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Chapter 14: Problem 46

For what values of the constant \(k\) does the Second Derivative Test guarantee that \(f(x, y)=x^{2}+k x y+y^{2}\) will have a saddle point at (0,0)? A local minimum at (0, 0)? For what values of \(k\) is the Second Derivative Test inconclusive? Give reasons for your answers.

Short Answer

Expert verified
Saddle point: \(|k| > 2\); local minimum: \(|k| < 2\); inconclusive: \(|k| = 2\).

Step by step solution

01

Define Partial Derivatives

First, calculate the first partial derivatives of the function \( f(x, y) = x^2 + kxy + y^2 \). These are: \( f_x = 2x + ky \) and \( f_y = kx + 2y \). The critical point occurs when \( f_x = 0 \) and \( f_y = 0 \). Testing for \( x = 0 \) and \( y = 0 \) at the origin gives us a critical point.
02

Calculate Second Partial Derivatives

Next, compute the second partial derivatives. These are \( f_{xx} = 2 \), \( f_{yy} = 2 \), and \( f_{xy} = f_{yx} = k \).
03

Apply the Second Derivative Test

The Second Derivative Test for functions of two variables uses the determinant of the Hessian matrix \( D = f_{xx}f_{yy} - (f_{xy})^2 \). Substituting the second partial derivatives, we have \( D = 2\cdot2 - k^2 = 4 - k^2 \).
04

Determine Values for Saddle Point

A saddle point occurs when \( D < 0 \). Thus, the function has a saddle point at \((0,0)\) when \(4 - k^2 < 0\), which simplifies to \( |k| > 2 \).
05

Determine Values for a Local Minimum

For a local minimum at \((0, 0)\), \( D > 0 \) and \( f_{xx} > 0 \). Since \( f_{xx} = 2 > 0 \), a local minimum occurs when \( 4 - k^2 > 0 \), leading to \( |k| < 2 \).
06

Determine Inconclusive Values

The Second Derivative Test is inconclusive when \( D = 0 \). Therefore, \( 4 - k^2 = 0 \) or \( |k| = 2 \) are the values of \( k \) where the test is inconclusive.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
Understanding partial derivatives is crucial when studying functions with two or more variables. For the function \( f(x, y) = x^2 + kxy + y^2 \), partial derivatives allow us to explore how the function changes in different directions: specifically along the axes (each variable held fixed).
  • The partial derivative with respect to \(x\), denoted \( f_x \), treats \(y\) as a constant. For our function, this results in: \( f_x = 2x + ky \).
  • The partial derivative with respect to \(y\), denoted \( f_y \), treats \(x\) as constant. Calculating this gives: \( f_y = kx + 2y \).
Finding the critical point involving equating these first derivatives to zero. Here, we solve \( f_x = 0 \) and \( f_y = 0 \). When evaluated at the origin (x=0, y=0), these equations confirm that (0, 0) is a critical point. Partial derivatives help us understand the behavior and characteristics of functions of multiple variables at these points.
Saddle Point
A saddle point in a function is where the point is stable in one direction and unstable in another, resembling a saddle surface. In the context of our exercise with \( f(x, y) = x^2 + kxy + y^2 \), whether a saddle point occurs at a critical point can be determined with the Second Derivative Test and understanding when its result is negative.
  • The Hessian matrix determinant, \( D = f_{xx}f_{yy} - (f_{xy})^2 \), helps us decide this. For \( (0,0) \) to be a saddle point, \( D < 0 \).
  • Given our problem, \( D = 4 - k^2 \), and so a saddle point occurs if \( 4 - k^2 < 0 \).
  • This simplifies to \(|k| > 2\), meaning that the parameter \(k\) must have an absolute value greater than 2 to manifest a saddle point at (0, 0).
Saddle points are essential in multivariable calculus as they represent interesting points of inflection and can indicate areas of transition on a surface.
Critical Point
A critical point is a location on a function where the gradient (or first derivatives) of the function vanish. It can be a point of local maxima, minima, or saddle points. In the second derivative test, we identify these points by analyzing the first partial derivatives of the function.
  • For \( f(x, y) = x^2 + kxy + y^2 \), the critical point (0,0) emerges when \( f_x = 2x + ky = 0 \) and \( f_y = kx + 2y = 0 \). Solving these at the origin yields our critical point.
Critical points signify points of interest where we must investigate further, using second derivatives, to understand the nature or type of point it indeed is. When paired with the second derivative information and tests, these points reveal vital data about the behavior of functions and their surfaces in calculus.
Hessian Matrix
The Hessian matrix is a vital tool in multivariable calculus used to categorize critical points. For our function, it is a 2x2 matrix formed by the second partial derivatives:
  • Main diagonal elements are the pure second derivatives (\( f_{xx} \) and \( f_{yy} \)).
  • Off-diagonal elements are the mixed derivatives (\( f_{xy} \) and \( f_{yx} \) which are equal).
The Hessian for our function becomes:\[H = \begin{bmatrix} 2 & k \k & 2 \end{bmatrix}\]
The Determinant of the Hessian matrix, \( D = f_{xx}f_{yy} - (f_{xy})^2 \), helps assess the nature of critical points:
  • If \( D > 0 \) and \( f_{xx} > 0 \), a local minimum is confirmed.
  • If \( D < 0 \), the point is a saddle point.
  • When \( D = 0 \), the test does not provide conclusive information about the nature of the critical point.
In our exercise, understanding the Hessian and its determinant guides us to deciding how the point (0,0) behaves in terms of being a local minimum, saddle point, or inconclusive.

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Most popular questions from this chapter

When we try to fit a line \(y=m x+b\) to a set of numerical data points \(\left(x_{1}, y_{1}\right)\) \(\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right),\) we usually choose the line that minimizes the sum of the squares of the vertical distances from the points to the line. In theory, this means finding the values of \(m\) and \(b\) that minimize the value of the function $$w=\left(m x_{1}+b-y_{1}\right)^{2}+\cdots+\left(m x_{n}+b-y_{n}\right)^{2}. \quad (1)$$ (See the accompanying figure.) Show that the values of \(m\) and \(b\) that do this are $$m=\frac{\left(\sum x_{k}\right)\left(\sum y_{k}\right)-n \sum x_{k} y_{k}}{\left(\sum x_{k}\right)^{2}-n \sum x_{k}^{2}},\quad (2)$$ $$b=\frac{1}{n}\left(\sum y_{k}-m \sum x_{k}\right),\quad(3)$$ with all sums running from \(k=1\) to \(k=n\). Many scientific calculators have these formulas built in, enabling you to find \(m\) and \(b\) with only a few keystrokes after you have entered the data. The line \(y=m x+b\) determined by these values of \(m\) and \(b\) is called the least squares line, regression line, or trend line for the data under study. Finding a least squares line lets you 1\. summarize data with a simple expression, 2\. predict values of \(y\) for other, experimentally untried values of \(x\), 3\. handle data analytically. We demonstrated these ideas with a variety of applications in Section 1.4.

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