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Magazine Chapter 11: Problem 8
Find the stationary points of the function $$ z=\left(x^{2}+y^{2}\right)^{2}-8\left(x^{2}-y^{2}\right) $$ and determine their nature.
Short Answer
Expert verified
Stationary points are: \((0,0)\) (saddle point), \((2,0)\) and \((-2,0)\) (local minima).
Step by step solution
01
Compute the Partial Derivatives
Calculate the partial derivatives of the function with respect to both x and y. Let the function be \(z = f(x, y) = (x^2 + y^2)^2 - 8(x^2 - y^2)\). Find \(f_x = \frac{\partial z}{\partial x}\) and \(f_y = \frac{\partial z}{\partial y}\). 1. \(f_x = \frac{\partial}{\partial x} [(x^2 + y^2)^2 - 8(x^2 - y^2)] = 4x(x^2 + y^2) - 16x\)2. \(f_y = \frac{\partial}{\partial y} [(x^2 + y^2)^2 - 8(x^2 - y^2)] = 4y(x^2 + y^2) + 16y\)
02
Set the Partial Derivatives to Zero
To find stationary points, set the partial derivatives equal to zero and solve for x and y.1. \(f_x = 4x(x^2 + y^2) - 16x = 0\)2. \(f_y = 4y(x^2 + y^2) + 16y = 0\)For \(f_x = 0\): \(4x(x^2 + y^2 - 4) = 0\) leading to \(x = 0\) or \(x^2 + y^2 = 4\). For \(f_y = 0\): \(4y(x^2 + y^2 + 4) = 0\) leading to \(y = 0\). Consider the solutions: x = 0 and y = 0 or \(x^2 + y^2 = 4\) and \(y = 0\).
03
Solve System of Equations
Combine the results from Step 2. Solve the system for points \((x, y)\). Possible solutions are:1. \(x = 0\) and \(y = 0\)2. \(x^2 + y^2 = 4\) with \(y = 0\), giving \(x^2 = 4\) which results in \(x = 2\) or \(x = -2\). The stationary points are \((2,0)\) and \((-2,0)\). Therefore, the stationary points are \((0,0)\), \((2,0)\), and \((-2,0)\).
04
Determine the Nature of Stationary Points
To determine whether each stationary point is a maximum, minimum or saddle point, compute the second order partial derivatives and use the Hessian determinant.1. \(f_{xx} = \frac{\partial^2 z}{\partial x^2} = 12x^2 + 4y^2 - 16\)2. \(f_{yy} = \frac{\partial^2 z}{\partial y^2} = 12y^2 + 4x^2 + 16\)3. \(f_{xy} = \frac{\partial^2 z}{\partial x \partial y} = 8xy\)Compute the Hessian determinant: \(H = f_{xx}f_{yy} - (f_{xy})^2\)For \((0,0)\):1. \(f_{xx} = -16\), \(f_{yy} = 16\), \(f_{xy} = 0\)2. \(H = (-16)(16) - (0)^2 = -256\), which is negative, hence \((0,0)\) is a saddle point.For \((2,0)\) and \((-2,0)\):1. \(f_{xx} = 32\), \(f_{yy} = 48\), \(f_{xy} = 0\)2. \(H = 32 \times 48 - 0 = 1536\), which is positive, hence \((2,0)\) and \((-2,0)\) are local minima.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivatives
In multivariable calculus, partial derivatives are used to measure how a function changes as each individual variable is varied, keeping the other variables constant. To find the stationary points of a function like \(z = (x^2 + y^2)^2 - 8(x^2 - y^2)\), we first compute its partial derivatives with respect to both x and y.
The partial derivative with respect to x, denoted as \(f_x\), shows the rate of change of z in the x-direction. Similarly, the partial derivative with respect to y, denoted as \(f_y\), shows the rate of change of z in the y-direction.
For the given function:
These derivatives are used to find critical points by setting them equal to zero.
The partial derivative with respect to x, denoted as \(f_x\), shows the rate of change of z in the x-direction. Similarly, the partial derivative with respect to y, denoted as \(f_y\), shows the rate of change of z in the y-direction.
For the given function:
- \(f_x = \frac{\partial z}{\partial x} = 4x(x^2 + y^2) - 16x\)
- \(f_y = \frac{\partial z}{\partial y} = 4y(x^2 + y^2) + 16y\)
These derivatives are used to find critical points by setting them equal to zero.
Critical Points
Critical points, also known as stationary points, occur where the partial derivatives of a function are zero. Solving \(f_x = 0\) and \(f_y = 0\) simultaneously allows us to determine the critical points.
From the given function's partial derivatives:
We then solve these equations to determine that our critical points are \((0,0)\), \((2,0)\), and \((-2,0)\). These points are referred to as critical points because at these locations, the function has zero slope in all directions, indicating a possible maximum, minimum, or saddle point.
From the given function's partial derivatives:
- \(4x(x^2 + y^2 - 4) = 0\) leads to \(x = 0\) or \(x^2 + y^2 = 4\)
- \(4y(x^2 + y^2 + 4) = 0\) leads to \(y = 0\)
We then solve these equations to determine that our critical points are \((0,0)\), \((2,0)\), and \((-2,0)\). These points are referred to as critical points because at these locations, the function has zero slope in all directions, indicating a possible maximum, minimum, or saddle point.
Hessian Determinant
To classify the nature of critical points, we use the Hessian determinant. The Hessian matrix is a square matrix of second-order partial derivatives, and its determinant provides valuable information about the function鈥檚 curvature at critical points.
For the given function, we calculate:
The Hessian determinant, H, is given by: \[ H = f_{xx}f_{yy} - (f_{xy})^2 \]
This determinant, evaluated at each critical point, helps determine if the point is a local minimum, local maximum, or a saddle point.
For the given function, we calculate:
- \( f_{xx} = \frac{\partial^2 z}{\partial x^2} = 12x^2 + 4y^2 - 16\)
- \( f_{yy} = \frac{\partial^2 z}{\partial y^2} = 12y^2 + 4x^2 + 16\)
- \( f_{xy} = \frac{\partial^2 z}{\partial x \partial y} = 8xy\)
The Hessian determinant, H, is given by: \[ H = f_{xx}f_{yy} - (f_{xy})^2 \]
This determinant, evaluated at each critical point, helps determine if the point is a local minimum, local maximum, or a saddle point.
Local Minima
A local minimum occurs at a critical point if the Hessian determinant is positive, and the second partial derivatives indicate a positive curvature in all directions. Simply put, the function forms a valley at this point.
From our earlier calculations, points \((2,0)\) and \((-2,0)\) both have a Hessian determinant of 1536, which is positive. The second derivatives at these points indicate positive curvature:
Thus, points \((2,0)\) and \((-2,0)\) are local minima for the function.
From our earlier calculations, points \((2,0)\) and \((-2,0)\) both have a Hessian determinant of 1536, which is positive. The second derivatives at these points indicate positive curvature:
- \(f_{xx} = 32\)
- \(f_{yy} = 48\)
- \(f_{xy} = 0\)
Thus, points \((2,0)\) and \((-2,0)\) are local minima for the function.
Saddle Points
A saddle point occurs at a critical point where the Hessian determinant is negative, indicating that the function behaves like a saddle 鈥� it has both upward and downward curvatures.
At the critical point \((0,0)\), the Hessian determinant is calculated as \(-256\)
Since the Hessian is negative, \((0,0)\) is a saddle point. Here, the function decreases in some directions and increases in others, resembling a horse saddle. This mixed curvature differentiates a saddle point from a maximum or minimum.
At the critical point \((0,0)\), the Hessian determinant is calculated as \(-256\)
- \(f_{xx} = -16\)
- \(f_{yy} = 16\)
- \(f_{xy} = 0\)
Since the Hessian is negative, \((0,0)\) is a saddle point. Here, the function decreases in some directions and increases in others, resembling a horse saddle. This mixed curvature differentiates a saddle point from a maximum or minimum.
