91Ó°ÊÓ

When we talk about partial derivatives, we are focusing on the change of a function with respect to one variable while keeping the others constant. This is particularly useful when dealing with functions of more than one variable, like functions of two variables usually denoted as \(f(x, y)\).

In the provided exercise, the function is \(f(x, y) = x^{2} - y^{2} - 2x + 4y + 6\). To find where it changes in the \(x\) and \(y\) direction, we compute the partial derivatives.

• The partial derivative with respect to \(x\) is determined by differentiating \(f\) while considering \(y\) constant, resulting in \(f_x = 2x - 2\).
• The partial derivative with respect to \(y\) is found similarly by treating \(x\) as a constant, giving us \(f_y = -2y + 4\).

These partial derivatives help us determine the critical points of the function by identifying where the slope equals zero (i.e., flat spots on the graph). In this case, setting these partial derivatives to zero led to the solution \((x, y) = (1, 2)\).

By finding the critical points, we seek to identify potential maxima, minima, or saddle points of the function.

The second derivative test is a valuable tool for analyzing critical points found via partial derivatives. Its purpose is to determine the nature of these critical points—whether they're local maxima, minima, or saddle points.

First, we calculate the second partial derivatives, which involve differentiating the first partial derivatives again:

• \(f_{xx}\), the second partial derivative with respect to \(x\), is \(2\).
• \(f_{yy}\), the second partial derivative with respect to \(y\), is \(-2\).
• \(f_{xy}\), the mixed derivative, is \(0\).

These derivatives are then used to calculate the value \(D\), known as the determinant of the Hessian matrix, using the formula:\[D = f_{xx} f_{yy} - (f_{xy})^2\]For this exercise, \(D = (2)(-2) - (0)^2 = -4\).

The sign of \(D\) helps us discern the nature of the critical point:- 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.

Here, since \(D < 0\), the critical point \((1, 2)\) is a saddle point.

The Hessian matrix is a square matrix composed of second-order partial derivatives of a function. It is a crucial component when employing the second derivative test to examine the convexity or concavity characteristics of a multivariable function at a point.

For a function \(f(x, y)\), the Hessian matrix is structured as follows:\[H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{xy} & f_{yy}\end{bmatrix}\]

In our specific problem:

• \(f_{xx} = 2\)
• \(f_{yy} = -2\)
• \(f_{xy} = 0\)

Accordingly, the Hessian matrix is:\[H = \begin{bmatrix} 2 & 0 \ 0 & -2\end{bmatrix}\]
The determinant of the Hessian matrix, calculated as \(D = f_{xx} f_{yy} - (f_{xy})^2\), is a critical number used to apply the second derivative test.

The Hessian helps us understand the local behavior of the surface described by \(f(x, y)\) at a critical point. In this exercise, the determinant is \(-4\), leading us to conclude that the critical point \((1, 2)\) is a saddle point. Understanding the Hessian provides a more comprehensive insight into how the function behaves around critical points.