Problem 209 - Calculus 3

Question

Find the linear approximation of each function at the indicated point. $$f(x, y)=\arctan (x+2 y), P(1,0)$$

Step-by-Step Solution

Answer

Linear approximation: \( L(x, y) = \frac{\pi}{4} - \frac{1}{2} + \frac{1}{2}x + y \).

Step 1: Identify the point and function

The given function is \[ f(x, y) = \arctan(x + 2y) \]and we are interested in finding the linear approximation at the point \( P(1, 0) \). We will use the formula for the linear approximation, which is \[ L(x, y) = f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b) \]where \((a, b)\) is the point \((1, 0)\).

Step 2: Calculate partial derivatives

The partial derivatives of \(f(x,y) = \arctan(x + 2y)\) are:- \( f_x(x, y) = \frac{1}{1 + (x + 2y)^2} \) using the chain rule.- \( f_y(x, y) = \frac{2}{1 + (x + 2y)^2} \) because the derivative of \(2y\) is \(2\).These derivatives will be evaluated at the point \((1, 0)\).

Step 3: Evaluate function and derivatives at the point

At the point \((1, 0)\), - \( f(1, 0) = \arctan(1 + 2 \times 0) = \arctan(1) = \frac{\pi}{4} \).- \( f_x(1, 0) = \frac{1}{1 + (1 + 2 \times 0)^2} = \frac{1}{2} \).- \( f_y(1, 0) = \frac{2}{1 + (1 + 2 \times 0)^2} = 1 \).

Step 4: Write the linear approximation

Insert the evaluated values into the formula for the linear approximation:\[ L(x, y) = \frac{\pi}{4} + \frac{1}{2}(x - 1) + 1(y - 0) \].Simplify this to get:\[ L(x, y) = \frac{\pi}{4} + \frac{1}{2}x - \frac{1}{2} + y \].Combine the constants:\[ L(x, y) = \frac{\pi}{4} - \frac{1}{2} + \frac{1}{2}x + y \].

Key Concepts

Partial Derivatives

When dealing with functions of multiple variables, partial derivatives help us understand how the function changes with respect to each variable individually. For a function like \( f(x, y) = \arctan(x + 2y) \), partial derivatives \( f_x \) and \( f_y \) are calculated. Here, \( f_x(x, y) \) represents how \( f \) changes as \( x \) changes, with \( y \) held constant. Similarly, \( f_y(x, y) \) shows the change in \( f \) as \( y \) changes, keeping \( x \) constant.

Partial derivatives are crucial in linear approximation because they determine the slope of the tangent plane to the surface at a specified point. Understanding how to compute these derivatives, such as \( f_x(x, y) = \frac{1}{1 + (x + 2y)^2} \) and \( f_y(x, y) = \frac{2}{1 + (x + 2y)^2} \), allows us to predict how small changes in \( x \) or \( y \) will affect the function value.

Chain Rule

The chain rule is a powerful tool in calculus, enabling you to find the derivative of composite functions. In the context of the multi-variable function \( f(x, y) = \arctan(x + 2y) \), using the chain rule is essential to correctly compute the partial derivatives. We treat \( x + 2y \) as a single variable and derive it step by step.

When calculating \( f_x \), for instance, the chain rule allows us to differentiate \( \arctan(u) \) with respect to \( u \), yielding \( \frac{1}{1 + u^2} \), and then differentiate \( u = x + 2y \) with respect to \( x \), which is simply 1. This interconnected method provides the full derivative \( f_x(x, y) = \frac{1}{1 + (x + 2y)^2} \).
The same applies to \( f_y \) where the additional factor from differentiating \( 2y \) gives us the extra factor of 2 in \( f_y(x, y) = \frac{2}{1 + (x + 2y)^2} \). These steps are often condensed but are central to understanding how each part of the function contributes to the overall rate of change.

Function Evaluation

Once partial derivatives are determined, evaluating both the function and its derivatives at a particular point provides specific numeric values pertinent to the approximation. In our problem, at the point \( P(1, 0) \):

• The original function \( f(1, 0) = \arctan(1) = \frac{\pi}{4} \)
• The derivative \( f_x(1, 0) = \frac{1}{2} \)
• The derivative \( f_y(1, 0) = 1 \)

Easier said, evaluating these allows us to plug numbers into the linear approximation formula, transforming abstract derivatives into tangible slopes aiding in sketching the tangent planes and approximating values.

Tangent Plane

The tangent plane is akin to touching the surface of a 3D object with a flat plane. In linear approximation, the tangent plane closely approximates a surface near a given point \( P(a, b) \). At \( P(1, 0) \), our function \( f(x, y) \) is approximated by the equation \( L(x, y) = \frac{\pi}{4} + \frac{1}{2}(x - 1) + 1(y - 0) \).

This simplified description provides a linearly flat, yet intimately connected understanding that helps us visualize and compute changes around \( P(1, 0) \). The approximation reflects the behavior of the original curved surface but is manageable with simple arithmetic, offering a practical way to assess small changes without exhaustive calculations on the actual function's curvature. Understanding the tangent plane helps demystify how calculus transforms complex surfaces into linear narratives. Engage the tangent plane as a bridge to better predict and understand the layered structure of multi-variable functions.