Problem 42 - Calculus Early Transcendentals

Question

At the beginning of this section we considered the function $$f(x, y)=\frac{\sin \left(x^{2}+y^{2}\right)}{x^{2}+y^{2}}$$ and guessed that \(f(x, y) \rightarrow 1\) as \((x, y) \rightarrow(0,0)\) on the basis of numerical evidence. Use polar coordinates to confirm the value of the limit. Then graph the function.

Step-by-Step Solution

Answer

Using polar coordinates, the limit is 1 as \((x, y) \rightarrow (0,0)\).

Step 1: Convert to Polar Coordinates

To evaluate the limit of the function \( f(x, y) = \frac{\sin(x^2 + y^2)}{x^2 + y^2} \) as \((x, y) \to (0, 0)\), we convert the Cartesian coordinates \((x, y)\) to polar coordinates. In polar coordinates, \( x = r\cos(\theta) \) and \( y = r\sin(\theta) \). Thus, \( x^2 + y^2 = r^2 \). The function becomes: \( f(r, \theta) = \frac{\sin(r^2)}{r^2} \).

Step 2: Analyze the Limit as r Approaches Zero

In polar coordinates, the limit becomes \( \lim_{r \to 0} \frac{\sin(r^2)}{r^2} \). This limit is known from single-variable calculus, the limit of \( \frac{\sin(t)}{t} \) as \( t \to 0 \) is \( 1 \). Since \( r^2 \to 0 \) as \( r \to 0 \), \( \lim_{r \to 0} \frac{\sin(r^2)}{r^2} = 1 \).

Step 3: Conclusion About the Limit

Thus, using polar coordinates, we find that \( f(x, y) \to 1 \) as \( (x, y) \to (0, 0) \), confirming our numerical evidence. The function behaves the same regardless of the direction \( (x, y) \to (0, 0) \).

Step 4: Graphing the Function

To graph the function, we consider plotting \( f(x, y) \) over a region around the origin, avoiding the undefined point \((0,0)\). The function resembles a surface that is smooth near the origin, with values approaching 1 as shown on the graph.

Key Concepts

Polar Coordinates

Polar coordinates are a versatile and powerful tool, especially in multivariable calculus. They provide an alternative to the regular Cartesian coordinate system by expressing points on a plane using a radius and angle. Instead of identifying a point by

• Horizontal (x) and vertical (y) distances.
• We use the distance (r) from the origin and the angle (θ) from the positive x-axis.

This transformation is particularly helpful for dealing with circular symmetry. In polar coordinates, any point (x, y) is expressed by

• The equations: \( x = r \cos(\theta) \) and \( y = r \sin(\theta) \).
• Where \( r \) is the radius or distance from the origin, and \( \theta \) is the angle from the positive x-axis.

This system allows us to simplify expressions like \( x^2 + y^2 \) into \( r^2 \), reducing complexity in limit evaluation by focusing directly on the distance from the origin without worrying about the angle changes.

Limit Evaluation

Evaluating limits in calculus is a fundamental skill, which determines how a function behaves as an input approaches a certain value. In the context of multivariable calculus, assessing limits involves understanding how the function behaves as a point approaches a specific target in 2D or 3D space.
For the function \( f(x, y) = \frac{\sin(x^2 + y^2)}{x^2 + y^2} \), we wanted to determine what value it tends towards as \( (x, y) \rightarrow (0,0) \).
The neat trick learned here is using polar coordinates to simplify the target. By converting to polar coordinates, the problem boils down to evaluating the limit \( \lim_{r \to 0} \frac{\sin(r^2)}{r^2} \).
In single-variable calculus, we know

• That \( \lim_{t \to 0} \frac{\sin(t)}{t} = 1 \).

This evaluation is validated for \( r^2 \rightarrow 0 \) as well, affirming that the limit is indeed 1. This process showcases how powerful limit evaluation can be for verifying behaviors of multivariable functions without guesswork.

Single-Variable Calculus

Single-variable calculus skills often extend to multivariable contexts, acting as a foundational toolset. It allows us to handle more complex functions by breaking them down into simpler, more manageable components.
For example, the expression \( \frac{\sin(t)}{t} \) is a classic example where single-variable calculus shines. This expression is notably important because as \( t \) approaches zero, its limit approaches one, serving as a theorem utilized often.
In our given function \( \frac{\sin(r^2)}{r^2} \), which surfaces after converting to polar coordinates, we see the principle directly at play. This connection helps extend solutions and understanding from single-variable to more complex multivariable problems effortlessly.
By grasping these fundamental skills in single-variable calculus, it's easier to navigate,

• Evaluate complex multivariable limits.
• Identify patterns using known results and theorems.
Overall, single-variable calculus is a crucial part of understanding broader calculus applications.

Graphing Functions

Graphing functions helps visualize mathematical expressions, making it easier to comprehend and interpret their behavior.
In multivariable calculus, graphing typically involves visualizing surfaces in 3D rather than simple lines or curves. This provides insight into the function's behavior over a region around a specified point, like the origin.
For the function \( f(x, y) = \frac{\sin(x^2 + y^2)}{x^2 + y^2} \), graphed over a surface,

• We omit the undefined point at (0,0) to understand how it behaves nearby.
• The graph reveals a smooth surface approaching a value of 1 as it nears the origin.

Visualizing functions with multivariable components like this can:

• Aid in confirming numerical evidence.
• Provide a more intuitive grasp of the function’s changes over different regions.
• Help predict how alterations might impact overall behavior.

Making graphs a valuable tool in interpreting and confirming multivariable calculus concepts.