Problem 52 - Calculus Early Transcendentals

Question

At what points of \(\mathbb{R}^{2}\) are the following functions continuous? $$f(x, y)=\left\\{\begin{array}{ll}\frac{1-\cos \left(x^{2}+y^{2}\right)}{x^{2}+y^{2}} & \text { if }(x, y)\neq(0,0)\\\0 & \text { if }(x, y)=(0,0)\end{array}\right.$$

Step-by-Step Solution

Answer

Function: \(f(x, y) = \left\\{\begin{array}{ll}\frac{1-\cos \left(x^{2}+y^{2}\right)}{x^{2}+y^{2}} & \text { if }(x, y)\neq(0,0)\\0 & \text { if }(x, y)=(0,0)\end{array}\right.\) Answer: Yes, the function is continuous at all points in \(\mathbb{R}^2\), including the point \((0,0)\).

Step 1: Determine the definition of the function at different points

The function \(f(x, y)\) is defined as: $$ f(x, y)=\left\\{\begin{array}{ll}\frac{1-\cos \left(x^{2}+y^{2}\right)}{x^{2}+y^{2}} & \text { if }(x, y)\neq(0,0)\\\0 & \text { if }(x, y)=(0,0)\end{array}\right. $$ For all points \((x, y) \neq (0,0)\), the function is well-defined. We need to determine the continuity of the function at point \((0,0)\).

Step 2: Look at the limit of the function as it approaches \((0,0)\) from different directions

The function will be continuous at \((0,0)\) if the limit exists and it's equal to the function's value at this point. We'll work on the limit in polar coordinates, since it will be easier to calculate the limit this way. Transform the coordinates \((x, y)\) into polar coordinates \((r, \theta)\) using the equations \(x = r\cos(\theta)\) and $y = r\sin(\theta): $$ \lim_{(x,y) \to (0,0)} \frac{1 - \cos(x^2 + y^2)}{x^2 + y^2} = \lim_{r \to 0} \frac{1-\cos (r^2)}{r^2} $$

Step 3: Apply L'Hôpital's rule to find the limit

Since the limit has the indeterminate form \(\frac{0}{0}\), L'Hôpital's rule can be applied: $$ \lim_{r \to 0} \frac{1 - \cos(r^2)}{r^2} = \lim_{r \to 0} \frac{d(1-\cos(r^2))}{dr} \times \frac{1}{\frac{d(r^2)}{dr}} $$ Compute the derivatives: $$ \frac{d(1-\cos(r^2))}{dr} = -2r\sin(r^2) $$ $$ \frac{d(r^2)}{dr} = 2r $$ Plug the derivatives back into the limit: $$ \lim_{r \to 0} \frac{1 - \cos(r^2)}{r^2} = \lim_{r \to 0} \frac{-2r\sin(r^2)}{2r} $$ Now we can cancel out the \(2r\) terms: $$ \lim_{r \to 0} \frac{1 - \cos(r^2)}{r^2} = \lim_{r \to 0} \sin(r^2) $$ Since \(\lim_{r \to 0} \sin(r^2) = 0\), which is equal to the function's value at \((0,0)\), the function is continuous at \((0,0)\).

Step 4: Conclusion

The function \(f(x, y)\) is continuous at all points in \(\mathbb{R}^2\), including the point \((0,0)\).

Key Concepts

Limits in Polar Coordinates

Understanding continuity of a function in multiple dimensions can be perplexing when different paths approach a point. That's where polar coordinates come in handy as a tool that simplifies the process. By changing the Cartesian coordinates

• \((x, y)\)
• to polar form as \((r, \theta)\)
• using \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\),

we focus on the radial distance \(r\) from the origin. This means we're checking how the function behaves when moving closer to the origin from any direction. Moving along different angles \(\theta\) can help ensure the limit doesn't depend on path choice. This is particularly useful for testing continuity at a critical point like the origin: if the limit exists and equals the function's value at

• \((0,0)\),

the function is continuous there. Polar coordinates simplify checking continuity because you only need to evaluate the limit as \(r\) approaches zero.

L'Hôpital's Rule

Sometimes function limits encounter the challenge of indeterminate forms such as \(\frac{0}{0}\). That's where L'Hôpital's Rule becomes the key. This rule acts as a mathematical lifeline, allowing us to compute limits that would otherwise seem elusive. The rule states that by taking the derivative of the numerator and the denominator separately, and then calculating the limit, you can resolve the indeterminate form.In practice:

• For the given problem, we initially find the limit
• This takes the form \(\lim_\{r \to 0\}\frac{1 - \cos(r^2)}{r^2}\).
• Recognizing the \(\frac{0}{0}\) state,
• we differentiate
• \(1 - \cos(r^2)\) and \(r^2\)
• respectively

Applying L'Hôpital’s Rule reveals the process as

• \(\lim_\{r \to 0\} \frac{-2r\sin(r^2)}{2r}\) which simplifies to \(\lim_\{r \to 0\} \sin(r^2)\).
• At this point,
• as \(r\) approaches zero, \(\sin(r^2)\) naturally approaches zero too.

Thus, L'Hôpital’s Rule resolves the limit problem elegantly, proving its value in complex calculus scenarios.

Multivariable Calculus

Multivariable calculus expands calculus concepts into multiple dimensions, dealing with functions that rely on more than one variable. It allows for a deeper understanding of changes over planes and in space.
For continuity in multivariable functions, we need to be sure that as

• one approaches a point,
• the function's value tends to stabilize and aligns with the function value at that point.

In practical terms: This involves lurking into partial derivatives, gradients, and limits.
When defining continuity,
Each direction must yield the same limit for it to exist and match the function’s value at that point.
Utilizing methods like converting to polar coordinates helps examine limits in a simplified manner, especially around critical points like the origin.

• For problem-solving,
• embedding these concepts helps
• analyze complex functions comprehensively, confirming
• continuity over actionable regions within their domain.

Overall, multivariable calculus opens doors to evaluating and understanding the behavior of functions in higher dimensions, driving further mathematical exploration.