Critical point of two variable function

In summary, the problem involves determining the type of critical point at (0,0) for two given functions with critical points at (0,0) and second derivatives of 0 at that point. The first function appears to be a saddle point, while the second function requires further analysis, potentially involving third derivatives.
  • #1
gertrudethegr
6
0

Homework Statement




Each of the functions f have a critical point at (0,0), however at this point the second derivatives are all zero. Determine te type of critical point at (0,0) in this case

1) f(x,y)=x2y+xy2
2)f(x,y) = x4+2*x3y+x2y2+y4





The Attempt at a Solution




For the first part by plotting it on Wofram I saw that it was a saddle, and i substituted the values x=y and x=-y to show that there is a saddle point in one direction and a straight line in the other, hence a saddle point, is this sufficient

On 2) I know I must use taylors expansion, but up to how many terms, and what do I do once i get there?
 
Physics news on Phys.org
  • #2
gertrudethegr said:

Homework Statement




Each of the functions f have a critical point at (0,0), however at this point the second derivatives are all zero. Determine te type of critical point at (0,0) in this case

1) f(x,y)=x2y+xy2
2)f(x,y) = x4+2*x3y+x2y2+y4





The Attempt at a Solution




For the first part by plotting it on Wofram I saw that it was a saddle, and i substituted the values x=y and x=-y to show that there is a saddle point in one direction and a straight line in the other, hence a saddle point, is this sufficient
I doubt it very much. I don't see anything in your work that indicates you took partial derivatives. Your textbook should have some examples of categorizing critical points by the use of partials.
gertrudethegr said:
On 2) I know I must use taylors expansion, but up to how many terms, and what do I do once i get there?
I don't think a Taylor series has anything to do with this problem. The comment I made before applies here as well.
 
  • #3
The only examples I found using partials considered the second derivative at the given point, however as the questions states the second derivative at the points are all 0, hence all the examples I have found give an inconclusive solution... that is why I have become unstuck, but if I take third derivatives at the point (0,0) there are some non zero solutions, but I am unsure how to identify them as max, min, saddle...
 

What is the critical point of a two variable function?

The critical point of a two variable function is a point where both partial derivatives are equal to zero. This means that the slope in both the x and y directions is zero, indicating a potential extremum or saddle point.

How do you find the critical point of a two variable function?

To find the critical point, you must first take the partial derivatives of the function with respect to each variable. Next, set both derivatives equal to zero and solve the resulting system of equations to find the values of x and y at the critical point.

What is the significance of the critical point in a two variable function?

The critical point is significant because it can help determine the maximum, minimum, or saddle points of the function. It is also used in optimization problems to find the optimal value of the function.

Can a two variable function have more than one critical point?

Yes, a two variable function can have multiple critical points. This can happen when the function has multiple extrema or when there is a saddle point in addition to an extremum.

How does the critical point relate to the second derivative test?

The second derivative test is used to determine the nature of the critical point (maximum, minimum, or saddle point). It involves taking the second derivatives of the function at the critical point and evaluating them to determine the concavity of the function. This test is only applicable when both second derivatives are nonzero.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
798
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
825
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
1K
  • Calculus and Beyond Homework Help
Replies
8
Views
348
  • Calculus and Beyond Homework Help
Replies
1
Views
416
  • Calculus and Beyond Homework Help
Replies
22
Views
2K
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
Back
Top