Calculus III - Multivariate Continuity

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SUMMARY

The discussion focuses on the continuity of the function f(x,y) = (x^3 - y^3)/(x^2 + y^2) at the point (0,0) and the continuity of its partial derivatives f_x and f_y. It is established that f is continuous at (0,0) when expressed in polar coordinates, as the limit approaches zero regardless of the angle θ. However, the derivatives ∂f/∂x and ∂f/∂y are not continuous at this point, indicating a discrepancy between the continuity of the function and its partial derivatives.

PREREQUISITES
  • Understanding of multivariable calculus concepts, specifically continuity.
  • Familiarity with polar coordinates and their application in limits.
  • Knowledge of partial derivatives and their significance in multivariable functions.
  • Ability to analyze limits in multiple dimensions.
NEXT STEPS
  • Study the concept of continuity in multivariable functions in detail.
  • Learn how to convert Cartesian coordinates to polar coordinates for limit evaluation.
  • Investigate the implications of discontinuous partial derivatives on the overall function.
  • Explore examples of functions with similar characteristics to deepen understanding of continuity and differentiability.
USEFUL FOR

Students and educators in multivariable calculus, mathematicians analyzing continuity in functions of several variables, and anyone studying the properties of derivatives in relation to continuity.

Jonmundsson
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Homework Statement


Let
[itex]\begin{equation*}<br /> f(x,y) = \begin{cases} \dfrac{x^3 - y^3}{x^2 + y^2}, \hspace{1.1em} (x, y) \neq (0,0) \\ 0, \hspace{4em} (x,y) = (0,0) \end{cases}<br /> \end{equation*}[/itex]

Is [itex]f[/itex] continuous at the point [itex](0,0)[/itex]? Are [itex]f_x[/itex] og [itex]f_y[/itex] continuous at the point [itex](0,0)[/itex]?

Homework Equations



Polar coords

The Attempt at a Solution



If you convert f to polar it's easy to see that it is continuous (since it doesn't depend on [itex]\theta[/itex]. I'm just wondering if the derivative of the polar function is dependent on [itex]\theta[/itex] then it isn't continuous and therefore neither are [itex]f_x[/itex] and [itex]f_y[/itex]

Steps:

[itex]\displaystyle \lim _{(x,y) \to (0,0)} \dfrac{x^3 - y^3}{x^2 + y^2} = \lim _{r \to 0} \hspace{0.3em} r (cos^3 \theta - sin^3 \theta) = 0[/itex]

Define [itex]g(r) = r (cos^3 \theta - sin^3 \theta)[/itex] then [itex]g'(r) = cos^3 \theta - sin^3 \theta[/itex] and [itex]\displaystyle \lim _{r \to 0} cos^3 \theta - sin^3 \theta[/itex] doesn't exist.

Cheers.
 
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Jonmundsson said:

Homework Statement


Let
[itex]\begin{equation*}<br /> f(x,y) = \begin{cases} \dfrac{x^3 - y^3}{x^2 + y^2}, \hspace{1.1em} (x, y) \neq (0,0) \\ 0, \hspace{4em} (x,y) = (0,0) \end{cases}<br /> \end{equation*}[/itex]

Is [itex]f[/itex] continuous at the point [itex](0,0)[/itex]? Are [itex]f_x[/itex] og [itex]f_y[/itex] continuous at the point [itex](0,0)[/itex]?

Homework Equations



Polar coords

The Attempt at a Solution



If you convert f to polar it's easy to see that it is continuous (since it doesn't depend on [itex]\theta[/itex]. I'm just wondering if the derivative of the polar function is dependent on [itex]\theta[/itex] then it isn't continuous and therefore neither are [itex]f_x[/itex] and [itex]f_y[/itex]

Steps:

[itex]\displaystyle \lim _{(x,y) \to (0,0)} \dfrac{x^3 - y^3}{x^2 + y^2} = \lim _{r \to 0} \hspace{0.3em} r (cos^3 \theta - sin^3 \theta) = 0[/itex]

Define [itex]g(r) = r (cos^3 \theta - sin^3 \theta)[/itex] then [itex]g'(r) = cos^3 \theta - sin^3 \theta[/itex] and [itex]\displaystyle \lim _{r \to 0} cos^3 \theta - sin^3 \theta[/itex] doesn't exist.

Cheers.
How about ∂f/∂x and ∂f/∂y? Are the continuous at (0, 0) ?
 
They aren't. I'm asking whether it is a coincident that the derivative of g and the partials of f aren't continuous.
 

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