changes to \((1.05, 2.1)\):\[ z = 5(1.05)^2 + (2.1)^2 \]\[ z = 5(1.1025) + 4.41 \]\[ z = 5.5125 + 4.41 = 9.9225 \]

Step 3: Compute Actual Change in z ( abla z)

Compute the actual change in \( z \), denoted by \( \Delta z \), which is the difference between the new value and the initial value:\[ \Delta z = 9.9225 - 9 = 0.9225 \]

Step 4: Calculate Partial Derivatives

Calculate the partial derivatives \( \frac{{\partial z}}{{\partial x}} \) and \( \frac{{\partial z}}{{\partial y}} \) at \( (x, y) = (1, 2) \):\[ \frac{{\partial z}}{{\partial x}} = 10x \rightarrow 10 \times 1 = 10 \]\[ \frac{{\partial z}}{{\partial y}} = 2y \rightarrow 2 \times 2 = 4 \]

Step 5: Compute Differential (dz)

Now compute the differential \( dz \) using:\[ dz = \frac{{\partial z}}{{\partial x}} \Delta x + \frac{{\partial z}}{{\partial y}} \Delta y \]where \( \Delta x = 1.05 - 1 = 0.05 \) and \( \Delta y = 2.1 - 2 = 0.1 \).\[ dz = 10 \times 0.05 + 4 \times 0.1 = 0.5 + 0.4 = 0.9 \]

Step 6: Compare abla z and dz

Compare \( \Delta z = 0.9225 \) and \( dz = 0.9 \). Note that they are very close, indicating that the differential \( dz \) is a good approximation of \( \Delta z \) for small changes.

Key Concepts

Partial Derivatives

Understanding partial derivatives is essential when dealing with functions of multiple variables, like our function here with respect to variables \(x\) and \(y\). Essentially, a partial derivative measures how a function changes as one of the variables changes while keeping the others constant. In the context of our problem, we calculated the partial derivatives of \(z\) with respect to \(x\) and \(y\). This process involves differentiating \(z = 5x^2 + y^2\) with respect to one variable at a time:

• \(\frac{\partial z}{\partial x} = 10x\), giving the rate of change of \(z\) as \(x\) changes.
• \(\frac{\partial z}{\partial y} = 2y\), indicating how \(z\) changes with \(y\).

Evaluating these at our initial point \((1, 2)\), we found \(\frac{\partial z}{\partial x} = 10\) and \(\frac{\partial z}{\partial y} = 4\). These values tell us, momentarily, the influence of each variable on \(z\) at that specific point.

Delta z and dz Comparison

The comparison of \(\Delta z\) and \(dz\) helps us understand how differential and actual changes in a function relate to each other. \(\Delta z\) is the actual change in the function value as \((x, y)\) moves from \((1, 2)\) to \((1.05, 2.1)\). In this context, it results in:\[ \Delta z = z(1.05, 2.1) - z(1, 2) = 9.9225 - 9 = 0.9225 \]\(dz\), known as the differential, approximates this change using the gradient provided by the partial derivatives. It is calculated as:\[ dz = \frac{\partial z}{\partial x} \Delta x + \frac{\partial z}{\partial y} \Delta y = 10 \times 0.05 + 4 \times 0.1 = 0.9 \]Though \(\Delta z\) and \(dz\) are not identical, they are very close, with \(dz\) slightly underestimating the actual change. This highlights how differentials provide a good approximation, especially for small changes.

Differential Approximation

Differential approximation is a powerful concept in calculus, serving as a tool for estimating changes in function values. By using the formula \[ dz = \frac{\partial z}{\partial x} \Delta x + \frac{\partial z}{\partial y} \Delta y \]we are essentially using a linear approximation of the function near a point. This method assumes the function behaves linearly based on the tangent plane at the initial point, which is appropriate when the changes in \(x\) and \(y\) are small.This technique is beneficial because:

• It simplifies calculations, avoiding recalculating the entire function for new values.
• It provides insights into the sensitivity of the function concerning each variable.
• It serves as a foundation for more advanced topics such as Taylor series expansion.

In practice, if the changes are small enough, \(dz\) closely follows \(\Delta z\), allowing predictions and assessments without extensive computation.