Problem 9 - 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 \(3 .\) \(z=2 x^{2} y^{3} ; P(1,1), Q(0.99,1.02)\)

Step-by-Step Solution

Answer

The approximate change using dz is 0.08, and the exact change Δz is approximately 0.0804.

Step 1: Find Partial Derivatives

To use the total differential, we first need the partial derivatives of the function. Given \(z = 2x^2y^3\), calculate \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) at the point \(P(1,1)\).\- \(\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(2x^2y^3) = 4xy^3\).\- \(\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(2x^2y^3) = 6x^2y^2\).\At \(P(1, 1)\): \- \(\frac{\partial z}{\partial x} = 4 \cdot 1 \cdot 1^3 = 4\). \- \(\frac{\partial z}{\partial y} = 6 \cdot 1^2 \cdot 1^2 = 6\).

Step 2: Calculate Total Differential dz

Using the total differential formula \(dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy\), evaluate \(dz\).\The change in \(x\) is \(dx = 0.99 - 1 = -0.01\).The change in \(y\) is \(dy = 1.02 - 1 = 0.02\).Substitute these into the total differential: \\(dz = 4(-0.01) + 6(0.02) = -0.04 + 0.12 = 0.08\).

Step 3: Calculate Exact Change Δz

To find the exact change \(\Delta z\), evaluate \(z\) at both points \(P\) and \(Q\).\- At \(P(1,1)\): \ \(z = 2 \cdot 1^2 \cdot 1^3 = 2\).\- At \(Q(0.99,1.02)\): \ \(z = 2 \cdot (0.99)^2 \cdot (1.02)^3\). Calculate using a calculator: - \(0.99^2 = 0.9801\)\- \(1.02^3 \approx 1.061208\)\- Therefore, \(z = 2 \cdot 0.9801 \cdot 1.061208 \approx 2.0804\). Now calculate \(\Delta z = z(Q) - z(P) = 2.0804 - 2 = 0.0804\).

Key Concepts

Partial Derivatives

In calculus, particularly in multivariable calculus, understanding partial derivatives is essential. Partial derivatives measure how a function changes with respect to one variable while keeping other variables constant. For a function of two variables, like the one given, the partial derivative with respect to one variable gives us the rate of change of the function in the direction of that variable. Given a function such as \( z = 2x^2y^3 \), we find the partial derivatives regarding \( x \) and \( y \) separately.

• Partial derivative with respect to \( x \) is \( \frac{\partial z}{\partial x} = 4xy^3 \).
• Partial derivative with respect to \( y \) is \( \frac{\partial z}{\partial y} = 6x^2y^2 \).

These derivatives help in evaluating the total differential, which approximates changes in the function due to small variations in \( x \) and \( y \).

Exact Change in Calculus

Exact change in calculus refers to the precise change in the value of a function as we move from one point to another. In the given problem, this is represented by \( \Delta z \), the difference in the function \( z \) evaluated at two different points. Calculating this exact change gives us an accurate measure, unlike the approximate measure provided by the total differential.To find \( \Delta z \), calculate \( z \) at the start point \( P \) and the endpoint \( Q \):

• At \( P(1,1) \): \( z = 2 \cdot 1^2 \cdot 1^3 = 2 \)
• At \( Q(0.99, 1.02) \): \( z = 2 \cdot (0.99)^2 \cdot (1.02)^3 \approx 2.0804 \)

The change \( \Delta z \) is then \( 2.0804 - 2 = 0.0804 \). This accurately reflects how \( z \) changes as \( x \) and \( y \) vary from \( P \) to \( Q \).

Differential Approximation

Differential approximation uses the concept of total differential to estimate the change in a function. It is a technique that provides a way to approximate small changes in a function's output based on changes in its inputs. For the function \( z = 2x^2y^3 \), the total differential \( dz \) is calculated using:\[ dz = \frac{\partial z}{\partial x} \, dx + \frac{\partial z}{\partial y} \, dy \]Here, \( dx \) and \( dy \) are small changes in \( x \) and \( y \), respectively.In our example:

• \( dx = 0.99 - 1 = -0.01 \)
• \( dy = 1.02 - 1 = 0.02 \)

Calculating \( dz \):\( dz = 4(-0.01) + 6(0.02) = -0.04 + 0.12 = 0.08 \)This approximation is often close to the exact change, especially for small variations in \( x \) and \( y \), and is a useful tool in many practical applications to quickly estimate changes without requiring complex calculations.