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Chapter 14: Problem 16

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. $$f(x, y)=e^{y}\left(y^{2}-x^{2}\right)$$

Short Answer

Expert verified
The function has a saddle point at (0, -2); the test is inconclusive at (0, 0).

Step by step solution

01

Find the First Partial Derivatives

First, we need to find the partial derivatives \(f_x\) and \(f_y\) of the function \(f(x, y) = e^y (y^2 - x^2)\). The partial derivative with respect to \(x\) is:\[ f_x = \frac{\partial}{\partial x}(e^y (y^2 - x^2)) = -2x e^y \]The partial derivative with respect to \(y\) is:\[ f_y = \frac{\partial}{\partial y}(e^y (y^2 - x^2)) = e^y (2y) + e^y (y^2 - x^2) = e^y (2y + y^2 - x^2) \]
02

Solve for Critical Points

To find the critical points, set both partial derivatives to zero.Set \( f_x = -2x e^y = 0 \):Since \(e^y eq 0\), we have \(x = 0\).Set \( f_y = e^y (2y + y^2 - x^2) = 0 \):\(e^y (2y + y^2 - x^2) = 0\) implies \(2y + y^2 - x^2 = 0\).Using \(x = 0\):\(2y + y^2 = 0\)Solving \(y(y+2) = 0\) gives \(y = 0\) or \(y = -2\).Thus, the critical points are \((0, 0)\) and \((0, -2)\).
03

Use Second Derivative Test

To classify these critical points, compute the second partial derivatives. Calculate:- \(f_{xx} = \frac{\partial}{\partial x}(-2x e^y) = -2e^y\)- \(f_{yy} = \frac{\partial}{\partial y}(e^y (2y + y^2 - x^2)) = e^y(2 + 2y) + e^y(2y + y^2 - x^2) = e^y(2 + 4y + y^2 - x^2)\)- \(f_{xy} = \frac{\partial}{\partial y}(-2x e^y) = -2xe^y\)The determinant of the Hessian matrix is given by:\[ H = f_{xx}f_{yy} - (f_{xy})^2 \]
04

Analyze the Critical Points

Evaluate at \((0, 0)\):\[f_{xx}(0, 0) = -2, \quad f_{yy}(0, 0) = 0, \quad f_{xy}(0, 0) = 0\]\[H(0, 0) = (-2)(0) - (0)^2 = 0\]When \(H = 0\), the test is inconclusive at \((0, 0)\).Evaluate at \((0, -2)\):\[f_{xx}(0, -2) = -2e^{-2}, \quad f_{yy}(0, -2) = e^{-2}\]\[H(0, -2) = (-2e^{-2})(e^{-2}) - (0)^2 = -2e^{-4} < 0\]Since \(H(0, -2) < 0\), \((0, -2)\) is a saddle point.

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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, a critical point of a function is a point where the gradient (or all first partial derivatives) is zero or undefined. For a function of two variables, such as \[ f(x, y) = e^{y}(y^2 - x^2) \], finding critical points involves setting both partial derivatives with respect to each variable equal to zero.
  • The partial derivative with respect to \(x\), \( f_x \), provides the slope of the function in the \(x\) direction.
  • The partial derivative with respect to \(y\), \( f_y \), represents the slope in the \(y\) direction.
To locate the critical points, solve the equations \( f_x = 0 \) and \( f_y = 0 \) simultaneously. For the function given, the calculations reveal critical points at \((0, 0)\) and \((0, -2)\). Understanding critical points is crucial because they include potential locations for local maxima, minima, or saddle points.
Partial Derivatives
Partial derivatives are crucial in multivariable calculus. They indicate how a function changes as one of the variables changes, while all other variables are held constant. In a function of two variables like \[ f(x, y) = e^y(y^2 - x^2) \], you will find:
  • \( f_x \), the partial derivative with respect to \(x\), computed by differentiating while considering \(y\) as constant.
  • \( f_y \), the partial derivative with respect to \(y\), found by differentiating while keeping \(x\) constant.
For example, \( f_x = -2x e^y \) describes the rate of change in the \(x\) direction. The partial derivative \( f_y \) is more complex, combining multiple terms as shown in \[ f_y = e^y(2y + y^2 - x^2) \]. Calculating these derivatives is the first critical step to analyze functions in the context of local extrema and saddle points.
Hessian Matrix
The Hessian matrix is a square matrix of second-order partial derivatives of a scalar-valued function. It's essential for determining the nature of critical points. For a function \( f(x, y) \), the Hessian matrix \( H \) is structured as:\[H = \begin{bmatrix}f_{xx} & f_{xy} \f_{yx} & f_{yy}\end{bmatrix}\]
  • \( f_{xx} \) and \( f_{yy} \) are the second partial derivatives of \( f \) with respect to \( x \) and \( y \) respectively.
  • \( f_{xy} = f_{yx} \) is the mixed partial derivative.
Evaluating this matrix at critical points helps us apply the second derivative test, determining whether these points are minima, maxima, or saddle points.
Second Derivative Test
The second derivative test in multivariable calculus is a method used to classify critical points of a function. It involves the Hessian matrix and helps decide if a critical point is a local minimum, local maximum, or a saddle point.For each critical point, compute the determinant of the Hessian matrix \( H \). This is given by \( H = f_{xx}f_{yy} - (f_{xy})^2 \).
  • If \( H > 0 \) and \( f_{xx} > 0 \), the point is a local minimum.
  • If \( H > 0 \) and \( f_{xx} < 0 \), the point is a local maximum.
  • If \( H < 0 \), the point is a saddle point.
  • If \( H = 0 \), the test is inconclusive.
For our function, evaluating at \((0, 0)\) left us inconclusive due to \( H = 0 \). However, at \((0, -2)\), we found \( H < 0 \), indicating a saddle point. This test is vital for fully understanding the behavior of functions at their critical points.

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Most popular questions from this chapter

Find the absolute maximum and minimum values of \(f\) on the set \(D .\) $$\begin{array}{l}{f(x, y)=3+x y-x-2 y, \quad D \text { is the closed triangular }} \\ {\text { region with vertices }(1,0),(5,0), \text { and }(1,4)}\end{array}$$

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. $$f(x, y)=\left(x^{2}+y^{2}\right) e^{y^{2}-x^{2}}$$

Verify that the conclusion of Clairaut's Theorem holds, that is, \(u_{x y}=u_{y x}\) $$u=x^{4} y^{2}-2 x y^{5}$$

Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. $$f(x, y)=x y+\frac{1}{x}+\frac{1}{y}$$

Use a graphing device as in Example 4 (or Newton's method or a rootfinder) to find the critical points of \(f\) correct to three decimal places. Then classify the critical points and find the highest or lowest points on the graph. $$f(x, y)=2 x+4 x^{2}-y^{2}+2 x y^{2}-x^{4}-y^{4}$$
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