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

Approximate \( \Delta z \) is 0.08, exact \( \Delta z \) is approx. 0.0798.

Step 1: Identify the given function

We start with the function given by the equation \( z = 2x^2y^3 \). We need to determine how \( z \) changes as the point moves from \( P(1,1) \) to \( Q(0.99, 1.02) \).

Step 2: Calculate the partial derivatives

Compute the partial derivatives of \( z \) with respect to \( x \) and \( y \):- \( \frac{\partial z}{\partial x} = 4xy^3 \),- \( \frac{\partial z}{\partial y} = 6x^2y^2 \).

Step 3: Evaluate the partial derivatives at the initial point

Substitute the coordinates of point \( P(1, 1) \) into the partial derivatives:- \( \frac{\partial z}{\partial x} \big|_{(1, 1)} = 4 \cdot 1 \cdot 1^3 = 4 \),- \( \frac{\partial z}{\partial y} \big|_{(1, 1)} = 6 \cdot 1^2 \cdot 1^2 = 6 \).

Step 4: Use the total differential to approximate \( \Delta z \)

The total differential \( dz \) is given by \[ dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy \].The changes in \( x \) and \( y \) from \( P \) to \( Q \) are \( dx = 0.99 - 1 = -0.01 \) and \( dy = 1.02 - 1 = 0.02 \).Substitute these values: \[ dz = 4(-0.01) + 6(0.02) = -0.04 + 0.12 = 0.08 \].

Step 5: Calculate the exact change \( \Delta z \) using a calculator

First, calculate the value of \( z \) at both points.- At \( P(1,1) \): \( z = 2(1)^2(1)^3 = 2 \).- At \( Q(0.99, 1.02) \): \( z \approx 2(0.99)^2(1.02)^3 \).Using a calculator, find: \( z \approx 2 \times 0.9801 \times 1.061208 \approx 2.0798 \).Thus, \( \Delta z = z(0.99, 1.02) - z(1, 1) \approx 2.0798 - 2 = 0.0798 \).

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They allow us to examine how a multivariable function changes with respect to one variable while keeping the other variables constant. Consider the function \( z = 2x^2y^3 \). To find the partial derivative of \( z \) with respect to \( x \), we differentiate treating \( y \) as a constant. This results in \( \frac{\partial z}{\partial x} = 4xy^3 \). Similarly, for \( y \), we differentiate treating \( x \) as a constant, giving \( \frac{\partial z}{\partial y} = 6x^2y^2 \).

• These partial derivatives provide the rate of change of \( z \) relative to each variable.
• By evaluating these at a point, such as \( P(1, 1) \), we can examine specific changes.

An understanding of these derivatives is crucial for exploring how functions behave in a multivariable setting.

Approximation of Change

The total differential is a valuable tool for approximating how functions change. When you can't easily calculate the exact change, the total differential provides a reasonable approximation. Using partial derivatives, we express the total change \( dz \) in terms of changes in \( x \) and \( y \): \[ dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy \]In our exercise, this translates to \( dz = 4(-0.01) + 6(0.02) = 0.08 \).

• Total differential simplifies the process by considering the linear approximation.
• It's especially useful when moving from point \( P \) to a nearby point \( Q \).

Though not perfect, it gives a decent idea of how much the function value changes.

Calculus Problem Solving

In calculus problem solving, particularly with multivariable functions, it's key to break down the problem into manageable parts. For example, with the function \( z = 2x^2y^3 \), our task was to determine the change from \( P(1, 1) \) to \( Q(0.99, 1.02) \). First, we identified the function and its partial derivatives. Then, we evaluated these derivatives at the point of interest.

• Calculating small adjustments called \( dx \) and \( dy \) helped us use the total differential formula.
• This approach streamlined finding \( dz \), the estimated change.

The step-by-step nature of calculus allows you to tackle even complex functions with confidence.

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions of several variables. It requires understanding derivatives in a new dimension, such as partial derivatives. These become crucial in analyzing how each variable uniquely affects the function. In our case, for \( z = 2x^2y^3 \), calculating how changes in \( x \) and \( y \) impact \( z \) provides insights.

• It's important to examine the behavior of functions in higher-dimensional spaces.
• Multivariable calculus empowers us to model and solve real-world problems involving multiple changing factors.

Grasping the concepts of multivariable calculus can unlock new ways to approach complex, dynamic systems.