Problem 333 - Calculus Volume 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. $$f(x, y)=e^{-\left(x^{2}+y^{2}+2 x\right)}$$

Step-by-Step Solution

Answer

The critical point \((-1, 0)\) is a local maximum.

Step 1: Find the First Partial Derivatives

To apply the second derivative test, we first need to find the critical points by finding where the first partial derivatives are equal to zero. Compute the partial derivative of \( f(x, y) \) with respect to \( x \), denoted as \( f_x \), and with respect to \( y \), denoted as \( f_y \).\[f_x = \frac{\partial}{\partial x} e^{-(x^2 + y^2 + 2x)} = e^{-(x^2 + y^2 + 2x)}(-2x - 2)\]\[f_y = \frac{\partial}{\partial y} e^{-(x^2 + y^2 + 2x)} = e^{-(x^2 + y^2 + 2x)}(-2y)\]

Step 2: Solve for Critical Points

Set the first partial derivatives equal to zero and solve for \( x \) and \( y \):\[-2x - 2 = 0 \quad \Rightarrow \quad x = -1\]\[-2y = 0 \quad \Rightarrow \quad y = 0\]Thus, the critical point is \( (-1, 0) \).

Step 3: Compute the Second Partial Derivatives

Next, calculate the second partial derivatives to use in the second derivative test:\[f_{xx} = \frac{\partial}{\partial x}(-2x - 2)e^{-(x^2 + y^2 + 2x)} = e^{-(x^2 + y^2 + 2x)}((4x + 4)(-1) + (-2))\]\[f_{yy} = \frac{\partial}{\partial y}(-2y)e^{-(x^2 + y^2 + 2x)} = e^{-(x^2 + y^2 + 2x)}((4y)(-1) + (-2))\]\[f_{xy} = \frac{\partial}{\partial y}(-2x - 2)e^{-(x^2 + y^2 + 2x)} = (4y)e^{-(x^2 + y^2 + 2x)}\]

Step 4: Evaluate the Second Derivative Test at the Critical Point

Evaluate the second derivatives at the critical point \((-1,0)\). The test involves the determinant of the Hessian matrix:\[D = f_{xx}(-1, 0) \cdot f_{yy}(-1, 0) - (f_{xy}(-1, 0))^2\]Calculate each term:\[f_{xx}(-1, 0) = e^{-1} \cdot (4 \cdot (-1) + 4 \cdot 0 + (-2)) = e^{-1} \cdot (-6)\]\[f_{yy}(-1, 0) = e^{-1} \cdot (4 \cdot 0 + (-2)) = e^{-1} \cdot (-2)\]\[f_{xy}(-1, 0) = 0\]Thus,\[D = e^{-1} \cdot (-6) \cdot e^{-1} \cdot (-2) - 0 = 12 \, e^{-2}\]

Step 5: Classify the Critical Point

From the second derivative test, since \( D > 0 \) and \( f_{xx}(-1, 0) < 0 \), the critical point \((-1, 0)\) is a local maximum of the function.

Key Concepts

Second Derivative Test

In multivariable calculus, the second derivative test is an important tool used to classify critical points of functions of two variables. These critical points are places where the function might have a local maximum, local minimum, or saddle point. To apply this test, we first find the critical points by setting the first partial derivatives of the function to zero. At these points, we examine the second partial derivatives. The test involves calculating the determinant ( \( D \) ) of the Hessian matrix, which is composed of these second partial derivatives. If \( D > 0 \) and \( f_{xx} \) at the critical point is less than zero, it indicates a local maximum. If \( f_{xx} \) is greater than zero, it indicates a local minimum. If \( D < 0 \) , it's a saddle point, and if \( D = 0 \), the test is inconclusive.

Understanding how to use the second derivative test allows you to discern important characteristics about the behavior of functions and their graphical representations.

Critical Points

Critical points are locations on the function where its gradient (or derivative) is zero. For a function of two variables, this means both partial derivatives with respect to each variable must be zero. Finding these points is the first step in determining where the function can potentially have maximum or minimum values. At critical points, the slope of the function is flat, making them prime candidates for local extrema.

In context, for the function \( f(x, y) = e^{-(x^2 + y^2 + 2x)} \), solving \( f_x = 0 \) and \( f_y = 0 \) gives us the critical point \( (-1, 0) \). This process involves setting the expressions obtained from the partial differentiations equal to zero and solving for the variables involved.

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They measure how the function changes as each independent variable changes, while keeping others constant. For example, if we have a function \( f(x, y) \), the partial derivative \( f_x \) represents the rate of change of \( f \) as \( x \) changes, while \( y \) remains unchanged.

To find critical points, the first partial derivatives are typically set to zero. This is because, at these points, the change in the function along each axis is zero, indicating potential maxima, minima, or saddle points. Calculating partial derivatives accurately is crucial and often requires using rules similar to those in single-variable calculus, such as the chain rule and product rule.

Hessian Matrix

The Hessian matrix is a square matrix composed of the second-order partial derivatives of a scalar-valued function. It is essential in the second derivative test for analyzing the concavity of functions and determining the nature of critical points. For a function \( f(x, y) \), the Hessian matrix \( H \) is given by:
\[H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{bmatrix}\]
The determinant of this matrix, denoted as \( D \), is used to classify critical points:

• If \( D > 0 \), and \( f_{xx} > 0 \), the point is a local minimum.
• If \( D > 0 \), and \( f_{xx} < 0 \), the point is a local maximum.
• If \( D < 0 \), the point is a saddle point.
• If \( D = 0 \), the test is inconclusive.

Understanding the Hessian matrix helps in visualizing how a function "curves" around its critical points, and thus, is invaluable in multivariable optimization and calculus.