Problem 332 - Calculus 3

Question

For the following exercises, use the second derivative test to identify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these. $$ f(x, y)=x^{2}+y-e^{y} $$

Step-by-Step Solution

Answer

The critical point at \((0, 0)\) is a saddle point.

Step 1: Find the First Derivatives

First, determine the partial derivatives \(f_x\) and \(f_y\). Calculate \(f_x = \frac{\partial}{\partial x} (x^2 + y - e^y) = 2x\).Calculate \(f_y = \frac{\partial}{\partial y} (x^2 + y - e^y) = 1 - e^y\).

Step 2: Identify Critical Points

To find the critical points, set the first derivatives \(f_x\) and \(f_y\) equal to zero:1. \(2x = 0 \Rightarrow x = 0\)2. \(1 - e^y = 0 \Rightarrow e^y = 1 \Rightarrow y = 0\)Thus, the only critical point is at \((x, y) = (0, 0)\).

Step 3: Find the Second Derivatives

Next, calculate the second partial derivatives to use in the second derivative test:1. \(f_{xx} = \frac{\partial^2}{\partial x^2} f = 2\)2. \(f_{yy} = \frac{\partial^2}{\partial y^2} f = -e^y\) (at critical point, \(y = 0\), so \(f_{yy} = -1\))3. \(f_{xy} = \frac{\partial^2}{\partial x \partial y} f = 0 \) as there is no \(xy\) term present.

Step 4: Apply the Second Derivative Test

Use the second derivative test:Calculate \(D = f_{xx}f_{yy} - (f_{xy})^2\).At \((0, 0)\), \(f_{xx} = 2\), \(f_{yy} = -1\), and \(f_{xy} = 0\).Thus, \(D = 2(-1) - 0^2 = -2\).

Step 5: Determine the Nature of the Critical Point

Since \(D = -2 < 0\), the critical point \((0, 0)\) is a saddle point.

Key Concepts

Critical Points

When working with multivariable functions, like our function \( f(x, y) = x^2 + y - e^y \), critical points play a vital role in understanding its behavior. They are the points in the domain where the gradient is zero or undefined. These are the points where the function can potentially have a local maximum, a local minimum, or a saddle point.
To find critical points in a function of two variables, we first calculate the partial derivatives with respect to each variable. In our example, we have the partial derivatives as \( f_x = 2x \) and \( f_y = 1 - e^y \).

• We then set these partial derivatives equal to zero to find the critical points.
• For \( f_x = 0 \), we get \( x = 0 \).
• For \( f_y = 0 \), solving \( 1 - e^y = 0 \) results in \( y = 0 \).

Hence, the critical point for the function is \((0, 0)\). Once identified, we can perform further analyses, like the second derivative test, to determine its nature.

Partial Derivatives

Partial derivatives extend the concept of derivatives to functions with multiple variables. They measure how a function changes as one specific variable changes, keeping the other variables constant.
For a function \( f(x, y) \), the partial derivatives are denoted as \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). In our example,

• The partial derivative with respect to \(x\) is \( f_x = 2x \), which tells us how the function changes as \( x \) changes.
• The partial derivative with respect to \(y\) is \( f_y = 1 - e^y \), showing the change in the function with variations in \( y \).

These partial derivatives are crucial for determining the slopes of the function in the \( x \) and \( y \) directions. They lead us to the critical points by setting \( f_x \) and \( f_y \) to zero. Moreover, they provide the foundation for calculating second partial derivatives used in the second derivative test.

Saddle Point

A saddle point, as seen in our function \( f(x, y) = x^2 + y - e^y \), is a type of critical point with interesting properties. It is neither a local maximum nor a local minimum. Instead, it resembles a saddle, where the surface curves upwards in one direction and downwards in another.
To determine if a critical point is a saddle point, we employ the second derivative test. This involves calculating the second partial derivatives and using them to compute the determinant:\[ D = f_{xx} f_{yy} - (f_{xy})^2. \]
In our problem:

• \( f_{xx} = 2 \) and \( f_{yy} = -e^y \) (which is \(-1\) at \( y = 0 \)).
• \( f_{xy} = 0 \).
• Therefore, \( D = 2 \times (-1) - 0^2 = -2 \).

The determinant \( D \) being negative ensures that our critical point \((0, 0)\) is indeed a saddle point. This test helps us differentiate saddle points from other types of critical points efficiently.