Problem 36 - Calculus Early Transcendentals

Question

At what points of \(\mathbb{R}^{2}\) are the following functions continuous? $$S(x, y)=\frac{4 x^{2} y^{2}}{x^{2}+y^{2}}$$

Step-by-Step Solution

Answer

If not, specify the points where it is not continuous. Answer: Yes, the function S(x, y) is continuous everywhere in \(\mathbb{R}^2\).

Step 1: Find the domain of the function

Identify the points (x,y) for which the function is well-defined. The denominator should not be equal to zero, so we have the following equation: $$x^2 + y^2 \neq 0$$ This is true for all (x, y) except the point (0, 0). Thus, the function is well-defined for all points in \(\mathbb{R}^2\) except (0, 0).

Step 2: Verify continuity away from the origin

Since the function is a rational function and the denominator is non-zero for all (x, y) ≠ (0,0), S(x, y) is continuous everywhere except possibly at the origin, (0, 0).

Step 3: Investigate the limit as (x, y) approaches (0, 0)

To verify the continuity at (0,0), first find the limit as (x, y) approaches (0, 0). We can try to rewrite the function to see if we can find this limit: $$\lim_{(x, y) \to (0, 0)} \frac{4x^2 y^2}{x^2 + y^2} = \lim_{(x, y) \to (0, 0)} \frac{4x^2 y^2}{(x^2 + y^2)}$$ We can rewrite this limit using polar coordinates: $$r^2 = x^2 + y^2$$ $$\theta = \arctan\frac{y}{x} $$ Our function becomes: $$S(r, \theta) = \frac{4r^2 \cos^2(\theta) r^2 \sin^2(\theta)}{r^2} = 4r^2 \cos^2(\theta) \sin^2(\theta)$$ So now the limit is: $$\lim_{r \to 0}4r^2 \cos^2(\theta) \sin^2(\theta) = 0$$ Since the limit exists and is equal to zero, the function S(x, y) is continuous at the origin (0, 0).

Step 4: Conclusion

The function S(x, y) is continuous everywhere in \(\mathbb{R}^2\), including the origin as we have shown the limit at (0, 0) exists.

Key Concepts

Polar Coordinates

The concept of polar coordinates is incredibly useful in multivariable calculus, especially when dealing with problems involving circles or rotational symmetry. In polar coordinates, every point in a plane is defined by a distance from a reference point and an angle from a reference direction. This is different from Cartesian coordinates, which use horizontal and vertical distances. A point in polar coordinates is designated as \[(r, \theta)\]. Here:

• \(r\) represents the radial distance from the origin (radial coordinate).
• \(\theta\) is the angle measured from the positive x-axis (angular coordinate).

Polar coordinates can be transformed into Cartesian coordinates with the equations:

• \(x = r \cos(\theta)\)
• \(y = r \sin(\theta)\)

By using polar coordinates, functions that seem complex in Cartesian coordinates can become simpler. In the given exercise, converting the function into polar coordinates helped in analyzing its behavior near the origin, transforming a complex quotient into an expression involving \(r\) that is easier to evaluate.

Limits in Multivariable Calculus

In single-variable calculus, finding a limit as a variable approaches a point is straightforward because we have only one path along which we approach. In multivariable calculus, however, things get more interesting as there are infinitely many paths one could take to reach a point.To determine continuity or the existence of limits in two-variable functions such as \(S(x, y)\), analysts often evaluate the function's behavior as it approaches the point from different directions.For instance, evaluating a limit at the origin (0,0) might involve approaching along the x-axis, y-axis, or a path described by \(y=mx\). In the given problem, the polar coordinate transformation streamlined this analysis, since any path can be represented with a combination of \(r\) (the radial distance) and various angles \(\theta\).If a limit is consistent across all paths, it's convincing evidence that the limit exists at that point. Here, converting the original function to polar coordinates showed that as \(r \to 0\), regardless of \(\theta\), led to a limit of zero, thereby confirming the function's continuity at the origin.

Rational Functions

Rational functions are expressions that represent the quotient of two polynomials. In general, they have the form \(f(x) = \frac{g(x)}{h(x)}\), where both \(g(x)\) and \(h(x)\) are polynomials. One key issue with rational functions is that they can be undefined when the denominator equals zero.The exercise presented a function: \(S(x,y) = \frac{4x^2 y^2}{x^2+y^2}\). For rational functions in two variables, the domain usually excludes points that make the denominator zero. In this case, that occurs at the origin \((0,0)\). Away from the origin, the function behaves nicely, and continuity is straightforward.To understand continuity at \((0,0)\), potential discontinuities or undefined regions are crucial since the limit's existence there resolves it. The successful conversion to polar coordinates demonstrated that the rational function approaches a single value (zero) as \((x,y)\) approach \((0,0)\), therefore proving continuity across its domain, despite the initial complication at the origin.