Problem 21 - Calculus for Biology and Medicine

Question

Find the linearization of \(f(x, y)\) at the indicated point \(\left(x_{0}, y_{0}\right).\) $$ f(x, y)=\tan (x+y) ;(0,0) $$

Step-by-Step Solution

Answer

The linearization of \( f(x,y) = \tan(x+y) \) at (0,0) is \( L(x, y) = x + y \).

Step 1: Understand the Concept of Linearization

Linearization of a function of two variables, such as \( f(x, y) \), around a point \((x_0, y_0)\) involves approximating the function by its tangent plane at that point. The formula for the linear approximation or linearization 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 of \( f \) with respect to \( x \) and \( y \), respectively.

Step 2: Calculate the Value of the Function at the Point

Calculate \( f(x_0, y_0) \) where \( x_0 = 0 \) and \( y_0 = 0 \). Substitute these values into the function: \[f(0, 0) = \tan(0 + 0) = \tan(0) = 0\]

Step 3: Find the Partial Derivative with Respect to x

Determine the partial derivative of \( f(x, y) = \tan(x+y) \) with respect to \( x \). The derivative of \( \tan(u) \) with respect to \( u \) is \( \sec^2(u) \), so: \[f_x(x, y) = \sec^2(x+y) \cdot 1 = \sec^2(x+y)\]To evaluate at \( (0, 0) \):\[f_x(0, 0) = \sec^2(0) = 1\]

Step 4: Find the Partial Derivative with Respect to y

Determine the partial derivative of \( f(x, y) = \tan(x+y) \) with respect to \( y \). Similarly, using the chain rule:\[f_y(x, y) = \sec^2(x+y) \cdot 1 = \sec^2(x+y)\]To evaluate at \( (0, 0) \):\[f_y(0, 0) = \sec^2(0) = 1\]

Step 5: Use the Linearization Formula

Substitute the calculated values into the linearization formula: \[L(x, y) = f(0, 0) + f_x(0, 0)(x - 0) + f_y(0, 0)(y - 0)\]This simplifies given our values:\[L(x, y) = 0 + 1\cdot x + 1\cdot y = x + y\]

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus, especially when dealing with functions of multiple variables. When you have a function like \( f(x, y) \), which depends on two variables, you can explore how the function changes as each variable varies, while keeping the other one constant. This is where partial derivatives come into play.

• Partial Derivative with respect to x: Denoted as \( f_x(x, y) \), reflects how the function changes as only \( x \) varies. In the context of our function \( \tan(x+y) \), \( f_x(x, y) \) is found by differentiating \( \tan(x+y) \) with respect to \( x \), resulting in \( \sec^2(x+y) \).
• Partial Derivative with respect to y: Similarly, \( f_y(x, y) \) captures the changes with respect to \( y \), also yielding \( \sec^2(x+y) \) when differentiating \( \tan(x+y) \) with respect to \( y \).

By evaluating these partial derivatives at a specific point, like \((0,0)\) in our example, you can understand the local behavior of the function around this point. Here, both partial derivatives equal 1, setting up a straightforward linear relationship between small changes in \( x \) and \( y \) and the resultant change in \( f \).

Tangent Plane

The concept of the tangent plane extends the idea of a tangent line from single-variable calculus to functions of multiple variables. For a function \( f(x, y) \), the tangent plane at a point \((x_0, y_0)\) serves as a linear approximation to the surface defined by the function. Think of it as the best flat surface that just "touches" the surface of \( f(x, y) \) at that point.

• Equation of the Tangent Plane: This is given by \( L(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) \). It involves the function value at the point and the partial derivatives, providing a plane that approximates the function around \((x_0, y_0)\).
• Slope and Changes: The partial derivatives \( f_x(x_0, y_0) \) and \( f_y(x_0, y_0) \) act like the slopes in different directions, giving you an idea of how steep or flat the plane is relative to the \( x \) and \( y \) axes.

In our exercise, the tangent plane simplistically simplifies to \( x + y \), implying an equal rate of change in both \( x \) and \( y \).

Multivariable Calculus

Multivariable calculus involves the calculus of functions with more than one variable. It's like expanding on the concepts of derivatives and integrals from single-variable calculus to functions of several variables, such as \( f(x, y) \). This branch of calculus attempts to deal with real-world problems where things are not dependent on just one factor, making it more robust and applicable.

• Understanding Change: In the world of multivariable functions, you get a fuller picture of how a function behaves by examining all its partial derivatives. This helps you see how closely or how far you are from an exact value when using approximations, like tangent planes.
• Applications: These concepts find usage in fields like engineering, economics, and sciences, providing insights that single-variable calculus could not.

In our example, by using linearization and the tangent plane, we are actually employing techniques of multivariable calculus to approximate the behavior of \( f(x, y) = \tan(x+y) \) near the point \((0,0)\). This approximative approach makes complex problems easier to handle, offering clearer insights into multifaceted systems.