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Chapter 13: Problem 52

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

Short Answer

Expert verified
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 by step solution

01

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 \((0,0)\).
02

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): $$ \lim_{(x,y) \to (0,0)} \frac{1 - \cos(x^2 + y^2)}{x^2 + y^2} = \lim_{r \to 0} \frac{1-\cos (r^2)}{r^2} $$
03

Apply L'H么pital's rule to find the limit

Since the limit has the indeterminate form \(\frac{0}{0}\), L'H么pital'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 \((0,0)\).
04

Conclusion

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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'H么pital's Rule
Sometimes function limits encounter the challenge of indeterminate forms such as \(\frac{0}{0}\). That's where L'H么pital'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'H么pital鈥檚 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'H么pital鈥檚 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鈥檚 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.

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Most popular questions from this chapter

Find an equation of the plane passing through the point (3,2,1) that slices off the region in the first octant with the least volume.

Consider the following functions \(f.\) a. Is \(f\) continuous at (0,0)\(?\) b. Is \(f\) differentiable at (0,0)\(?\) c. If possible, evaluate \(f_{x}(0,0)\) and \(f_{y}(0,0)\) d. Determine whether \(f_{x}\) and \(f_{y}\) are continuous at \((0,0).\) e. Explain why Theorems 5 and 6 are consistent with the results in parts \((a)-(d).\) $$f(x, y)=\sqrt{|x y|}$$

Find the dimensions of the rectangular box with maximum volume in the first octant with one vertex at the origin and the opposite vertex on the ellipsoid \(36 x^{2}+4 y^{2}+9 z^{2}=36\).

Use the definition of the gradient (in two or three dimensions), assume that \(f\) and \(g\) are differentiable functions on \(\mathbb{R}^{2}\) or \(\mathbb{R}^{3},\) and let \(c\) be a constant. Prove the following gradient rules. a. Constants Rule: \(\nabla(c f)=c \nabla f\) b. Sum Rule: \(\nabla(f+g)=\nabla f+\nabla g\) c. Product Rule: \(\nabla(f g)=(\nabla f) g+f \nabla g\) d. Quotient Rule: \(\nabla\left(\frac{f}{g}\right)=\frac{g \nabla f-f \nabla g}{g^{2}}\) e. Chain Rule: \(\nabla(f \circ g)=f^{\prime}(g) \nabla g,\) where \(f\) is a function of one variable

Limits at (0,0) may be easier to evaluate by converting to polar coordinates. Remember that the same limit must be obtained as \(r \rightarrow 0\) along all paths to (0,0) Evaluate the following limits or state that they do not exist. $$\lim _{(x, y) \rightarrow(0,0)} \frac{(x-y)^{2}}{\left(x^{2}+y^{2}\right)^{3 / 2}}$$
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