Problem 13 - Calculus Early Transcendentals

Question

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. $$f(x, y)=e^{x} \cos y$$

Step-by-Step Solution

Answer

The function has saddle points at \( (x, y) \) with \( y = (2k+1)\frac{\pi}{2} \). No local maxima or minima exist.

Step 1: Find Partial Derivatives

First, we need to find the partial derivatives of the function \( f(x, y) = e^x \cos y \). The partial derivative with respect to \( x \) is \( f_x = \frac{\partial f}{\partial x} = e^x \cos y \), and the partial derivative with respect to \( y \) is \( f_y = \frac{\partial f}{\partial y} = -e^x \sin y \).

Step 2: Solve for Critical Points

To find critical points, we set the partial derivatives equal to zero: \( f_x = e^x \cos y = 0 \) and \( f_y = -e^x \sin y = 0 \). From \( f_x = 0 \), we get \( \cos y = 0 \). This occurs when \( y = (2n+1)\frac{\pi}{2} \) for integers \( n \). From \( f_y = 0 \), we have \( \sin y = 0 \), which gives us \( y = m\pi \) for integers \( m \). These are satisfied when \( y = (2k+1)\pi/2 \), but no \( y \) satisfies both conditions simultaneously. Hence, the critical points' solution is when \( x = x \) and \( y = (2k+1)\frac{\pi}{2} \).

Step 3: Second Derivative Test

To classify these critical points, we calculate the second partial derivatives: \( f_{xx} = e^x \cos y \), \( f_{yy} = -e^x \cos y \), and \( f_{xy} = -e^x \sin y \). The Hessian determinant \( H \) at any critical point is \( H = f_{xx}f_{yy} - (f_{xy})^2 = (e^x \cos y)(-e^x \cos y) - (e^x \sin y)^2 = -e^{2x} (\cos^2 y + \sin^2 y) = -e^{2x} \). Since \( H eq 0 \) and is negative all over domain due to \( e^{2x} > 0 \), it implies that the critical points are saddle points.

Step 4: Conclusion and Graphing

Based on the second derivative test, there are no local maxima or minima in the function; all critical points are saddle points. If graphing software is used, setting different values for \( x \) and a series for \( y = (2k+1)\frac{\pi}{2} \) will help visualize the saddle points.

Key Concepts

Partial Derivatives

Partial derivatives are a way to analyze how a multivariable function changes as we alter one variable, keeping the others constant. For any function of two variables like our function \( f(x, y) = e^x \cos y \), finding the partial derivatives means computing the derivative with respect to each variable separately.

• The partial derivative with respect to \( x \), denoted as \( f_x \), is found by holding \( y \) constant and differentiating \( e^x \cos y \) with respect to \( x \), resulting in \( f_x = e^x \cos y \).
• The partial derivative with respect to \( y \), denoted as \( f_y \), is found by holding \( x \) constant and differentiating \( e^x \cos y \) with respect to \( y \), leading to \( f_y = -e^x \sin y \).

Partial derivatives are pivotal in identifying critical points, as these points occur where the partial derivatives vanish (i.e., are equal to zero).
They provide insights into how the function behaves along different axes, which set the stage for further analysis.

Hessian Matrix

The Hessian matrix is a square matrix of second-order partial derivatives of a scalar-valued function, providing a comprehensive view of the curvature around a point. For our function \( f(x, y) = e^x \cos y \), the Hessian matrix \( H \) is:

• \( f_{xx} = e^x \cos y \)
• \( f_{yy} = -e^x \cos y \)
• \( f_{xy} = f_{yx} = -e^x \sin y \)

The Hessian matrix is then given as:\[H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{bmatrix} = \begin{bmatrix} e^x \cos y & -e^x \sin y \ -e^x \sin y & -e^x \cos y \end{bmatrix}\]
Using the Hessian, we conclude that the nature of the critical point depends on its determinant \( H = f_{xx}f_{yy} - (f_{xy})^2 = -e^{2x} \).
The negative determinant in this problem hints towards the presence of saddle points.

Saddle Points

Saddle points are critical points where the surface curves upwards in one direction and downwards in another, resembling a saddle. They're neither local maxima nor minima. For our function \( f(x, y) = e^x \cos y \), all critical points are saddle points because the Hessian determinant is negative:

• A critical point becomes a saddle point when the Hessian matrix determinant is negative, indicating mixed curvatures in different directions.
• Since \( H = -e^{2x} < 0 \) for all \( x \), each critical point behaves as a saddle point.

Understanding saddle points is crucial in fields like economics and physics, where such points can represent equilibrium states that are stable against small perturbations in some directions.

Critical Points

Critical points are locations on a graph where the gradient (the vector of partial derivatives) is zero. For the function \( f(x, y) = e^x \cos y \), to find these points, we solved the conditions \( f_x = 0 \) and \( f_y = 0 \):

• From \( f_x = 0 \), we derived that \( \cos y = 0 \), leading to points where \( y = (2n+1)\frac{\pi}{2} \).
• From \( f_y = 0 \), we got \( \sin y = 0 \), resulting in points where \( y = m\pi \).

No \( y \) satisfies both equations, so critical points occur at any \( x \) value with \( y = (2k+1)\frac{\pi}{2} \), yielding pairs of \( (x, y) \) without unique values at a particular \( y \).
Analyzing critical points is foundational for understanding the behavior of functions in calculus, including determining where maxima, minima, or saddle points may exist.