Problem 17 - 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. $$f(x, y)=y^{2}-2 y \cos x, \quad 1 \leqslant x \leqslant 7$$

Step-by-Step Solution

Answer

Critical points are local minima and maxima depending on \(x\).

Step 1: Find Partial Derivatives

To identify critical points, we first find the partial derivatives of the function with respect to both variables. The partial derivative with respect to \(x\) is \(f_x(x, y) = 2y \sin x\). The partial derivative with respect to \(y\) is \(f_y(x, y) = 2y - 2\cos x\).

Step 2: Set Partial Derivatives to Zero

To find critical points, we set the partial derivatives equal to zero. Thus, \(f_x(x, y) = 2y \sin x = 0\) and \(f_y(x, y) = 2y - 2\cos x = 0\). This gives us the conditions \(y \sin x = 0\) and \(y = \cos x\).

Step 3: Solve the Equations

Solving \(y \sin x = 0\), we get \(y=0\) or \(\sin x = 0\). For \(y=0\), substituting in \(y = \cos x\), we get \(y = \cos x = 0\), which doesn't give new information. Thus, considering \(\sin x = 0\), we get \(x = n\pi\) for \(n = 1, 2, ...\) within the given interval. Substituting into \(y = \cos x\), we find critical points at \((x, y) = (n\pi, (-1)^n))\) where \(x\) is in the interval \([1, 7]\).

Step 4: Classify the Critical Points

To determine local maxima, minima, or saddle points, we use the second derivative test. Compute the second partial derivatives: \(f_{xx} = 2y \cos x\), \(f_{yy} = 2\), and \(f_{xy} = 2 \sin x\). Calculate the determinant of the Hessian matrix \(H = f_{xx} f_{yy} - (f_{xy})^2\) evaluated at each critical point.

Step 5: Evaluate the Hessian Determinant

At critical points \((n\pi, (-1)^n)\), substitute values into Hessian determinants. For example, at \((2\pi, 1)\), calculate \(H = 2(2 \cdot 1) - (2\sin(2\pi))^2 = 4\). The determinant is positive, and \(f_{xx} > 0\) indicating local minima. At \((\pi, -1)\) and \((3\pi, -1)\), apply similar calculations to establish their nature.

Key Concepts

Critical Points

In multivariable calculus, critical points are essential to understand a function's behavior. For a function of two variables like our example, critical points are found by setting the partial derivatives with respect to each variable equal to zero. These points indicate where the function might have a local maximum, a local minimum, or a saddle point.

For the function given, we calculate the partial derivatives with respect to both variables, denoted as \( f_x(x, y) \) and \( f_y(x, y) \). Setting these derivatives to zero helps in locating potential critical points.

In our problem, \( f_x(x, y) = 2y \sin x \) and \( f_y(x, y) = 2y - 2\cos x \). By solving the equations \( y \sin x = 0 \) and \( y = \cos x \), we determine the critical points \((x, y) = (n\pi, (-1)^n))\), where \( x \) lies between 1 and 7, aligning with the boundaries of the function.

Partial Derivatives

Partial derivatives are the foundation of finding critical points in multivariable calculus. They measure how a function changes as its variables change. Partial derivatives with respect to one variable take all other variables as constants.

For the equation \( f(x, y) = y^2 - 2y \cos x \, \)finding the partial derivative with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \) or \( f_x \), means treating \( y \) as a constant. Similarly, finding \( \frac{\partial f}{\partial y} \) or \( f_y \) means treating \( x \) as a constant.

The calculated partial derivatives for our function are \( f_x(x, y) = 2y \sin x \) and \( f_y(x, y) = 2y - 2\cos x \). Setting these to zero was crucial in identifying the function's critical points.

Hessian Matrix

The Hessian matrix is an organized array of second-order partial derivatives of a function. It helps in analyzing the curvature near critical points. This information allows us to determine whether a critical point is a local maximum, local minimum, or saddle point.

For functions with two variables, the Hessian is a 2x2 matrix constructed as follows:

• \( f_{xx} \) - second partial derivative with respect to the first variable \( x \)
• \( f_{yy} \) - second partial derivative with respect to the second variable \( y \)
• \( f_{xy} \) and \( f_{yx} \) - second mixed partial derivatives

The Hessian matrix for our function is:
\[ H(x, y) = \begin{bmatrix} 2y \cos x & 2 \sin x \ 2 \sin x & 2 \end{bmatrix} \] Determining the sign of its determinant tells us about the nature of critical points.

Local Maxima and Minima

Local maxima and minima are points at which a function reaches its highest or lowest value in a certain neighborhood. They are crucial for understanding the topography of functions in multivariable calculus.

Once critical points are identified, we classify them using the determinant of the Hessian matrix, \( H \). If \( H > 0 \):

• and \( f_{xx} > 0 \), it indicates a local minimum.
• and \( f_{xx} < 0 \), it indicates a local maximum.

Conversely, if \( H < 0 \), the point is a saddle point, showing a change in direction or curvature.

For instance, at critical points like \((2\pi, 1)\), a positive Hessian determinant with \( f_{xx} > 0 \) confirms a local minimum. The evaluation of other points such as \((\pi, -1)\) would show whether they are maxima, minima, or saddle points.