Math Problem Statement

Odredi neprekidnost

Odredi neprekidnost

Solution

To determine the continuity of the given function f(x,y)f(x, y), we need to check if the function is continuous at all points in its domain, including the origin (0,0)(0, 0).

Steps to Determine Continuity:

  1. Away from (0,0)(0, 0): The function f(x,y)f(x, y) is defined as f(x,y)=x4x3y2x2+y2,when x2+y20.f(x, y) = \frac{x^4 - x^3y^2}{x^2 + y^2}, \quad \text{when } x^2 + y^2 \neq 0. For x2+y20x^2 + y^2 \neq 0, the denominator is non-zero, and the function is well-defined. Since it is a rational function and continuous wherever the denominator is non-zero, f(x,y)f(x, y) is continuous for all points where x2+y20x^2 + y^2 \neq 0.

  2. At the Origin (0,0)(0, 0): At (0,0)(0, 0), the function is defined as f(0,0)=1f(0, 0) = 1. To check continuity at this point, we calculate the limit of f(x,y)f(x, y) as (x,y)(0,0)(x, y) \to (0, 0) and compare it with f(0,0)=1f(0, 0) = 1.

    We analyze the limit: lim(x,y)(0,0)f(x,y)=lim(x,y)(0,0)x4x3y2x2+y2.\lim_{(x, y) \to (0, 0)} f(x, y) = \lim_{(x, y) \to (0, 0)} \frac{x^4 - x^3y^2}{x^2 + y^2}.

    • Using polar coordinates: Let x=rcosθx = r\cos\theta and y=rsinθy = r\sin\theta, where r=x2+y2r = \sqrt{x^2 + y^2}. Substituting: f(x,y)=(rcosθ)4(rcosθ)3(rsinθ)2r2.f(x, y) = \frac{(r\cos\theta)^4 - (r\cos\theta)^3(r\sin\theta)^2}{r^2}. Simplify the numerator: f(x,y)=r4cos4θr5cos3θsin2θr2=r2(cos4θrcos3θsin2θ).f(x, y) = \frac{r^4 \cos^4\theta - r^5 \cos^3\theta \sin^2\theta}{r^2} = r^2 (\cos^4\theta - r\cos^3\theta \sin^2\theta).

      As r0r \to 0, the term r2(cos4θrcos3θsin2θ)0r^2 (\cos^4\theta - r\cos^3\theta \sin^2\theta) \to 0. Thus: lim(x,y)(0,0)f(x,y)=0.\lim_{(x, y) \to (0, 0)} f(x, y) = 0.

    Since the limit lim(x,y)(0,0)f(x,y)=0\lim_{(x, y) \to (0, 0)} f(x, y) = 0 does not equal f(0,0)=1f(0, 0) = 1, the function f(x,y)f(x, y) is not continuous at (0,0)(0, 0).

Conclusion:

The function f(x,y)f(x, y) is continuous everywhere except at the origin (0,0)(0, 0), where it is discontinuous.


Do you have any further questions, or would you like to delve into more detailed aspects of this result?

Questions for Exploration:

  1. Why does the polar coordinate substitution simplify the limit calculation?
  2. How do we confirm the continuity for x2+y20x^2 + y^2 \neq 0?
  3. What role does the definition of f(x,y)f(x, y) at (0,0)(0, 0) play in determining continuity?
  4. Could modifying f(0,0)f(0, 0) make the function continuous everywhere?
  5. What are other approaches to test continuity in two-variable functions?

Tip:

When analyzing continuity of multivariable functions at a point, polar coordinates can help simplify cases with radial symmetry or where terms depend on rr.

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Math Problem Analysis

Mathematical Concepts

Continuity of Multivariable Functions
Limits in Multivariable Calculus
Polar Coordinates

Formulas

f(x, y) = \frac{x^4 - x^3y^2}{x^2 + y^2}, \text{for } x^2 + y^2 \neq 0
Limit: \lim_{(x, y) \to (0, 0)} f(x, y)

Theorems

Continuity Definition for Multivariable Functions

Suitable Grade Level

Undergraduate (Calculus Level)