Problem 208 - Calculus Volume 3

Question

Find the linear approximation of each function at the indicated point. \(\quad f(x, y)=e^{x} \cos y ; P(0,0)\)

Step-by-Step Solution

Answer

The linear approximation is \(L(x,y) = 1 + x\).

Step 1: Understand the Formula for Linear Approximation

The linear approximation of a function \(f(x,y)\) at a point \((a,b)\) is given by the formula: \[ L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b) \]Here, \(f_x\) and \(f_y\) are the partial derivatives of \(f\) with respect to \(x\) and \(y\), respectively.

Step 2: Calculate the Function Value at the Point

Evaluate the function \(f(x, y) = e^x \cos y\) at the point \((0, 0)\):\[ f(0, 0) = e^0 \cos(0) = 1 \cdot 1 = 1 \]

Step 3: Compute Partial Derivative with respect to \(x\)

Find the partial derivative of \(f\) with respect to \(x\):\[ f_x(x, y) = \frac{\partial}{\partial x}[e^x \cos y] = e^x \cos y \] Evaluate it at \((0,0)\):\[ f_x(0, 0) = e^0 \cos(0) = 1\]

Step 4: Compute Partial Derivative with respect to \(y\)

Find the partial derivative of \(f\) with respect to \(y\):\[ f_y(x, y) = \frac{\partial}{\partial y}[e^x \cos y] = -e^x \sin y \] Evaluate it at \((0,0)\):\[ f_y(0, 0) = -e^0 \sin(0) = 0\]

Step 5: Substitute Values into Linear Approximation Formula

Substitute \(f(0, 0) = 1\), \(f_x(0, 0) = 1\), and \(f_y(0, 0) = 0\) into the linear approximation formula:\[ L(x, y) = 1 + 1\cdot(x - 0) + 0\cdot(y - 0) = 1 + x \]

Key Concepts

Partial Derivatives

Partial derivatives are a foundational concept in multivariable calculus. They measure how a multivariable function changes as we adjust one variable while keeping others constant. If you have a function like \( f(x, y) = e^x \cos y \), you can find the partial derivative with respect to \( x \) while treating \( y \) as a constant. This derivation shows how \( f \) changes as \( x \) varies. Conversely, take the partial derivative with respect to \( y \) to see the effect on \( f \) as \( y \) changes.

For the given function, the partial derivative with respect to \( x \) is \( f_x(x, y) = e^x \cos y \). Calculating at the point \((0, 0)\) gives \( f_x(0, 0) = 1 \). The partial derivative with respect to \( y \) is \( f_y(x, y) = -e^x \sin y \), and evaluated at \((0, 0)\) results in \( f_y(0, 0) = 0 \).

• Understanding partial derivatives helps you analyze the slope or rate of change of a surface at a given point.
• This concept becomes crucial when applying methods like linear approximation to predict function behavior near a point of interest.

Multivariable Functions

Multivariable functions extend the concept of a function from single-variable Inputs to ones involving multiple variables. For example, \( f(x, y) = e^x \cos y \) is a multivariable function because it depends on both \( x \) and \( y \). These functions can represent surfaces in space, which makes their study important in understanding phenomena in physics, engineering, and economics.

One unique aspect of multivariable functions is how they allow for complex interactions between variables. Adjusting one variable can significantly affect the output, especially if the function isn't linear. Linear approximation helps simplify these interactions by focusing on changes near a specific point.

• Linear approximation involves finding a tangent plane, akin to finding a tangent line for single-variable functions, that best represents the surface near a point.
• Understanding these ideas is vital for visualizing and solving real-world problems where several factors interact simultaneously.

Calculus

Calculus provides the tools necessary to analyze change and accumulation. For multivariable functions, calculus expands into partial derivatives, integral calculus, and more.

With linear approximation, we use calculus principles to predict the values of functions near a given point. This adaptability makes calculus indispensable in fields that require precision and foresight, such as engineering and economics. Linear approximation uses derivatives to approximate values, yielding the tangent plane equation in multivariable settings.

• Calculus allows us to find the slope of the tangent plane using partial derivatives, providing clear insight into how the function behaves near the approximation point.
• By simplifying complex functions, linear approximation becomes powerful for analysis, reducing computational complexity.

Understanding calculus deeply enriches your problem-solving toolbox and empowers you to tackle complicated scenarios by breaking them into manageable parts.