Problem 94 - Calculus 3

Question

For the following exercises, determine the region in which the function is continuous. Explain your answer. $$ f(x, y)=\frac{x^{2} y}{x^{2}+y^{2}} $$

Step-by-Step Solution

Answer

The function is continuous everywhere on \( \mathbb{R}^2 \).

Step 1: Understanding the Problem

We are given a function of two variables, \( f(x, y) = \frac{x^2 y}{x^2 + y^2} \), and we need to determine where it is continuous. A function is continuous where it does not have any indeterminate forms or discontinuities.

Step 2: Identify Potential Points of Discontinuity

The function \( f(x, y) = \frac{x^2 y}{x^2 + y^2} \) might be discontinuous where the denominator \( x^2 + y^2 = 0 \). Since \( x^2 + y^2 = 0 \) only at the point \((x, y) = (0, 0)\), this is our candidate for a point of discontinuity.

Step 3: Examine Continuity at Critical Points

We need to examine if \( f(x, y) \) is continuous at \((0, 0)\). At other points, it is continuous as the denominator is non-zero. At \((0, 0)\), we check the limit: \[ \lim_{(x, y) \to (0, 0)} \frac{x^2 y}{x^2 + y^2}. \]

Step 4: Compute the Limit

To compute the limit \( \lim_{(x, y) \to (0, 0)} \frac{x^2 y}{x^2 + y^2} \), consider the limit along various paths. Along the x-axis (\( y=0 \)), the limit is 0, and along the y-axis (\( x=0 \)), the limit is also 0. Checking along the line \( y = mx \), the limit still evaluates to 0: \( \lim_{x \to 0} \frac{x^2(mx)}{x^2 + (mx)^2} = \lim_{x \to 0} \frac{mx^3}{x^2(1+m^2)} = 0 \). This suggests the limit is 0. However, we still need to confirm its consistency from other paths.

Step 5: Conclusion on Path Independent Limits

By also considering another polar coordinate \( (r, \theta) \), \( x = r\cos(\theta), y = r\sin(\theta) \), \( \lim_{r \to 0} \frac{r^2 \cos^2(\theta) \cdot r \sin(\theta)}{r^2 \cos^2(\theta) + r^2 \sin^2(\theta)} = \lim_{r \to 0} r\cdot\cos^2(\theta) \sin(\theta) = 0 \). The limit evaluates to 0, indicating continuity at \((0,0)\).

Step 6: Final Continuity Conclusion

Since the limit at \((0, 0)\) is 0 and \( f(0, 0) = 0 \), the function is actually continuous everywhere for all \( (x, y) \). The function is therefore continuous across the entire plane \( \mathbb{R}^2 \).

Key Concepts

Continuity

In multivariable calculus, continuity of a function at a point means that the function's output changes smoothly as the inputs approach that point. For functions with two variables, this means that small changes in

• the x-direction,
• the y-direction,
• or both,

result in small changes in the value of the function. For the function given, we can say it is continuous everywhere in its domain if there are no abrupt jumps, holes, or points of infinity as the inputs vary. Specifically, for our function, it is continuous wherever the denominator \(x^2 + y^2 eq 0\). Hence, except potentially at the origin (0,0), we must check its behavior more closely there.

Limit Evaluation

Evaluating limits is crucial in determining the continuity at specific points. For our function, we particularly need to check the behavior as \((x,y)\) approaches \((0,0)\). This involves evaluating \(\lim_{(x, y) \to (0, 0)} \frac{x^2 y}{x^2 + y^2}\). One common method is to check different paths leading to (0,0):

• Along the x-axis (where \(y=0\)): the corresponding limit results in 0.
• Along the y-axis (where \(x=0\)): again, we find the limit is 0.
• Along lines where \(y=mx\) for any constant m: again, the limit concludes as 0.

Finally, by converting the coordinates to polar form, we observe that the limit is consistently 0, independent of direction, showing that the function is likely continuous at (0,0).

Discontinuity in Functions

Discontinuity in a function occurs when there is an abrupt change or interruption in the output values, generally at certain points called points of discontinuity. Typically, this happens when the function is undefined or when limits don’t agree from all paths. For this problem, the only possible point of discontinuity is at \((0,0)\), because it's defined everywhere else in the xy-plane. However, upon examining the behavior at this point and finding all directional limits agree and equal the value of the function at (0,0), we conclude no actual discontinuity exists at this point.

Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to functions of more than one variable. This involves studying limits, continuity, derivatives, and integrals in multiple dimensions. In the given function of two variables, understanding where it is continuous requires tools from multivariable calculus like path-dependent limits and polar coordinates transformations. Such techniques help evaluate how functions behave in a plane, rather than along just a line, providing a richer view of continuity and allowing us to deal with potential complications of multidimensional spaces.