Calculus III - Multivariate Continuity

In summary, the function f(x,y) is defined as (x^3 - y^3)/(x^2 + y^2) for all points except (0,0), where it is equal to 0. It is continuous at (0,0) when converted to polar coordinates, but its derivatives with respect to x and y are not continuous at this point. This is because the derivative of the polar function and the partial derivatives of f are not continuous.
  • #1
Jonmundsson
22
0

Homework Statement


Let
[itex]\begin{equation*}
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}
\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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  • #2
Jonmundsson said:

Homework Statement


Let
[itex]\begin{equation*}
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}
\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) ?
 
  • #3
They aren't. I'm asking whether it is a coincident that the derivative of g and the partials of f aren't continuous.
 

Related to Calculus III - Multivariate Continuity

1. What is multivariate continuity in calculus?

Multivariate continuity in calculus refers to the concept of a function being continuous in multiple variables. This means that the function remains unchanged when approached from any direction in the domain, and that the limit at a given point exists.

2. How is multivariate continuity different from single-variable continuity?

Single-variable continuity only considers the continuity of a function in one variable, while multivariate continuity takes into account multiple variables. This means that a function can be continuous in one variable but not in another, making it discontinuous in multivariate calculus.

3. What are some examples of multivariate continuous functions?

Examples of multivariate continuous functions include polynomials, rational functions, trigonometric functions, and exponential functions. These functions have smooth and continuous graphs in multiple variables.

4. How is multivariate continuity used in real-life applications?

Multivariate continuity is used in many real-life applications, such as in physics, engineering, and economics. It helps in analyzing the behavior of functions with multiple variables, making predictions and solving problems in these fields.

5. How can one test for multivariate continuity?

In order to test for multivariate continuity, one can use the definition of continuity, which involves taking limits and checking if they exist at a given point. Another method is to check for continuity along different paths approaching the point, and if the values of the function remain consistent, then the function is multivariate continuous.

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