Find the D value of critical point and the type of critical point

In summary: This is not just a quirk of one variable calculus, but a real phenomenon that can occur in higher dimensions as well. In summary, the function f(x,y) = [e^(-y^2)]cos(4x) has a critical point at (0,0) and the task is to find the D value at this point and determine its type (max, min, saddle, or none). To find the D value, the partial derivatives (fx, fy, fxx, fyy, fxy, and fyx) must be computed. The correct values of these derivatives at (0,0) are fx = 0, fy = 1, fxx = -4, fyy = -4
  • #1
coolusername
36
0

Homework Statement



The function f(x,y) = [e^(-y^2)]cos(4x) has a critical point (0,0)

Homework Equations



Find the D value at the critical point. What type of critical point is it? (max, min, saddle or none)

The Attempt at a Solution



I know that to find the D value I must compute the partial derivatives (fox, fey, fxy and fyx)

fx = [-4e^(-y^2)]sin(4x)

fy = [-2ye^(-y^2)]cos(4x)

fxx = [-16e^(-y^2)](cos(4x)

fyy = [4(y^2)e^(-y^2)]cos(4x)

fxy = [8ye^(-y^2)]sin(4x)

fxy = [8ye^(-y^2)]sin(4x)

D = (fxx)(fyy) - (fxy)^2
= [-16e^(-y^2)](cos(4x)[4(y^2)e^(-y^2)]cos(4x) - {[8ye^(-y^2)]sin(4x)}^2

evaluated at the critical point (0,0)

= (-16)(0) - (0)(0)
= 0

This means that the 2nd derivative test gave no info. But somehow it's not the right answer. Did I find the correct value of D? Is my approach correct?

Thanks
 
Physics news on Phys.org
  • #2
Having the discriminant be zero doesn't mean there is no max or min. In your case, ##f(0,0)=1##. Can you tell from the original equation whether that might be a max or min?
 
  • Like
Likes 1 person
  • #3
I see from the original equation that the max of f(x,y) is 1. However what would the value of D be? Did I calculate it correctly? I keep getting 0 but since the critical point has a max behaviour then shouldn't D be more than 0?
 
  • #4
coolusername said:

Homework Statement



The function f(x,y) = [e^(-y^2)]cos(4x) has a critical point (0,0)

Homework Equations



Find the D value at the critical point. What type of critical point is it? (max, min, saddle or none)

The Attempt at a Solution



I know that to find the D value I must compute the partial derivatives (fox, fey, fxy and fyx)

fx = [-4e^(-y^2)]sin(4x)

fy = [-2ye^(-y^2)]cos(4x)

fxx = [-16e^(-y^2)](cos(4x)

fyy = [4(y^2)e^(-y^2)]cos(4x)

fxy = [8ye^(-y^2)]sin(4x)

fxy = [8ye^(-y^2)]sin(4x)

D = (fxx)(fyy) - (fxy)^2
= [-16e^(-y^2)](cos(4x)[4(y^2)e^(-y^2)]cos(4x) - {[8ye^(-y^2)]sin(4x)}^2

evaluated at the critical point (0,0)

= (-16)(0) - (0)(0)
= 0

This means that the 2nd derivative test gave no info. But somehow it's not the right answer. Did I find the correct value of D? Is my approach correct?

Thanks

Your second derivative ##f_{yy}## is incorrect; it should not be 0 at (0,0), and ##D(0,0) \neq 0##.
 
  • Like
Likes 1 person
  • #5
Oh, I forgot to do the product rule on fy thus my fyy was incorrect.

Thanks LCKurtz & Ray Vickson!
 
  • #6
coolusername said:
I see from the original equation that the max of f(x,y) is 1. However what would the value of D be? Did I calculate it correctly? I keep getting 0 but since the critical point has a max behaviour then shouldn't D be more than 0?

Regardless of the calculation error that I overlooked, the discriminant can be zero even though there is a relative max or min. This is even true in one variable. Consider ##y=x^4##. This has an absolute min at ##x=0## yet ##y'=0## and ##y''=0## at ##x=0## so the second derivative test fails to discriminate the relative min.
 

Related to Find the D value of critical point and the type of critical point

1. What is the D value of a critical point?

The D value of a critical point is the determinant of the Hessian matrix, which is a square matrix containing the second-order partial derivatives of a multivariable function. It is used to determine if a critical point is a maximum, minimum, or saddle point.

2. How do you find the D value of a critical point?

To find the D value of a critical point, you first need to calculate the Hessian matrix by taking the second-order partial derivatives of the function. Then, find the determinant of the matrix. If the determinant is positive, the critical point is a minimum. If it is negative, the critical point is a maximum. If it is zero, further analysis is needed to determine the type of critical point.

3. What is the significance of the D value in determining the type of critical point?

The D value is significant because it helps determine the nature of the critical point. If the D value is positive, the critical point is a minimum, which means the function decreases in all directions. If the D value is negative, the critical point is a maximum, and the function increases in all directions. If the D value is zero, further analysis is needed to determine the type of critical point.

4. What does a positive D value indicate about a critical point?

A positive D value indicates that the critical point is a minimum. This means that the function decreases in all directions around the critical point and has a lower value than all nearby points.

5. How does the type of critical point affect the behavior of a function?

The type of critical point affects the behavior of a function because it determines the direction in which the function changes around that point. A minimum critical point means the function decreases in all directions, while a maximum critical point means the function increases in all directions. A saddle point has both increasing and decreasing directions, making it a point of inflection for the function.

Similar threads

  • Calculus and Beyond Homework Help
Replies
27
Views
370
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
913
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
Back
Top