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Chapter 15: Problem 37

At what points of \(\mathbb{R}^{2}\) are the following functions continuous? $$p(x, y)=\frac{4 x^{2} y^{2}}{x^{4}+y^{2}}$$

Short Answer

Expert verified
Answer: The function p(x,y) is continuous at all points in 鈩澛�.

Step by step solution

01

Definition of continuity for functions of two variables

Continuity of a function \(p(x,y)\) at a point \((a,b)\) means that the limit of \(p(x,y)\) as \((x,y) \to (a,b)\) exists and is equal to \(p(a,b)\). In other words, for a function to be continuous at a point, $$ \lim_{(x,y) \to (a,b)} p(x,y) = p(a,b). $$
02

Check continuity at the origin (0, 0)

We will first analyze the continuity of \(p(x,y)\) at the origin (0, 0). If we directly substitute the origin into the function, we have $$p(0,0) = \frac{4 \cdot 0^2 \cdot 0^2}{0^4 + 0^2} = 0.$$ Now we need to check the limit of the function as \((x,y) \to (0,0)\). Since the function has both \(x\) and \(y\) in the denominator, we need to find a way to simplify the limit as it approaches the origin. One common technique is to change to polar coordinates. So, we set $$ x = r \cos \theta,\ y=r \sin \theta. $$ Then, as \((x,y) \to (0,0)\), we have \(r \to 0\). So, the limit we need to evaluate is $$ \lim_{r \to 0} p(r \cos \theta, r \sin \theta). $$Substituting \(x = r \cos \theta\) and \(y = r \sin \theta\) in our function we have $$p(r \cos \theta, r \sin \theta) = \frac{4 (r \cos \theta)^2 (r \sin \theta)^2}{(r \cos \theta)^4 + (r \sin \theta)^2}.$$ Simplifying this expression we obtain $$ p(r \cos \theta, r \sin \theta) = \frac{4 r^4 \cos^2\theta \sin^2\theta}{r^4 (\cos^4\theta + \sin^2\theta)}. $$ Now, dividing both the numerator and denominator by \(r^4\), we have $$ p(r \cos \theta, r \sin \theta) = \frac{4 \cos^2\theta \sin^2\theta}{\cos^4\theta + \sin^2\theta}. $$As \(r \to 0\), this expression remains finite and continuous, so we have $$ \lim_{(x,y) \to (0,0)} p(x,y) = 0. $$ By the definition of continuity, since the limit exists and equals the value of the function at the origin, \(p(x,y)\) is continuous at (0, 0).
03

Check continuity outside the origin

At points where neither \(x\) nor \(y\) are zero, the function is a rational function and is continuous wherever the denominator is non-zero. The denominator of \(p(x,y)\) is \(x^4 + y^2\). This expression is always non-negative, and only zero when both \(x\) and \(y\) are zero. Since we have already shown that the function is continuous at the origin, we can conclude that the function is also continuous at all other points in \(\mathbb{R}^2\).
04

Conclusion

The function \(p(x,y) = \frac{4x^2y^2}{x^4+y^2}\) is continuous at all points in \(\mathbb{R}^2\).

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Key Concepts

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

Polar Coordinates
Imagine trying to describe the location of a point not through the usual x-y coordinates but with a different system, namely polar coordinates. In this system, any point on a plane is described by how far it is from the origin (radius) and the angle it makes with a reference direction. Here鈥檚 how it works:

  • The distance from the point to the origin is called the radius, denoted as \(r\).
  • The angle that the line from the origin to the point makes with the positive x-axis is the angle \(\theta\).
Converting from traditional x-y coordinates to polar coordinates involves the following:

  • \(x = r \cos \theta\)
  • \(y = r \sin \theta\)
This can be particularly useful in evaluating limits. When functions become complicated in rectangular coordinates, polar coordinates can simplify expressions, as was done with the function
\(p(x, y) = \frac{4x^2y^2}{x^4 + y^2}.\)By substituting \(x = r \cos \theta\) and \(y = r \sin \theta\), the problem becomes easier to handle as you focus on the radius \(r\) shrinking to zero while the angular component provides context.
Multivariable Calculus
Multivariable calculus allows us to explore functions that depend on more than one variable. Unlike single-variable calculus which deals with curves, multivariable functions describe surfaces or hypersurfaces. The concepts of limits, derivatives, and continuity extend naturally into this multidimensional world.

For a function of two variables, \(p(x, y)\), continuity means that as any pair \((x, y)\) approaches a point \((a, b)\), the function value \(p(x,y)\) should approach the value at that point, \(p(a,b)\). This was shown when checking the continuity of \(p(x, y)\) at the origin and other points.

The steps include:
  • Identifying ragged areas in the domain, such as denominators becoming zero.
  • Simplifying the function via coordinate transformations, like with polar coordinates.
  • Verifying the behavior around special points, checking if both finite and non-zero limits exist.
The journey through multivariable calculus enriches understanding of how tiny changes in inputs affect outputs over surfaces, not just along lines.
Limits in Two Variables
Limits determine how a function behaves as it approaches a particular point from any direction. This concept is crucial in ensuring that functions of two variables like \(p(x, y) = \frac{4x^2y^2}{x^4 + y^2}\) are well-defined and continuous over their domains.

When evaluating limits in two variables, the challenge is ensuring the same value is approached no matter the path taken towards the point. Here鈥檚 how:
  • From the Cartesian route: Approaching via straight lines like \(y = mx\).
  • From Polar coordinates: Using \(x = r\cos\theta\) and \(y = r\sin\theta\) can help to express complex expressions as functions of the single variable \(r\).
In the given solution, as \((x, y)\) approaches \((0, 0)\), using different paths leads to the same limit鈥攃onfirming the continuity at the origin. These concepts are indispensable in determining points of continuity and ensuring functions behave predictably in two-dimensional spaces.

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Most popular questions from this chapter

Find the points at which the following surfaces have horizontal tangent planes. $$z=\cos 2 x \sin y \text { in the region }-\pi \leq x \leq \pi,-\pi \leq y \leq \pi$$

Differentials with more than two variables Write the differential dw in terms of the differentials of the independent variables. $$w=f(x, y, z)=\sin (x+y-z)$$

Problems with two constraints Given a differentiable function \(w=f(x, y, z),\) the goal is to find its absolute maximum and minimum values (assuming they exist) subject to the constraints \(g(x, y, z)=0\) and \(h(x, y, z)=0,\) where \(g\) and \(h\) are also differentiable. a. Imagine a level surface of the function \(f\) and the constraint surfaces \(g(x, y, z)=0\) and \(h(x, y, z)=0 .\) Note that \(g\) and \(h\) intersect (in general) in a curve \(C\) on which maximum and minimum values of \(f\) must be found. Explain why \(\nabla g\) and \(\nabla h\) are orthogonal to their respective surfaces. b. Explain why \(\nabla f\) lies in the plane formed by \(\nabla g\) and \(\nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value. c. Explain why part (b) implies that \(\nabla f=\lambda \nabla g+\mu \nabla h\) at a point of \(C\) where \(f\) has a maximum or minimum value, where \(\lambda\) and \(\mu\) (the Lagrange multipliers) are real numbers. d. Conclude from part (c) that the equations that must be solved for maximum or minimum values of \(f\) subject to two constraints are \(\nabla f=\lambda \nabla g+\mu \nabla h, g(x, y, z)=0,\) and \(h(x, y, z)=0\)

A function of one variable has the property that a local maximum (or minimum) occurring at the only critical point is also the absolute maximum (or minimum) (for example, \(f(x)=x^{2}\) ). Does the same result hold for a function of two variables? Show that the following functions have the property that they have a single local maximum (or minimum), occurring at the only critical point, but the local maximum (or minimum) is not an absolute maximum (or minimum) on \(\mathbb{R}^{2}\). a. \(f(x, y)=3 x e^{y}-x^{3}-e^{3 y}\) $$ \text { b. } f(x, y)=\left(2 y^{2}-y^{4}\right)\left(e^{x}+\frac{1}{1+x^{2}}\right)-\frac{1}{1+x^{2}} $$ This property has the following interpretation. Suppose a surface has a single local minimum that is not the absolute minimum. Then water can be poured into the basin around the local minimum and the surface never overflows, even though there are points on the surface below the local minimum. (Source: Mathematics Magazine, May \(1985,\) and Calculus and Analytical Geometry, 2 nd ed., Philip Gillett, 1984 )

Find the absolute maximum and minimum values of the following functions over the given regions \(R .\) \(f(x, y)=x^{2}+y^{2}-2 y+1 ; R=\left\\{(x, y): x^{2}+y^{2} \leq 4\right\\}\) (This is Exercise \(47, \text { Section } 15.7 .)\)
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