Considering Existing of a Partial Derivative

Click For Summary
SUMMARY

The discussion focuses on evaluating the existence of a partial derivative for the function defined by the equation \(\frac{x^5 + y^6}{(x^2 + y^2)^\alpha}\) as \((x, y)\) approaches \((0, 0)\). It is established that using polar coordinates simplifies the evaluation, where \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). The limit converges to zero as long as \(\alpha < 5\), indicating that the function is continuous at the origin. Participants recommend directly calculating the partial derivatives with respect to \(x\) and \(y\) without focusing on continuity.

PREREQUISITES
  • Understanding of polar coordinates in calculus
  • Knowledge of limits and continuity in multivariable functions
  • Familiarity with partial derivatives
  • Basic proficiency in applying L'Hôpital's rule (though not recommended here)
NEXT STEPS
  • Study the conversion of Cartesian coordinates to polar coordinates in calculus
  • Learn about the conditions for continuity in multivariable functions
  • Explore the process of calculating partial derivatives for functions of two variables
  • Investigate the implications of different values of \(\alpha\) on the function's behavior
USEFUL FOR

Students and educators in calculus, particularly those focusing on multivariable calculus, as well as mathematicians interested in the behavior of functions near singular points.

makyol
Messages
17
Reaction score
0

Homework Statement



[PLAIN]http://www.netbookolik.com/wp-content/uploads/2010/07/q1.png



Homework Equations





The Attempt at a Solution



I thought, in order to take derivative function must be cont. so it would be nice to check limf as (x,y) goes to (0, 0) but it did not seem to be a correct solution because appliying l'hopital gets harder and harder. If anyone can help me to solve this problem?

Thanks!
 
Last edited by a moderator:
Physics news on Phys.org
Don't use L'hopital. Instead, convert to polar coordinates. That way, the single variable, r, measures the distance to the origin.

Since [itex]x= r cos(\theta)[/itex] and [itex]y= r sin(\theta)[/itex],

[tex]\frac{x^5+ y^6}{(x^2+ y^2)^\alpha}= \frac{r^5cos^5(\theta)+ r^6sin^6(\theta)}{r^\alpha}[/tex]
[tex]= \frac{r^5}{r^\alpha}(cos^5(\theta)+ rsin^6(\theta)}[/tex]

As long as [itex]\alpha< 5[/itex], that will go to 0 as r goes to 0 no matter what [itex]\theta[/itex] is and so the function will be continuous at (0, 0).

I would recommend you not worry about the continuity of the function itself but go ahead and take the derivatives with respect to x and y.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
5
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
4
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K