Problem 195 - Calculus 3

Question

Let \(z=f(x, y)=x e^{y} .\) Compute \(\Delta z\) from \(P(1,2)\) to \(Q(1.05,2.1)\) and then find the approximate change in \(z\) from point \(P\) to point \(Q .\) Recall \(\Delta z=f(x+\Delta x, y+\Delta y)-f(x, y),\) and \(d z\) and \(\Delta z\) are approximately equal.

Step-by-Step Solution

Answer

The approximate change \(\Delta z \approx 0.15 e^2\).

Step 1: Understand the Problem

We need to compute the change in the function \(z = f(x, y) = x e^y\) from point \(P(1,2)\) to point \(Q(1.05,2.1)\). This means calculating \(\Delta z = f(1.05, 2.1) - f(1, 2)\). We will also use differentials to approximate \(\Delta z\).

Step 2: Calculate \(f(P)\)

Evaluate the function at point \(P(1,2)\): \[f(1, 2) = 1 \cdot e^2 = e^2.\]

Step 3: Calculate \(f(Q)\)

Evaluate the function at point \(Q(1.05, 2.1)\):\[f(1.05, 2.1) = 1.05 \cdot e^{2.1}.\]

Step 4: Calculate \(\Delta z\)

Use the values from previous steps to find the exact change:\[\Delta z = f(1.05, 2.1) - f(1, 2) = 1.05 \cdot e^{2.1} - e^2.\] Evaluate using a calculator for numerical results.

Step 5: Use Differentials to Approximate \(\Delta z\)

Compute the partial derivatives: \(\frac{\partial z}{\partial x} = e^y\) and \(\frac{\partial z}{\partial y} = x e^y.\)At point \(P(1,2)\): \(\frac{\partial z}{\partial x} = e^2,\) \(\frac{\partial z}{\partial y} = 1 \cdot e^2 = e^2.\)The differential is:\[dz = e^2 \cdot \Delta x + e^2 \cdot \Delta y = e^2(\Delta x + \Delta y).\]

Step 6: Substitute and Approximate \(dz\)

Substitute \(\Delta x = 1.05 - 1 = 0.05\), \(\Delta y = 2.1 - 2 = 0.1\):\[dz = e^2 \cdot (0.05 + 0.1) = 0.15 e^2.\]This is the approximate change from \(P\) to \(Q\).

Key Concepts

Differential Calculus

In differential calculus, we focus on rates of change and slopes of curves. The concept of partial derivatives is crucial here because it helps us understand how a function changes as we tweak one variable, while keeping others constant. For functions of two variables like our example, partial derivatives symbolize the slope in the directions of each variable:

• Partial derivative with respect to x is found by treating y as constant.
• Partial derivative with respect to y is found by treating x as constant.

In our exercise, we compute:

• \( \frac{\partial z}{\partial x} = e^y \) at point \( P(1,2) \).
• \( \frac{\partial z}{\partial y} = xe^y \) at point \( P(1,2) \).

The partial derivatives tell us how the function changes infinitesimally with respect to each input variable, providing us with small yet vital insights into the behavior of the function.

Function Evaluation

Function evaluation is the process of finding the value of a function for particular input values. In this context, it involves plugging values into the function to obtain an output. As per the exercise, finding \( f(P) \) and \( f(Q) \) is a clear demonstration of this process:

• At \( P(1,2) \), the function value is evaluated as \( f(1, 2) = 1 \cdot e^2 \).
• At \( Q(1.05, 2.1) \), the function value is \( f(1.05, 2.1) = 1.05 \cdot e^{2.1} \).

By evaluating, we determine the actual value of the function at specified points in its domain. Function evaluation forms the backbone of many mathematical calculations, offering a grounded sense of how inputs translate to outputs.

Taylor Approximation

Taylor approximation is a method used to approximate the value of a function near a certain point. In the realm of multivariable calculus, it allows us to estimate the change in function values when moving slightly from a known point, using partial derivatives. For our problem:

• We approximate the change \(\Delta z\) using differentials.
• The differential \(dz\) can approximate \(\Delta z\) when the changes \(\Delta x\) and \(\Delta y\) are small.

The computation involves using the formula: \[ dz = \frac{\partial z}{\partial x}\Delta x + \frac{\partial z}{\partial y}\Delta y \]This employs the function and its derivatives's values at a specific point to approximate small changes. For point \(P\), we have: \[ dz = e^2(0.05 + 0.1) = 0.15e^2 \]This provides a handy estimate of how much \(z\) changes when moving from one nearby point to another.

Change in Variables

The concept of change in variables helps us explore how small shifts in input values can influence the outcome of a function. In our example, moving from point \(P(1,2)\) to \(Q(1.05,2.1)\) involves:

• \(\Delta x = 1.05 - 1 = 0.05\)
• \(\Delta y = 2.1 - 2 = 0.1\)

These small changes allow us to utilize differential calculus to approximate the resulting change in the function value. While calculations take into account these variations, using differentials gives us a more manageable way to handle complex changes, especially when dealing with nonlinear functions. The idea is that even small variations in \(x\) and \(y\) can be effectively captured through partial derivatives, helping us to predict the behavior of the function over tiny changes in its inputs.