Problem 95 - Calculus 3

Question

For the following exercises, determine the region in which the function is continuous. Explain your answer. $$ f(x, y)=\left\\{\begin{array}{ll}{\frac{x^{2} y}{x^{2}+y^{2}}} & {\text { if }(x, y) \neq(0,0)} \\ {0} & {\text { if }(x, y)=(0,0)}\end{array}\right\\} $$

Step-by-Step Solution

Answer

The function is continuous everywhere except at the point \((0,0)\).

Step 1: Define the function

The function in question is given by \( f(x, y) = \frac{x^2 y}{x^2 + y^2} \) for \( (x, y) eq (0, 0) \) and \( f(0, 0) = 0 \). It is continuous for all points except possibly at \( (0, 0) \).

Step 2: Analyze continuity for \((x, y) \neq (0, 0)\)

For any point \((x, y) eq (0, 0)\), the function is a rational function \( g(x, y) = \frac{x^2 y}{x^2 + y^2} \) where both the numerator and the denominator are polynomials. Polynomials are continuous everywhere, and the denominator \(x^2 + y^2\) is non-zero for \((x, y) eq (0, 0)\), so \(f(x, y)\) is continuous at all points other than \((0,0)\).

Step 3: Check continuity at \((0, 0)\) using limits

To check continuity at \((0, 0)\), we need \(\lim_{(x, y) \to (0, 0)} f(x, y) = f(0, 0) = 0\). Calculate this limit using polar coordinates where \( x = r \cos \theta \) and \( y = r \sin \theta \). The function becomes \( f(r, \theta) = \frac{r^2 \cos^2 \theta \sin \theta}{r^2} = \cos^2 \theta \sin \theta \). Since \( \cos^2 \theta \sin \theta \) can take any value between \(-1\) and \(1\) as \(\theta\) varies, the limit does not approach a single value, implying the limit does not exist.

Step 4: Conclude continuity of the function

Since the limit at \((0,0)\) does not exist, the function is not continuous at this point. Therefore, the function \(f(x, y)\) is continuous over all of \(\mathbb{R}^2\) except at \((0, 0)\).

Key Concepts

Continuous Function

A continuous function is a fundamental concept in calculus and analysis, describing functions that do not have abrupt changes, jumps, or gaps in their graphs. For a function to be continuous at a point, three criteria must be met:

• The function is defined at the point.
• The limit of the function as it approaches the point exists.
• The limit of the function equals the function's value at the point.

In the given exercise, these criteria are tested for the function \(f(x, y)\) defined as \(\frac{x^2 y}{x^2 + y^2}\) for all points except \((0,0)\), where it is defined as zero. Away from \((0, 0)\), the function involves a **rational function**, making it continuous everywhere except potentially at \((0, 0)\). The continuity test focuses heavily on this special point, using limits to determine behavior.

Polar Coordinates

Polar coordinates are a two-dimensional coordinate system where each point on a plane is determined by an angle and a distance from the origin. These are particularly useful when dealing with problems that have circular symmetry or when analyzing points around the origin.
For example, any point \((x, y)\) in Cartesian coordinates can be expressed in polar coordinates as \((r \cos \theta, r \sin \theta)\). Here, \(r\) is the distance from the origin and \(\theta\) is the angle from the positive x-axis.
The problem makes use of polar coordinates to examine the continuity of the function \(f(x, y)\) at the origin:\( (0,0) \). By converting the function into polar form, it simplifies the evaluation of limits as \((r, \theta) \to (0, \text{any } \theta)\). This transformation reveals that the behavior of \(f(x, y)\) depends significantly on \(\theta\), not converging to a specific value, signaling discontinuity at \((0,0)\).

Rational Function

Rational functions are composed of one polynomial divided by another. They can showcase simple behaviors or complex patterns, which are especially interesting when considering limits and continuity.

• The numerator and denominator in rational functions should both be polynomials.
• Discontinuities may arise from values making the denominator zero.

In the original exercise, the given function \(f(x, y)\) is a rational expression \(\frac{x^2 y}{x^2 + y^2}\). This setup indicates a continuous process for all points except where the denominator becomes zero. At \((0, 0)\), it's important to check carefully, as it's where the denominator equals zero, actually leading to a discontinuity.
For points other than the origin, the denominator \(x^2 + y^2\) never reaches zero in the real-plane scenario that excludes \((0,0)\). Hence, the rational function remains continuous across the rest of its defined domain.