Problem 332 - 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. $$f(x, y)=x^{2}+y-e^{y}$$

Step-by-Step Solution

Answer

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

Step 1: Find the First Derivatives

First, we need to find the partial derivatives of the function \(f(x, y) = x^2 + y - e^y\). Compute \(f_x\) and \(f_y\) as follows:\[\frac{\partial f}{\partial x} = f_x = 2x\]\[\frac{\partial f}{\partial y} = f_y = 1 - e^y\]

Step 2: Find Critical Points

Next, we set the first derivatives equal to zero to find the critical points. Solve the equations:\(2x = 0 \implies x = 0\)\(1 - e^y = 0 \implies e^y = 1 \implies y = 0\)Thus, the critical point is at \((0, 0)\).

Step 3: Find Second Derivatives

Now, we calculate the second derivatives, which are necessary for applying the second derivative test:\[\frac{\partial^2 f}{\partial x^2} = f_{xx} = 2\]\[\frac{\partial^2 f}{\partial y^2} = f_{yy} = -e^y\]\[\frac{\partial^2 f}{\partial x \partial y} = f_{xy} = 0\]

Step 4: Apply the Second Derivative Test

Evaluate the second derivative test at the critical point \((0, 0)\) using the determinant of the Hessian matrix. The Hessian is:\[D = f_{xx}f_{yy} - (f_{xy})^2\]Substitute the second derivative values:\[D = (2)(-e^0) - (0)^2 = 2(-1) = -2\]Since \(D < 0\), the critical point is a saddle point.

Key Concepts

Critical Points

Critical points are vital in determining the behavior of a function. A critical point is a location on a graph where the derivative is zero or undefined. For functions with two variables, like \( f(x, y) = x^2 + y - e^y \), critical points occur where both partial derivatives \( f_x \) and \( f_y \) are zero.

To find a critical point, solve the equations derived from setting these partial derivatives to zero. In our example, the critical point is found by solving \( 2x = 0 \) and \( 1 - e^y = 0 \). This gives us \( x = 0 \) and \( y = 0 \), so the critical point of the function \( f(x, y) \) is at \((0, 0)\).

Finding critical points is crucial because they act as candidates for local maxima, minima, or saddle points.

Partial Derivatives

Partial derivatives are the building blocks for finding critical points in multivariable calculus. They represent the rate of change of a function with respect to one of its variables, while keeping the others constant.

In the function \( f(x, y) = x^2 + y - e^y \), the partial derivative with respect to \( x \) (\( f_x \)) is \( 2x \) and with respect to \( y \) (\( f_y \)) is \( 1 - e^y \).

These derivatives help us understand how the function changes in different directions:

• The partial derivative \( f_x = 2x \) shows how the function changes with \( x \).
• The derivative \( f_y = 1 - e^y \) indicates how \( f(x, y) \) changes when \( y \) varies.

Finding where these partial derivatives equal zero allows us to identify critical points.

Hessian Matrix

The Hessian matrix is a square matrix composed of second order partial derivatives of a function, and is used in determining the nature of critical points.

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

• \( f_{xx} = 2 \) is the second partial derivative with respect to \( x \).
• \( f_{yy} = -e^y \) is the second partial derivative concerning \( y \).
• \( f_{xy} \) and \( f_{yx} \) typically represent mixed partial derivatives, which here are zero.

To apply the second derivative test, compute the determinant of the Hessian matrix \( D = f_{xx}f_{yy} - (f_{xy})^2 \). Evaluating at the critical point \((0, 0)\), we get \( D = -2 \).

Saddle Point

A saddle point is a special type of critical point where the function does not exhibit a local maximum or minimum but instead has a mixed type of behavior. In terms of topology, it resembles a horse saddle, curving up in one direction and down in the perpendicular direction.

In the second derivative test, determining whether a critical point is a saddle point relies on the determinant of the Hessian matrix.

• If \( D < 0 \), as in our example with \( D = -2 \), the critical point is a saddle point.

At a saddle point, the function's slope shifts from rising to falling (or vice versa), making them interesting in optimization problems but not suitable for max or min solutions.

Understanding saddle points helps in visualizing how a function behaves in different neighborhoods around the critical point, providing deeper insight into the function's nature.