Problem 19
Question
Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. \(f(x, y)=\ln \left(1+x^{2}+y^{2}\right)\)
Step-by-Step Solution
Verified Answer
The critical point of the function \(f(x, y) = \ln(1 + x^2 + y^2)\) is at (0, 0). Using the second derivative test, we determined that this point is a local minimum, with a value of f(0, 0) = 0.
1Step 1: Find the first partial derivatives
To begin, we will find the first partial derivative with respect to both x and y. These are given by:
\[\frac{\partial f}{\partial x} = \frac{d}{dx}\ln(1 + x^2 + y^2)\]
\[\frac{\partial f}{\partial y} = \frac{d}{dy}\ln(1 + x^2 + y^2)\]
Using the chain rule, we get:
\[\frac{\partial f}{\partial x} = \frac{2x}{1 + x^2 + y^2}\]
\[\frac{\partial f}{\partial y} = \frac{2y}{1 + x^2 + y^2}\]
2Step 2: Find the critical points
Critical points occur when both first partial derivatives equal zero. So, we need to solve the following system of equations:
\[\frac{2x}{1 + x^2 + y^2} = 0\]
\[\frac{2y}{1 + x^2 + y^2} = 0\]
The only solution for these equations is x = 0 and y = 0. Therefore, the critical point is at (0, 0).
3Step 3: Find the second partial derivatives and Hessian matrix
To classify the critical point, we need to compute the second partial derivatives and Hessian matrix. The second partial derivatives are given by:
\[\frac{\partial^2 f}{\partial x^2} = \frac{d^2}{dx^2}\ln(1+x^2+y^2)\]
\[\frac{\partial^2 f}{\partial x \partial y} = \frac{d^2}{dx dy}\ln(1+x^2+y^2)\]
\[\frac{\partial^2 f}{\partial y^2} = \frac{d^2}{dy^2}\ln(1+x^2+y^2)\]
Using implicit differentiation, we find that:
\[\frac{\partial^2 f}{\partial x^2} = \frac{2(1 - x^2 + y^2)}{(1 + x^2 + y^2)^2}\]
\[\frac{\partial^2 f}{\partial x \partial y} = \frac{d}{dy}(\frac{2x}{1 + x^2 + y^2}) = -\frac{4xy}{(1 + x^2 + y^2)^2}\]
\[\frac{\partial^2 f}{\partial y^2} = \frac{2(1 + x^2 - y^2)}{(1 + x^2 + y^2)^2}\]
Now, we need to calculate the Hessian matrix, which is given by:
\[H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\ \frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial y^2} \end{bmatrix}\]
At the critical point (0, 0), the Hessian matrix is:
\[H = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}\]
4Step 4: Apply the second derivative test
The second derivative test states that for a critical point (x, y), if the determinant of the Hessian matrix is positive, then the point is a local minimum if the second partial derivative with respect to x is positive, a local maximum if it is negative, and inconclusive if it is zero. If the determinant is negative, then the point is a saddle.
At (0, 0), the determinant of the Hessian matrix is:
\[|H| = (2)(2) - (0)(0) = 4\]
Since the determinant is positive and the second partial derivative with respect to x is positive, the critical point (0, 0) is a local minimum.
5Step 5: Identify the relative extrema
Based on the second derivative test, we found that the function \(f(x, y) = \ln(1 + x^2 + y^2)\) has a local minimum at the critical point (0, 0). Therefore, the relative extrema of the function is:
Local minimum: (0, 0) with f(0, 0) = \(\ln(1)\) = 0
Key Concepts
Critical PointsSecond Derivative TestHessian Matrix
Critical Points
In multivariable calculus, critical points are pivotal in understanding the behavior of functions with several variables. A critical point occurs where the function's gradient (vector of partial derivatives) is zero or undefined.
To find the critical points for a function like \(f(x, y)=\ln (1+x^{2}+y^{2})\), you first compute its first partial derivatives. For this function, the partial derivatives are:
\[\frac{\partial f}{\partial x} = \frac{2x}{1 + x^2 + y^2} \]
\[\frac{\partial f}{\partial y} = \frac{2y}{1 + x^2 + y^2} \]
Critical points occur where these derivatives simultaneously equal zero. Solving the equations \(\frac{2x}{1 + x^2 + y^2} = 0\) and \(\frac{2y}{1 + x^2 + y^2} = 0\), we find that \(x = 0\) and \(y = 0\) is the critical point of the function.
This indicates that the gradient is zero at \((0, 0)\), making it a candidate for a local extremum or saddle point.
To find the critical points for a function like \(f(x, y)=\ln (1+x^{2}+y^{2})\), you first compute its first partial derivatives. For this function, the partial derivatives are:
\[\frac{\partial f}{\partial x} = \frac{2x}{1 + x^2 + y^2} \]
\[\frac{\partial f}{\partial y} = \frac{2y}{1 + x^2 + y^2} \]
Critical points occur where these derivatives simultaneously equal zero. Solving the equations \(\frac{2x}{1 + x^2 + y^2} = 0\) and \(\frac{2y}{1 + x^2 + y^2} = 0\), we find that \(x = 0\) and \(y = 0\) is the critical point of the function.
This indicates that the gradient is zero at \((0, 0)\), making it a candidate for a local extremum or saddle point.
Second Derivative Test
The second derivative test is a technique used to classify critical points as local minima, maxima, or saddle points. After finding a critical point, you calculate the second derivatives to form the Hessian matrix.
The Hessian matrix \(H\) is a square matrix of all second-order partial derivatives. For a function \(f(x, y)\), it is represented as:
\[H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \ \frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial y^2} \end{bmatrix} \]
In this problem, the second derivatives for \(f(x, y)=\ln (1+x^{2}+y^{2})\) were computed as:
\[\frac{\partial^2 f}{\partial x^2} = \frac{2(1 - x^2 + y^2)}{(1 + x^2 + y^2)^2} \]
\[\frac{\partial^2 f}{\partial x \partial y} = -\frac{4xy}{(1 + x^2 + y^2)^2} \]
\[\frac{\partial^2 f}{\partial y^2} = \frac{2(1 + x^2 - y^2)}{(1 + x^2 + y^2)^2} \]
The Hessian matrix at the critical point \((0,0)\) is:
\[H = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix} \]
The determinants of \(H\) tells us the nature of the critical point:
The Hessian matrix \(H\) is a square matrix of all second-order partial derivatives. For a function \(f(x, y)\), it is represented as:
\[H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \ \frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial y^2} \end{bmatrix} \]
In this problem, the second derivatives for \(f(x, y)=\ln (1+x^{2}+y^{2})\) were computed as:
\[\frac{\partial^2 f}{\partial x^2} = \frac{2(1 - x^2 + y^2)}{(1 + x^2 + y^2)^2} \]
\[\frac{\partial^2 f}{\partial x \partial y} = -\frac{4xy}{(1 + x^2 + y^2)^2} \]
\[\frac{\partial^2 f}{\partial y^2} = \frac{2(1 + x^2 - y^2)}{(1 + x^2 + y^2)^2} \]
The Hessian matrix at the critical point \((0,0)\) is:
\[H = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix} \]
The determinants of \(H\) tells us the nature of the critical point:
- If the determinant is positive, the point is a local minimum or maximum.
- If it's negative, the point is a saddle point.
Hessian Matrix
The Hessian matrix is an essential tool in multivariable calculus to analyze the curvature of functions with respect to multiple variables. It provides insight into whether a function curves upwards or downwards at a given point, affecting the function's extremum classification.
For a function \(f(x, y)\), the Hessian matrix is constructed from its second partial derivatives:
\[H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \ \frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial y^2} \end{bmatrix} \]
It's crucial in deciding the type of critical points by evaluating its determinant and trace. In the example function \(f(x, y)=\ln (1+x^{2}+y^{2})\), the calculated Hessian matrix at the critical point \((0,0)\) is:
\[H = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix} \]
The determinant \(|H| = 4\) and both diagonal elements (trace) being positive indicate a local minimum. Understanding the structure and evaluation of the Hessian helps in determining the concavity of the function at any critical point, confirming whether the point is a potential extremum or saddle.
For a function \(f(x, y)\), the Hessian matrix is constructed from its second partial derivatives:
\[H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \ \frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial y^2} \end{bmatrix} \]
It's crucial in deciding the type of critical points by evaluating its determinant and trace. In the example function \(f(x, y)=\ln (1+x^{2}+y^{2})\), the calculated Hessian matrix at the critical point \((0,0)\) is:
\[H = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix} \]
The determinant \(|H| = 4\) and both diagonal elements (trace) being positive indicate a local minimum. Understanding the structure and evaluation of the Hessian helps in determining the concavity of the function at any critical point, confirming whether the point is a potential extremum or saddle.
Other exercises in this chapter
Problem 18
Find the first partial derivatives of the function. \(f(x, y)=x^{2} e^{y^{2}}\)
View solution Problem 18
Sketch the level curves of the function corresponding to each value of \(z\). \(h(u, v)=\sqrt{4-u^{2}-v^{2}}\)
View solution Problem 19
Sketch the level curves of the function corresponding to each value of \(z\). \(f(x, y)=2 x+3 y ; z=-2,-1,0,1,2\)
View solution Problem 20
Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relati
View solution