Problem 208 - Calculus 3

Question

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

Step-by-Step Solution

Answer

The linear approximation at \( P(0,0) \) is \( L(x, y) = 1 + x \).

Step 1: Understand the formula for linear approximation

The linear approximation, or the tangent plane approximation of a function of two variables at a point \(P(x_0, y_0)\), is given by \(L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)\), where \(f_x\) and \(f_y\) are the partial derivatives with respect to \(x\) and \(y\), respectively.

Step 2: Find the partial derivative with respect to x

To find \(f_x\), the partial derivative of \(f(x, y) = e^x \cos y\) with respect to \(x\), treat \(y\) as a constant. Thus, \(f_x = \frac{\partial}{\partial x}(e^x \cos y) = e^x \cos y\).

Step 3: Find the partial derivative with respect to y

To find \(f_y\), the partial derivative of \(f(x, y) = e^x \cos y\) with respect to \(y\), treat \(x\) as a constant. Thus, \(f_y = \frac{\partial}{\partial y}(e^x \cos y) = -e^x \sin y\).

Step 4: Evaluate the function and partial derivatives at P(0,0)

Evaluate \(f(0,0) = e^0 \cos 0 = 1\). For the partial derivatives, evaluate \(f_x(0,0) = e^0 \cos 0 = 1\) and \(f_y(0,0) = -e^0 \sin 0 = 0\).

Step 5: Write the linear approximation function

Substitute the evaluated values into the linear approximation formula: \(L(x, y) = 1 + 1 \times (x - 0) + 0 \times (y - 0) = 1 + x\). Therefore, the linear approximation of the function at the point \(P(0,0)\) is \(L(x, y) = 1 + x\).

Key Concepts

Multivariable Calculus

Multivariable calculus is the extension of calculus to functions of multiple variables, rather than just one. In single-variable calculus, we often deal with functions like \(f(x)\), which have one input and one output. However, in multivariable calculus, we study functions such as \(f(x, y)\), which depend on two or more inputs. This allows us to analyze systems and phenomena that are inherently multidimensional.
The focus of multivariable calculus is understanding how these variables interact and affect the overall function. For example, when investigating the function \(f(x, y) = e^x \cos y\) at a point \((x_0, y_0)\), we look at how changes in both \(x\) and \(y\) influence \(f(x, y)\).
These concepts are fundamental in fields like physics, engineering, and economics, where systems often depend on several parameters, hence making multivariable calculus an essential part of scientific and mathematical modeling.

• Function of multiple variables: \(f(x, y)\)
• Extension of single-variable calculus
• Models multidimensional phenomena

Partial Derivatives

Partial derivatives are a critical component of multivariable calculus, providing a way to understand how a function changes as one variable changes while keeping the others constant.
For a function \(f(x, y)\), the partial derivative with respect to \(x\), denoted as \(f_x(x, y)\), measures the rate of change of the function as \(x\) changes, holding \(y\) fixed. Similarly, the partial derivative with respect to \(y\), denoted as \(f_y(x, y)\), tells us how \(f\) changes as \(y\) changes, holding \(x\) fixed.
In the original exercise, we found the partial derivatives for \(f(x, y) = e^x \cos y\):

• \(f_x(x, y) = e^x \cos y\)
• \(f_y(x, y) = -e^x \sin y\)

This information becomes particularly useful in determining the local approximation of a function near a specific point using linear approximation techniques.

Tangent Plane Approximation

The tangent plane approximation, also known as linear approximation, involves using a plane to approximate the surface of a multivariable function near a specific point. This is analogous to using the tangent line to approximate a function in single-variable calculus.
The formula for the tangent plane, \(L(x, y)\), at a point \((x_0, y_0)\) is:

• \(L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)\)

In the exercise, this was used to find the linear approximation of \(f(x, y)=e^{x} \cos y\) at the point \(P(0,0)\).
This involves:

• Calculating \(f(0,0)\), \(f_x(0,0)\), and \(f_y(0,0)\),
• Plugging them into the tangent plane formula,
• The result: \(L(x, y) = 1 + x\), showing how locally the function behaves around \((0,0)\).

The tangent plane gives a simple linear representation of the complex behavior of a function, enabling easier analysis near the point of tangency.