Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

For the function \(f(x, y)=\tan \left(\left(x^{2}+y^{2}\right) / 64\right)\), find the second-order Taylor approximation based at \(\left(x_{0}, y_{0}\right)=(0,0)\). Then estimate \(f(0.2,-0.3)\) using (a) the first-order approximation, (b) the second-order approximation, and (c) your calculator directly.

Step-by-Step Solution

Answer

(a) 0, (b) 0.00203125, (c) 0.00203125

Step 1: Identify the components of the Taylor series

The second-order Taylor series approximation of a function \( f \) around a point \( (x_0, y_0) \) requires the function value \( f(x_0, y_0) \), the first partial derivatives \( f_x(x_0, y_0) \) and \( f_y(x_0, y_0) \), and the second partial derivatives \( f_{xx}(x_0, y_0) \), \( f_{yy}(x_0, y_0) \), and \( f_{xy}(x_0, y_0) \).

Step 2: Evaluate the function at the base point

Compute \( f(0, 0) = \tan((0^2 + 0^2) / 64) = \tan(0) = 0 \).

Step 3: Find the first partial derivatives of the function

First, compute \( f_x(x, y) \) and \( f_y(x, y) \). For \( f_x(x, y) \), use the chain rule: \ \( f_x = \frac{d}{dx} \tan\left(\frac{x^2 + y^2}{64}\right) = \sec^2\left(\frac{x^2 + y^2}{64}\right) \cdot \frac{2x}{64} = \frac{2x}{64}\sec^2\left(\frac{x^2 + y^2}{64}\right) \). \ Similarly, \( f_y = \frac{2y}{64}\sec^2\left(\frac{x^2 + y^2}{64}\right) \).

Step 4: Evaluate the first partial derivatives at the base point

At \( (0,0) \), both \( f_x(0, 0) = \frac{2 \cdot 0}{64} \cdot \sec^2(0) = 0 \) and \( f_y(0, 0) = \frac{2 \cdot 0}{64} \cdot \sec^2(0) = 0 \).

Step 5: Find the second partial derivatives

Compute \( f_{xx}(x, y) \), \( f_{yy}(x, y) \), and \( f_{xy}(x, y) \): \ \( f_{xx} = \frac{d}{dx}\left( \frac{2x}{64}\sec^2\left(\frac{x^2 + y^2}{64}\right) \right) = \frac{2}{64} \sec^2\left(\frac{x^2 + y^2}{64}\right)+ \frac{2x}{64} \cdot \frac{d}{dx}\left( \sec^2\left(\frac{x^2 + y^2}{64}\right) \right) \). \ Continue similarly for \( f_{yy} \) and \( f_{xy} \).

Step 6: Evaluate the second partial derivatives at the base point

At \( (0,0) \), these evaluations simplify to \( f_{xx}(0,0) = \frac{2}{64} = \frac{1}{32} \, f_{yy}(0,0) = \frac{1}{32} \, \text{and} \ f_{xy}(0,0) = 0 \).

Step 7: Write the second-order Taylor approximation

The second-order Taylor series is given by: \ \( f(x, y) \approx f(0, 0) + f_x(0,0)x + f_y(0, 0)y + \frac{1}{2}f_{xx}(0,0)x^2 + \frac{1}{2}f_{yy}(0,0)y^2 + f_{xy}(0,0)xy \), \ which simplifies to \( f(x, y) \approx \frac{1}{64}(x^2 + y^2) \).

Step 8: Estimate using first-order approximation

For the first-order approximation, neglect the second-degree terms, therefore: \ \( f(x,y) \approx f(0, 0) = 0 \).

Step 9: Estimate using second-order approximation

Use the expression from Step 7: \ \( f(0.2, -0.3) \approx \frac{1}{64}(0.2^2 + (-0.3)^2) = \frac{1}{64}(0.04 + 0.09) = \frac{1}{64} \times 0.13 = \frac{0.13}{64} \approx 0.00203125 \).

Step 10: Direct calculation using a calculator

Compute \( f(0.2, -0.3) = \tan((0.2^2 + (-0.3)^2)/64) = \tan(0.13/64) \approx \tan(0.00203125) \approx 0.00203125 \).

Key Concepts

Partial Derivatives

To understand partial derivatives, we need to delve into the realm of calculus, particularly when dealing with functions of multiple variables. Imagine we have a function, like our function \(f(x, y)=\tan \left((x^{2}+y^{2}) / 64\right)\), which depends on two variables, \(x\) and \(y\). Each variable influences the outcome of the function in its unique way. Partial derivatives allow us to focus on how the function changes with respect to one of these variables at a time, keeping the other constant.

Here's how it works:

• For the partial derivative with respect to \(x\), denoted by \(f_x(x, y)\), we are testing how \(f(x,y)\) changes as \(x\) changes, holding \(y\) constant.
• Similarly, the partial derivative with respect to \(y\), written as \(f_y(x, y)\), describes the change in \(f\) as \(y\) changes while \(x\) stays put.

By applying the rules of differentiation, such as the chain rule, we can calculate these derivatives. In our exercise, we found that \(f_x = \frac{2x}{64}\sec^2\left(\frac{x^2 + y^2}{64}\right)\) and \(f_y = \frac{2y}{64}\sec^2\left(\frac{x^2 + y^2}{64}\right)\).

These derivatives tell us how tiny changes in \(x\) or \(y\) impact \(f\) when we are near the point \((0,0)\). Partial derivatives are crucial for constructing Taylor approximations.

Calculus

Calculus is the mathematical toolkit used to study change. It's split mainly into two branches: differential and integral calculus. In our scenario, differential calculus is the star of the show, since we're interested in understanding how variables interact within a function using derivatives.

With Taylor series approximations, we use derivatives to approximate the values of complex functions near a given point. The concept hinges on the idea that you can explain a function using its derivatives at a specific point. For a second-order Taylor approximation of a function \(f(x, y)\), we use:

\[ f(x,y) \approx f(x_0, y_0) + f_x(x_0, y_0)(x-x_0) + f_y(x_0, y_0)(y-y_0) + \frac{1}{2}f_{xx}(x_0, y_0)(x-x_0)^2 + \frac{1}{2}f_{yy}(x_0, y_0)(y-y_0)^2 + f_{xy}(x_0, y_0)(x-x_0)(y-y_0) \]

This formula lets us build an approximate picture of the function's behavior around the base point by using the function's value, its first-order and second-order partial derivatives.

In our specific exercise, we explored these parts to find that the second-order approximation at \((0,0)\) simplifies to \(f(x, y) \approx \frac{1}{64}(x^2 + y^2)\). Such approximations make it easier to grasp how small shifts in \(x\) and \(y\) can affect \(f(x, y)\), especially when directly calculating \(f\) is difficult.

Multivariable Functions

Multivariable functions extend the idea of a regular function to more than one variable. Instead of \(f(x)\) which depends on a single variable, we now have \(f(x, y)\), which depends on two variables. Such functions are quite common in real-world scenarios, where outcomes often depend on several factors.

These functions take inputs, for example, \((x, y)\), and map them to an output value. A physical analogy could be the terrain on a map, where each point defined by \(x\) and \(y\) coordinates has a certain elevation.

Understanding these functions requires a grasp of how changes in each variable affect the overall function; this is where partial derivatives come into play. In our specific function \(f(x, y)=\tan \left((x^2 + y^2)/64\right)\), \((x^2+y^2)/64\) acts as a composite function that affects the angle input to the tangent function.

In the task, the multivariable function's behavior is examined through the lens of Taylor series approximation. By doing this, we build simplified models around a point \((x_0, y_0)\), making predictions about \(f\) without needing to compute complex trigonometric functions directly. Taylor series thereby serve as a bridge, turning the challenge of direct computation into a more manageable task using the algebraic manipulation of derivatives.