Problem 396 - Calculus Volume 3

Question

Determine whether the statement is true or false. Justify your answer with a proof . The linear approximation to the function of \(f(x, y)=5 x^{2}+x \tan (y) \quad\) at \(\quad(2, \pi)\) is given by \(L(x, y)=22+21(x-2)+(y-\pi)\)

Step-by-Step Solution

Answer

The statement is false, the linear approximation is incorrect.

Step 1: Identify the function and verify the point

The given function is \(f(x,y) = 5x^2 + x\tan(y)\). We are tasked to find its linear approximation at the point \((2, \pi)\). The linear approximation formula at a point \((a, b)\) is \(L(x,y) = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)\), where \(f_x\) and \(f_y\) are the partial derivatives with respect to \(x\) and \(y\).

Step 2: Calculate the value of function at the point

Calculate \(f(2, \pi) = 5(2)^2 + 2 \tan(\pi)\). Knowing that \(\tan(\pi) = 0\), we have \(f(2, \pi) = 5 \times 4 = 20\).

Step 3: Compute partial derivative with respect to \(x\)

The partial derivative \(f_x(x,y)\) is found by differentiating only \(x\) in the function: \(f_x = \frac{\partial}{\partial x}(5x^2 + x\tan(y)) = 10x + \tan(y)\). Evaluate at \((2, \pi)\): \(f_x(2, \pi) = 10(2) + \tan(\pi) = 20\).

Step 4: Compute partial derivative with respect to \(y\)

The partial derivative \(f_y(x,y)\) is obtained by differentiating only \(y\): \(f_y = \frac{\partial}{\partial y}(5x^2 + x\tan(y)) = x\sec^2(y)\). Evaluate at \((2, \pi)\): \(f_y(2, \pi) = 2\sec^2(\pi) = 2\times 1 = 2\), because \( \sec(\pi) = 1\).

Step 5: Formulate the linear approximation

Substitute the function value and the derivatives into the linear approximation formula: \(L(x, y) = 20 + 20(x-2) + 2(y-\pi)\). Simplifying gives \(L(x, y) = 20 + 20x - 40 + 2y - 2\pi = 20x + 2y - 2\pi - 20\).

Step 6: Verify the given linear approximation

Compare the derived linear approximation \(L(x, y) = 20x + 2y - 20 - 2\pi\) to the given \(L(x, y) = 22 + 21(x-2) + (y-\pi)\). Both are not equivalent, hence the given approximation is not correct.

Key Concepts

Partial Derivatives

Partial derivatives are an essential concept when dealing with functions of multiple variables. They represent the rate of change of the function with respect to one variable while keeping the other variables constant. In our problem, the function is \( f(x,y) = 5x^2 + x\tan(y) \), a function of two variables, \( x \) and \( y \). To find the linear approximation at a specific point, we need the partial derivatives with respect to each variable.

• The partial derivative with respect to \( x \), denoted as \( f_x \), is found by differentiating \( f \) while treating \( y \) as a constant. This gives \( f_x = 10x + \tan(y) \).
• For \( y \), the partial derivative, \( f_y \), is computed by treating \( x \) as a constant. It results in \( f_y = x\sec^2(y) \).

These derivatives provide a framework for understanding the behavior of \( f \) in the neighborhood of any point \( (a, b) \). They are crucial for forming the linear approximation, as they describe the sensitivity of the function in each direction. The specific values of these derivatives at the point \( (2, \pi) \) allow us to construct the linear map that best approximates \( f \) near that point.

Tangent Function

The tangent function, \( \tan(y) \), is a fundamental trigonometric function involved in our given example. It is important to understand its properties to accurately evaluate the function \( f(x, y) \) and its derivatives, especially at specific points such as \( y = \pi \).
The tangent function is periodic with a period of \( \pi \). This means that \( \tan(y) = \tan(y + n\pi) \) for any integer \( n \). Notably, at \( y = \pi \), \( \tan(\pi) = 0 \). This fact simplifies our computations significantly, as seen in the evaluation of \( f(2, \pi) \) and \( f_x(2, \pi) \).

• Understanding that the tangent of \( \pi \) is zero helps simplify the terms involved in derivatives where \( y = \pi \).
• This zero value impacts the slope of the tangent plane, meaning part of the change in the function when \( y \) changes from \( \pi \) will predominantly rely on the secant function in \( f_y \).

Thus, knowledge of the tangent function and its values at critical points such as \( \pi \) is key for formulating an accurate linear approximation.

Linearization Formula

The linearization formula helps us to create a simple first-degree polynomial that approximates a function near a given point. For a function \( f(x, y) \), the linearization at a point \( (a, b) \) is derived using the formula: \[ L(x, y) = f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b) \] This formula effectively shifts and changes the scale of the plane tangent to the surface at \( (a, b) \).
The application of this formula is straightforward once you have all the necessary components:

• The function's value at the point, \( f(2, \pi) = 20 \), comes from evaluating the original function at that point.
• The partial derivatives, \( f_x(2, \pi) = 20 \) and \( f_y(2, \pi) = 2 \), are slopes of the tangent plane in the \( x \) and \( y \) directions, respectively.
• These elements combine to form the plane \( L(x, y) = 20 + 20(x-2) + 2(y-\pi) \), which simplifies to \( L(x, y) = 20x + 2y - 20 - 2\pi \).

This linearization approximates \( f(x, y) \) effectively around \( (2, \pi) \), but differs from the proposed approximation \( L(x, y) = 22 + 21(x-2) + (y-\pi) \), clarifying the problem statement's error.