Problem 41 - Calculus Early Transcendentals

Question

Show that the Second Derivative Test is inconclusive when applied to the following functions at \((0,0) .\) Describe the behavior of the function at the critical point. $$f(x, y)=x^{4} y^{2}$$

Step-by-Step Solution

Answer

Short Answer: We computed the first and second partial derivatives of the function $$f(x, y) = x^4y^2$$, and found the critical point \((0,0)\). Then, we evaluated the Hessian matrix at this point and found it to be a zero matrix. Since the determinant of the Hessian matrix at \((0,0)\) is 0, the Second Derivative Test is inconclusive, and we cannot determine the behavior of the function at the critical point \((0,0)\) based on the Second Derivative Test alone.

Step 1: Compute the first partial derivatives

Compute the first partial derivatives with respect to x and y: $$\frac{\partial f}{\partial x} = \frac{\partial (x^4 y^2)}{\partial x} = 4x^{3}y^{2}$$ $$\frac{\partial f}{\partial y} = \frac{\partial (x^4 y^2)}{\partial y} = 2x^{4}y$$

Step 2: Check for critical points

To find critical points, set both first partial derivatives equal to zero and solve for x and y: $$4x^{3}y^{2} = 0$$ $$2x^{4}y = 0$$ Solving these equations, we find the critical point \((0,0)\).

Step 3: Compute the second partial derivatives and form a Hessian matrix

Compute the second partial derivatives with respect to x and y: $$\frac{\partial^2 f}{\partial x^2} = 12x^2y^2$$ $$\frac{\partial^2 f}{\partial y^2} = 2x^4$$ $$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = 8x^{3}y$$ The Hessian matrix will be of the form: $$H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} \end{bmatrix} = \begin{bmatrix} 12x^2y^2 & 8x^3y \\ 8x^3y & 2x^4 \end{bmatrix}$$ Evaluate the Hessian matrix at the critical point \((0,0)\): $$H(0,0) = \begin{bmatrix} 0 & 0 \\ 0 & 0\end{bmatrix}$$

Step 4: Apply the Second Derivative Test and analyze the critical point

The Second Derivative Test states that for a given critical point, if the determinant of the Hessian matrix is positive and the second partial derivative with respect to x is also positive, then the point is a local minimum. If the determinant is positive and the second partial derivative with respect to x is negative, then the point is a local maximum. If the determinant is negative, then the point is a saddle point. If the determinant is zero, the test is inconclusive. Calculate the determinant of the Hessian matrix at the critical point \((0,0)\): $$\det(H) = (0)(0) - (0)(0) = 0$$ Since the determinant of the Hessian matrix is zero, the Second Derivative Test is inconclusive. We cannot conclude if the point is a local minimum, local maximum, or a saddle point based on the Second Derivative Test.

Key Concepts

Hessian Matrix

The Hessian matrix is an important tool in multivariable calculus, especially when analyzing the nature of critical points. It is a square matrix composed of second-order partial derivatives of a scalar-valued function. For a function of two variables, like in our case, the Hessian matrix is a 2x2 matrix:

• The first entry \(\frac{\partial^2 f}{\partial x^2}\) represents the second partial derivative concerning the first variable, x.
• The second entry \(\frac{\partial^2 f}{\partial x \partial y}\) represents the mixed partial derivative between x and y.
• The third entry, equal to the second due to symmetry, \(\frac{\partial^2 f}{\partial y \partial x}\), and the final component \(\frac{\partial^2 f}{\partial y^2}\), is the second partial derivative with respect to y.

For the given function \ f(x, y) = x^4 y^2 \, the Hessian matrix at any point (x, y) is denoted by the matrix:\[H = \begin{bmatrix} 12x^2y^2 & 8x^3y \ 8x^3y & 2x^4 \end{bmatrix}\]At the critical point (0,0), this matrix becomes a zero matrix, indicating that it cannot provide clear information about the nature of the critical point.

Critical Points

Critical points of a function, especially in two or more variables, are where its first derivatives equal zero. These points are potential candidates for local maxima, minima, or saddle points. To find them, we solve the system of equations created by setting all first partial derivatives to zero.
For example, with the function \(f(x, y) = x^4 y^2\), the partial derivatives are:

• \(\frac{\partial f}{\partial x} = 4x^3y^2\)
• \(\frac{\partial f}{\partial y} = 2x^4y\)

Setting these equal to zero, we end up with:

• \(4x^3y^2 = 0\)
• \(2x^4y = 0\)

The simultaneous solution to these equations is the critical point (0,0). Here, however, the Hessian matrix does not help confirm the exact nature of the point due to its determinant being zero, leaving further analysis necessary.

Partial Derivatives

Partial derivatives are derivatives of a function of multiple variables taken with respect to one variable while keeping the other variables constant. They are foundational in understanding changes in multivariable functions.

In the context of this problem, the goal is to find the first partial derivatives to locate critical points:

• The derivative with respect to x is calculated by treating y as a constant: \(\frac{\partial}{\partial x}(x^4 y^2) = 4x^3 y^2\).
• Similarly, the derivative with respect to y is found by treating x as a constant: \(\frac{\partial}{\partial y}(x^4 y^2) = 2x^4 y\).

These derivatives are used to find values of x and y such that both derivatives are zero. These values define the critical points of the function, which are further explored using second derivatives and the Hessian matrix for definitive characteristics.

Determinant of a Matrix

The determinant is a scalar value that can be computed from the elements of a square matrix, offering insights into the matrix's properties. For a 2x2 matrix:

• \[A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\]
• The determinant is calculated as \(\det(A) = ad - bc\).;

In the Second Derivative Test, the determinant of the Hessian matrix helps us understand the nature of critical points:

• If positive, it suggests a local minimum or maximum, depending on another condition.
• If negative, the point is typically a saddle point.
• If zero, like in our problem, the Second Derivative Test becomes inconclusive.

For the Hessian at (0,0) in the function \(f(x, y) = x^4 y^2\), the determinant is zero, suggesting the test is inconclusive at this critical point. Thus, determining the nature of the critical point requires other techniques beyond the second derivative test.