f(x, y) = x³ + y² − 6x − 2y² + 2 - -
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![### Compute the Discriminant \( D(x, y) \) of the Function
Given the function:
\[ f(x, y) = x^3 + y^4 - 6x - 2y^2 + 2 \]
In order to compute the discriminant \( D(x, y) \), follow these steps outlined below:
1. **First, calculate the partial derivatives** of the function \( f(x, y) \).
- The first partial derivative with respect to \( x \) is:
\[
f_x = \frac{\partial}{\partial x}(x^3 + y^4 - 6x - 2y^2 + 2) = 3x^2 - 6
\]
- The first partial derivative with respect to \( y \) is:
\[
f_y = \frac{\partial}{\partial y}(x^3 + y^4 - 6x - 2y^2 + 2) = 4y^3 - 4y
\]
2. **Next, calculate the second partial derivatives** of \( f(x, y) \).
- The second partial derivative with respect to \( x \) is:
\[
f_{xx} = \frac{\partial}{\partial x}(3x^2 - 6) = 6x
\]
- The second partial derivative with respect to \( y \) is:
\[
f_{yy} = \frac{\partial}{\partial y}(4y^3 - 4y) = 12y^2 - 4
\]
- The mixed partial derivative \( f_{xy} \) (or \( f_{yx} \)) is:
\[
f_{xy} = \frac{\partial}{\partial y}(3x^2 - 6) = 0
\]
3. **Determine the discriminant \( D(x, y) \)** given by:
\[
D(x, y) = f_{xx} f_{yy} - (f_{xy})^2
\]
Substituting the values of the second partial derivatives, we get:
\[
D(x, y) = (6x)(12y^2 - 4) - (](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe132a23b-2bfe-4f64-9635-6f1845f8e4fa%2F5a2791fa-9893-4bd1-a70e-5fdeb715b08f%2Fquxk2wb_processed.png&w=3840&q=75)
Transcribed Image Text:### Compute the Discriminant \( D(x, y) \) of the Function
Given the function:
\[ f(x, y) = x^3 + y^4 - 6x - 2y^2 + 2 \]
In order to compute the discriminant \( D(x, y) \), follow these steps outlined below:
1. **First, calculate the partial derivatives** of the function \( f(x, y) \).
- The first partial derivative with respect to \( x \) is:
\[
f_x = \frac{\partial}{\partial x}(x^3 + y^4 - 6x - 2y^2 + 2) = 3x^2 - 6
\]
- The first partial derivative with respect to \( y \) is:
\[
f_y = \frac{\partial}{\partial y}(x^3 + y^4 - 6x - 2y^2 + 2) = 4y^3 - 4y
\]
2. **Next, calculate the second partial derivatives** of \( f(x, y) \).
- The second partial derivative with respect to \( x \) is:
\[
f_{xx} = \frac{\partial}{\partial x}(3x^2 - 6) = 6x
\]
- The second partial derivative with respect to \( y \) is:
\[
f_{yy} = \frac{\partial}{\partial y}(4y^3 - 4y) = 12y^2 - 4
\]
- The mixed partial derivative \( f_{xy} \) (or \( f_{yx} \)) is:
\[
f_{xy} = \frac{\partial}{\partial y}(3x^2 - 6) = 0
\]
3. **Determine the discriminant \( D(x, y) \)** given by:
\[
D(x, y) = f_{xx} f_{yy} - (f_{xy})^2
\]
Substituting the values of the second partial derivatives, we get:
\[
D(x, y) = (6x)(12y^2 - 4) - (
![### Which of these points are saddle points?
- [ ] \((\sqrt{2}, -1)\)
- [ ] \((\sqrt{2}, 0)\)
- [ ] \((-\sqrt{2}, 0)\)
- [ ] \((\sqrt{2}, 1)\)
- [ ] \((-\sqrt{2}, -1)\)
- [ ] \((-\sqrt{2}, 1)\)
**Explanation:**
This question is meant to determine which of the given points are saddle points. In a multivariable function, a saddle point occurs at a critical point where the function changes direction. That is, it is not a local minimum or maximum but rather a point where the surface curves upward in one direction and downward in another.
By evaluating the second partial derivatives and calculating the determinant of the Hessian matrix at each given point, one can determine the nature of each critical point. A point is a saddle point if the determinant of the Hessian matrix is negative at that point.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe132a23b-2bfe-4f64-9635-6f1845f8e4fa%2F5a2791fa-9893-4bd1-a70e-5fdeb715b08f%2Fa2i7xau_processed.png&w=3840&q=75)
Transcribed Image Text:### Which of these points are saddle points?
- [ ] \((\sqrt{2}, -1)\)
- [ ] \((\sqrt{2}, 0)\)
- [ ] \((-\sqrt{2}, 0)\)
- [ ] \((\sqrt{2}, 1)\)
- [ ] \((-\sqrt{2}, -1)\)
- [ ] \((-\sqrt{2}, 1)\)
**Explanation:**
This question is meant to determine which of the given points are saddle points. In a multivariable function, a saddle point occurs at a critical point where the function changes direction. That is, it is not a local minimum or maximum but rather a point where the surface curves upward in one direction and downward in another.
By evaluating the second partial derivatives and calculating the determinant of the Hessian matrix at each given point, one can determine the nature of each critical point. A point is a saddle point if the determinant of the Hessian matrix is negative at that point.
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