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Chapter 11: Problem 24

Examine the function for relative extrema and saddle points. $$ g(x, y)=x y $$

Short Answer

Expert verified
(0,0) is a saddle point of the function \(g(x, y)=x y\)

Step by step solution

01

Compute the First Partial Derivatives

For any function of two variables, start by finding its first-order partial derivatives with respect to x and y:The first partial derivative of \(g\) with respect to \(x\) is \(g_x(x, y) = y\). The first partial derivative of \(g\) with respect to \(y\) is \(g_y(x, y) = x.\)
02

Set Both First Partial Derivatives to Zero

Setting \(g_x = y = 0\) and \(g_y = x = 0\), any point that makes both of these true is a critical point. In this case, it's evident that \((0,0)\) is the only critical point.
03

Compute the Second Partial Derivatives

The second partial derivatives are:\(g_{xx}(x, y) = 0\), \(g_{yy}(x, y) = 0\), and \(g_{xy}=g_{yx} = 1\)
04

Apply the Second Partials Test

The second partials test uses the discriminant \(D = g_{xx}g_{yy} - g_{xy}^2\). If \(D > 0\) and \(g_{xx}\) (or \(g_{yy}\)) is positive at the critical point, the function has a local minimum. If \(D > 0\) and \(g_{xx}\) (or \(g_{yy}\)) is negative, the function has a local maximum. If \(D < 0\), the function has a saddle point. For the test to be inconclusive, \(D = 0\).For the critical point (0,0), \(D = g_{xx}(0,0)g_{yy}(0,0) - g_{xy}(0,0)^2 = 0(0) - 1^2 = -1<0\)
05

Determine the nature of the Critical Point

Because the discriminant \(D\) is less than zero at the critical point \((0,0)\), this means that \((0,0)\) is a saddle point of the function \(g(x, y)=x y\).

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Key Concepts

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

Partial Derivatives
Partial derivatives are a fundamental tool in multivariable calculus. They offer insights into the behavior of functions with multiple variables. In a function with two variables, such as our function \[g(x, y) = xy\], we calculate the partial derivative with respect to one variable while treating the other as a constant.
  • The partial derivative with respect to \(x\) is denoted as \(g_x(x, y)\). For \(g(x, y) = xy\), this results in \(g_x = y\), implying that the derivative with respect to \(x\) is equal to the constant \(y\).
  • Similarly, the partial derivative with respect to \(y\) is \(g_y(x, y)\), which results in \(g_y = x\), meaning the derivative with respect to \(y\) keeps \(x\) constant and is equal to \(x\).
These derivatives are essential in determining how the function changes as each variable varies individually. They play a critical role in locating critical points, which are potential locations for maxima, minima, or saddle points.
Critical Points
Critical points are the values of the variables where a function does not change, indicating potential maxima, minima, or saddle points. To find these in a function of two variables, we set both first partial derivatives to zero.For our function,
  • Set \(g_x = y = 0\) and \(g_y = x = 0\).
  • This provides the critical point at \((0, 0)\).
Critical points are crucial for analyzing the behavior of the function's graph. They indicate where the surface of the function levels out. Knowing these points helps to further explore the function's potential for local maxima, minima, or saddle points using the second partial derivatives.
Second Partials Test
The Second Partials Test, also known as the Hessian determinant test, is a method used to classify critical points of a function of two variables. After identifying the critical points, we compute second partial derivatives to form a determinant, which helps in determining the nature of these critical points.For the function \(g(x, y) = xy\):
  • Calculate the second derivatives: \(g_{xx} = 0\), \(g_{yy} = 0\), and \(g_{xy} = g_{yx} = 1\).
  • The discriminant \(D\) is calculated using \(D = g_{xx}g_{yy} - (g_{xy})^2\).
For our specific critical point \((0, 0)\), we get:\[D = (0)(0) - 1^2 = -1\]The test uses the value of \(D\) to determine:
  • \(D > 0\) and \(g_{xx} > 0\): local minimum.
  • \(D > 0\) and \(g_{xx} < 0\): local maximum.
  • \(D < 0\): saddle point.
  • \(D = 0\): test is inconclusive.
In this function, since \(D < 0\), it confirms the presence of a saddle point at (0,0).
Saddle Points
Saddle points are critical points where the function doesn't produce a local max or min. Instead, the surface symmetrically dips below and rises above the level of the saddle point, much like a horse saddle. They are characterized by the Second Partials Test when \(D < 0\). For our function \(g(x, y) = xy\) at the critical point \((0,0)\),
  • The second partials test returned \(D = -1\), confirming a saddle point.
  • This means around (0, 0), the surface neither strictly dips nor rises.
Saddle points are of particular interest in multivariable calculus because they represent intersections of different behaviors (incline versus decline) of a function's graph. Understanding saddle points helps in comprehensive analysis, showing transitions in the terrain of the function without extreme peaks or valleys.

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Most popular questions from this chapter

Find \(\partial w / \partial s\) and \(\partial w / \partial t\) by using the appropriate Chain Rule. \(w=x^{2}+y^{2}+z^{2}, \quad x=t \sin s, \quad y=t \cos s, \quad z=s t^{2}\)

Find \(d w / d t\) (a) using the appropriate Chain Rule and (b) by converting \(w\) to a function of \(t\) before differentiating. \(w=x^{2}+y^{2}+z^{2}, \quad x=e^{t} \cos t, \quad y=e^{t} \sin t, \quad z=e^{t}\)

Consider the function \(w=f(x, y),\) where \(x=r \cos \theta\) and \(y=r \sin \theta .\) Prove each of the following. (a) \(\frac{\partial w}{\partial x}=\frac{\partial w}{\partial r} \cos \theta-\frac{\partial w}{\partial \theta} \frac{\sin \theta}{r}\) \(\frac{\partial w}{\partial y}=\frac{\partial w}{\partial r} \sin \theta+\frac{\partial w}{\partial \theta} \frac{\cos \theta}{r}\) (b) \(\left(\frac{\partial w}{\partial x}\right)^{2}+\left(\frac{\partial w}{\partial y}\right)^{2}=\left(\frac{\partial w}{\partial r}\right)^{2}+\left(\frac{1}{r^{2}}\right)\left(\frac{\partial w}{\partial \theta}\right)^{2}\)

Find \(d w / d t\) (a) using the appropriate Chain Rule and (b) by converting \(w\) to a function of \(t\) before differentiating. \(w=x y z, \quad x=t^{2}, \quad y=2 t, \quad z=e^{-t}\)

Ideal Gas Law The Ideal Gas Law is \(p V=m R T,\) where \(R\) is a constant, \(m\) is a constant mass, and \(p\) and \(V\) are functions of time. Find \(d T / d t,\) the rate at which the temperature changes with respect to time.
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