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Magazine Chapter 15: Problem 64
Are the statements true or false? Give reasons for your answer. The function \(f(x, y)=x^{2}-y^{2}\) has a local minimum at the origin.
Short Answer
Expert verified
False, the function has a saddle point at the origin.
Step by step solution
01
Identify the Nature of the Function
The function in question is given by \(f(x, y) = x^2 - y^2\). This is a well-known example of a quadratic function in two variables.
02
Find Critical Points
To find critical points, we need to take the partial derivatives of \(f(x, y)\) and set them to zero. Compute \(f_x(x, y) = \frac{\partial}{\partial x}(x^2 - y^2) = 2x\) and \(f_y(x, y) = \frac{\partial}{\partial y}(x^2 - y^2) = -2y\). Set \(f_x = 0\) and \(f_y = 0\) which gives solutions \(x = 0\) and \(y = 0\), so the critical point is at the origin \((0,0)\).
03
Use the Second Derivative Test
The second derivative test for functions of two variables uses the Hessian matrix, which is\[H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2}\end{bmatrix} = \begin{bmatrix}2 & 0 \0 & -2 \end{bmatrix}.\]The determinant of \(H\), \(\text{det}(H) = (2)(-2) - (0)(0) = -4\), is less than zero.
04
Analyze Determinant of the Hessian
Since the determinant of the Hessian matrix \(\text{det}(H)\) is negative at \((0,0)\), according to the second derivative test, this implies that the critical point is a saddle point. Saddle points indicate neither a local minimum nor a local maximum.
05
Conclude the Nature at the Origin
Given that \((0, 0)\) is a saddle point, it is not a local minimum. This means the initial statement about the function having a local minimum at the origin is false.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Critical Points
In multivariable calculus, critical points are the points on a function where the derivative is zero or not defined. For functions of two variables like the one in our problem,
For the function \(f(x, y) = x^2 - y^2\), we find the partial derivatives: \(f_x = 2x\) and \(f_y = -2y\). Setting \(2x = 0\) and \(-2y = 0\) leads us to the critical point at the origin, \((0,0)\). This point is fundamental for determining the nature of the function at that location.
- Calculate the partial derivatives
- Set these derivatives equal to zero simultaneously
For the function \(f(x, y) = x^2 - y^2\), we find the partial derivatives: \(f_x = 2x\) and \(f_y = -2y\). Setting \(2x = 0\) and \(-2y = 0\) leads us to the critical point at the origin, \((0,0)\). This point is fundamental for determining the nature of the function at that location.
Hessian Matrix
The Hessian matrix is a square matrix of second-order partial derivatives of a scalar-valued function. For a function \(f(x, y)\), the Hessian is given by:\[\begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} \end{bmatrix}\]
For our function \(f(x, y) = x^2 - y^2\), the Hessian matrix turns out to be:\[\begin{bmatrix} 2 & 0 \0 & -2 \end{bmatrix}\]
Here, each element of the matrix represents a second derivative, crucial for analyzing the curvature properties that determine the type of critical point.
For our function \(f(x, y) = x^2 - y^2\), the Hessian matrix turns out to be:\[\begin{bmatrix} 2 & 0 \0 & -2 \end{bmatrix}\]
Here, each element of the matrix represents a second derivative, crucial for analyzing the curvature properties that determine the type of critical point.
Quadratic Functions
Quadratic functions in two variables, such as \(f(x, y) = x^2 - y^2\), are fundamental in finding critical points and understanding their nature.
These contain terms up to the second degree, often forming parabolas or hyperbolas. The function structure in our example is a hyperbolic paraboloid, typically creating a saddle point rather than a minimum or maximum.
This is confirmed through analyzing the Hessian, as quadratic functions make it simple to compute derivatives and identify the function's shape and curvature.
These contain terms up to the second degree, often forming parabolas or hyperbolas. The function structure in our example is a hyperbolic paraboloid, typically creating a saddle point rather than a minimum or maximum.
This is confirmed through analyzing the Hessian, as quadratic functions make it simple to compute derivatives and identify the function's shape and curvature.
Saddle Point
A saddle point is a type of critical point that resembles a saddle on a horse, having a higher value in one direction and a lower value in another.
- It indicates neither a local minimum nor a local maximum.
- Often identified through a negative determinant in the Hessian matrix.
Second Derivative Test
The second derivative test helps determine the nature of a critical point for functions of multiple variables by using the Hessian matrix.
In our problem, the determinant \(\text{det}(H) = -4\) is negative, establishing that \((0,0)\) is a saddle point and confirming that there is neither a local minimum nor maximum there.
- If the determinant of the Hessian is positive and the first entry is also positive, it's a local minimum.
- If the determinant is positive and the first entry negative, it's a local maximum.
- If the determinant is negative, the point is a saddle point.
In our problem, the determinant \(\text{det}(H) = -4\) is negative, establishing that \((0,0)\) is a saddle point and confirming that there is neither a local minimum nor maximum there.
