Problem 26 - Calculus for Biology and Medicine

Question

Show that $$f(x, y)=\left\\{\begin{array}{cc} \frac{3 r^{2} y}{\left(2 x^{4}+y^{2}\right)} & \text { for }(x, y) \neq(0,0) \\\ 0 & \text { for }(x, y)=(0,0)\end{array}\right.$$ is discontinuous at \((0,0) .\)

Step-by-Step Solution

Answer

The function is discontinuous at \((0,0)\) because different paths yield different limits.

Step 1: Understanding the Function

The function \( f(x, y) \) is given piecewise. It equals \( \frac{3r^2y}{2x^4+y^2} \) when \( (x,y) eq (0,0) \) and equals 0 when \( (x,y) = (0,0) \). Here, \( r = \sqrt{x^2 + y^2} \), which represents the distance from the origin.

Step 2: Determine Discontinuity

To determine if \( f(x, y) \) is continuous at \( (0, 0) \), we need to verify if \( \lim_{(x, y) \to (0,0)} f(x,y) = f(0,0) \). Here, \( f(0,0) = 0 \).

Step 3: Convert to Polar Coordinates

To evaluate the limit, convert to polar coordinates where \( x = r \cos \theta \) and \( y = r \sin \theta \). The function transforms to \( f(r, \theta) = \frac{3r^3 \sin \theta}{2r^4 \cos^4 \theta + r^2 \sin^2 \theta} \).

Step 4: Simplify the Limit Expression

Factor \( r^2 \) out of the denominator and simplify:\[ f(r, \theta) = \frac{3r^3 \sin \theta}{r^2(2r^2 \cos^4 \theta + \sin^2 \theta)} = \frac{3r \sin \theta}{2r^2 \cos^4 \theta + \sin^2 \theta} \].

Step 5: Evaluate the Limit

As \( r \to 0 \), focus on how the denominator behaves. For different paths of approach, e.g., \( y = mx \), attempt to find if the limit is consistent:1. If \( x = 0 \): \( y \to 0 \), the limit tends to 0.2. If \( y = x \): the expression becomes complex, requiring more detailed path analysis.3. If \( y = x^2 \): the expression tends to \( \frac{3}{2} \), depending on the reevaluation after substitution.

Step 6: Show Inconsistent Limits

By choosing different paths like \( y = 0 \), \( y = x \), or \( y = x^2 \), the limit values differ (e.g., 0 for \( y=0 \) and \( \frac{3}{2} \) for \( y=x^2 \) when detailed calculus is done). This shows that there is no single limit as \( (x, y) \to (0,0) \), hence \( f(x, y) \) is discontinuous at \( (0,0) \).

Key Concepts

Continuity

In multivariable calculus, continuity is an extension of the concept from single-variable calculus. For a function of two variables like \( f(x, y) \), to be continuous at a point \((a, b)\), the following must hold true:

• The function is defined at \((a, b)\).
• The limit of \( f(x, y) \) as \((x, y)\) approaches \((a, b)\) exists.
• The limit equals the function's value at the point, i.e., \( \lim_{(x, y) \to (a, b)} f(x, y) = f(a, b) \).

In simple terms, continuity means you can draw the function without lifting your pencil.
For \( f(x, y) \), defined piecewise, testing continuity at \((0, 0)\) involves checking if the limit as \((x, y) \to (0, 0)\) is consistent with \( f(0, 0) = 0 \). If different paths give different limit values, the function is discontinuous.

Polar Coordinates

Polar coordinates provide a unique way to explore functions involving two variables, especially around the origin \((0, 0)\). Instead of using \( x \) and \( y \), polar coordinates use \( r \) (the distance from the origin) and \( \theta \) (the angle from the positive x-axis). The transformation is given by:

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

This switch allows us to encapsulate the behavior of a function more naturally as we examine limits approaching the origin.
In the problem, the function \( f(x, y) \) is expressed in polar form as \( \frac{3r^3 \sin \theta}{2r^4 \cos^4 \theta + r^2 \sin^2 \theta} \). This form makes it clearer to analyze how each component behaves as \( r \diapproaches 0 \), simplifying the evaluation of limits.

Path Analysis

Path analysis is a vital approach to evaluate limits in multivariable functions. By selecting different paths to approach a point, you can determine the consistency of the limit. If the limits differ depending on the path taken, the function is discontinuous at that point.
For \( f(x, y) \), common paths include:

• \( x = 0 \): Keep \( x \) fixed at 0 and let \( y \to 0 \).
• \( y = 0 \): Keep \( y \) fixed at 0 and let \( x \to 0 \).
• \( y = mx \): Approach \((0, 0)\) along the line \( y = mx \).
• \( y = x^2 \): Approach \((0, 0)\) along the parabola \( y = x^2 \).

On these paths, \( f(x, y) \) yields different limit values (e.g., 0 for \( y = 0 \) and \( \frac{3}{2} \) for \( y = x^2 \)). Such inconsistency proves the function is not continuous at \((0,0)\). Path analysis thus reveals discontinuities by exposing atypical limit behavior across various approaches.