Problem 9 - Calculus for Biology and Medicine

Question

The tangent plane at the indicated point \(\left(x_{0}, y_{0}, z_{0}\right)\) exists. Find its equation. $$ f(x, y)=\ln (x+y) ;(2,-1,0) $$

Step-by-Step Solution

Answer

The equation of the tangent plane is \( z = x + y - 1 \).

Step 1: Understand the Tangent Plane Formula

The equation of a tangent plane to a surface \( z = f(x, y) \) at a point \((x_0, y_0, z_0)\) is given by: \[ z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) \] where \( f_x \) and \( f_y \) are the partial derivatives of \( f \) with respect to \( x \) and \( y \), respectively.

Step 2: Calculate Partial Derivatives

First, find \( f_x(x, y) \) and \( f_y(x, y) \) for the function \( f(x, y)=\ln(x+y) \). By using the chain rule: \[ f_x(x, y) = \frac{\partial}{\partial x}\ln(x+y) = \frac{1}{x+y} \times 1 = \frac{1}{x+y} \] \[ f_y(x, y) = \frac{\partial}{\partial y}\ln(x+y) = \frac{1}{x+y} \times 1 = \frac{1}{x+y} \]

Step 3: Evaluate at the Given Point

Calculate the partial derivatives at the point \((2, -1)\). Since \( (x_0, y_0) = (2, -1) \), we get:\[ f_x(2, -1) = \frac{1}{2-1} = 1 \] \[ f_y(2, -1) = \frac{1}{2-1} = 1 \]

Step 4: Write the Tangent Plane Equation

Substitute \( x_0 = 2 \), \( y_0 = -1 \), \( z_0 = 0 \), \( f_x(2, -1) = 1 \), and \( f_y(2, -1) = 1 \) into the tangent plane formula:\[ z = 0 + 1 \cdot (x - 2) + 1 \cdot (y + 1) \]Simplify to get:\[ z = x - 2 + y + 1 \]Thus, the equation of the tangent plane is:\[ z = x + y - 1 \]

Key Concepts

Calculus

Calculus is a branch of mathematics that studies how things change. It's all about understanding the rates at which quantities change, which makes it incredibly useful in fields like physics, engineering, and economics. Calculus can be divided into two main branches:

• Differential Calculus: This involves the study of derivatives and how they help determine the rate of change of a function.
• Integral Calculus: This branch focuses on the concept of area under curves, which is often related to accumulation of quantities.

In the context of finding a tangent plane, we focus on differential calculus. Understanding derivatives is key since they represent the slope of a function at a point. When this is done in more than one dimension, as in our exercise, we deal with partial derivatives. These allow us to understand how a function changes as one variable changes while keeping others constant. As a result, we can find the best fitting plane that just touches the surface of a 3D graph at a particular point.

Partial Derivatives

Partial derivatives play a crucial role in finding the equation of a tangent plane. Regular derivatives give us the rate of change of a function with respect to one variable. However, when a function depends on multiple variables, such as our example function, we use partial derivatives.Let's break it down: a function like \[ f(x, y) = \ln(x+y) \]depends on both \(x\) and \(y\). The partial derivative with respect to \(x\), denoted as \(f_x(x, y)\), describes how \(f\) changes as \(x\) changes, while \(y\) remains constant. Similarly, \(f_y(x, y)\) represents the change in \(f\) as \(y\) changes, while \(x\) is held constant.In our exercise, we calculated:

• \( f_x(x, y) = \frac{1}{x+y} \)
• \( f_y(x, y) = \frac{1}{x+y} \)

Evaluating these at the given point \((2, -1)\), we find both derivatives equal to 1, indicating that the function increases equally in both directions \(x\) and \(y\). This reflection of symmetry simplifies understanding the tangent plane's geometry.

Equation of a Plane

Establishing the equation of a plane is all about representing its position and slope in space. When working with functions of two variables, as shown in our exercise, the equation for the tangent plane is derived using the point and the partial derivatives.The general formula for a tangent plane is: \[ z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0) \]Here:

• \((x_0, y_0, z_0)\) is the point of tangency; it defines a specific place on the surface.
• \(f_x(x_0, y_0)\) and \(f_y(x_0, y_0)\) are the slopes of the surface in the \(x\) and \(y\) directions at that point.

For our function, substituting the given values, we calculated:\[ z = 0 + 1 \cdot (x - 2) + 1 \cdot (y + 1) \]After simplifying, this yields:\[ z = x + y - 1 \]This equation not only represents the surface but also gives insight into how small changes in \(x\) and \(y\) affect \(z\). It specifically ensures the plane 9\u00bbraces the surface at just one point, representing its tangent at the designated location.