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Chapter 15: Problem 50

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

Short Answer

Expert verified
Answer: The function is continuous at all points inside or on the circle with radius 2 centered at the origin: \(\{(x, y) \ | \ x^2 + y^2 \leq 4\}\).

Step by step solution

01

Identify the function

The function given is $$f(x, y)=\sqrt{4-x^{2}-y^{2}}.$$
02

Determine the domain of the function

In order for the function to be continuous, the expression inside the square root must be non-negative. That is, $$4-x^{2}-y^{2} \geq 0.$$ Solving the inequality, we obtain the following inequality: $$x^2 + y^2 \leq 4.$$ This is the equation of a circle with radius 2 and centered at the origin. Thus, the domain of the function consists of all points (x, y) inside or on the circle, including the circle's boundary: $$\{(x, y) \ | \ x^2 + y^2 \leq 4\}.$$
03

Check the continuity of the function in its domain

Since the square root function is continuous on its domain (i.e., for non-negative values of its argument), the only concern left is the continuity at the points on the boundary of the circle. On the boundary, the expression inside the square root becomes zero (\(4 - x^2 - y^2 = 0\)), and the square root of zero is also zero. So, the function is continuous at the boundary points as well.
04

Conclusion

The function $$f(x, y)=\sqrt{4-x^{2}-y^{2}}$$ is continuous at all points in its domain, which is all points inside or on the circle with radius 2 centered at the origin: $$\{(x, y) \ | \ x^2 + y^2 \leq 4\}.$$

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Key Concepts

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

Domain of a function
Understanding the domain of a function is crucial because it defines all the possible inputs for which the function is defined. In mathematical terms, the domain of a function is the set of all possible values that 'x' can take without causing any mathematical errors such as division by zero or taking the square root of a negative number.
  • For the function \(f(x, y)=\sqrt{4-x^2-y^2}\), the expression inside the square root \(4-x^2-y^2\) must be greater than or equal to zero.
  • This requirement ensures that the values under the square root are non-negative, making the function valuable in real numbers.
  • Solving the inequality \(4-x^2-y^2 \geq 0\) leads to \(x^2+y^2 \leq 4\), defining the domain as all points within or on the circle of radius 2 centered at the origin.
Thus, the domain is a critical concept to ensure a function behaves well within its defined range.
Square root function
The square root function, often denoted as \(\sqrt{x}\), is defined only for non-negative values of 'x'. This is because the square root of a negative number is not defined in the set of real numbers.
  • The output of a square root function for a positive or zero input is always a real number.
  • When considering a function like \(f(x, y) = \sqrt{4-x^2-y^2}\), the square root function exhibits continuity for all values where its argument is non-negative.
  • At the boundary condition, where \(4-x^2-y^2 = 0\), the square root returns zero, thus maintaining continuity.
Overall, the square root function's behavior is key in ensuring the continuity of more complex functions where it forms a part.
Circle in coordinate geometry
In coordinate geometry, a circle is defined by its center and radius. The standard form of a circle's equation is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) represents the center, and \(r\) is the radius.
  • For the function \(f(x, y) = \sqrt{4-x^2-y^2}\), the expression \(x^2 + y^2 \leq 4\) describes a circle centered at the origin \((0, 0)\) with a radius of 2.
  • The inequality \(x^2 + y^2 \leq 4\) signifies that every point inside or on the circle is part of the domain.
  • This geometric view is helpful in visualizing the area where the function is defined and continuous.
Understanding the circle in terms of its equation helps in identifying key features of the function's domain, especially in problems involving spatial contexts such as this one.

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Most popular questions from this chapter

Suppose the elevation of Earth's surface over a 16 -mi by 16 -mi region is approximated by the function \(z=10 e^{-\left(x^{2}+y^{2}\right)}+5 e^{-\left((x+5)^{2}+(y-3)^{2}\right) / 10}+4 e^{-2\left((x-4)^{2}+(y+1)^{2}\right)}\) a. Graph the height function using the window \([-8,8] \times[-8,8] \times[0,15]\) b. Approximate the points \((x, y)\) where the peaks in the landscape appear. c. What are the approximate elevations of the peaks?

Use what you learned about surfaces in Sections 13.5 and 13.6 to sketch a graph of the following functions. In each case, identify the surface and state the domain and range of the function. $$G(x, y)=-\sqrt{1+x^{2}+y^{2}}$$

A snapshot of a water wave moving toward shore is described by the function \(z=10 \sin (2 x-3 y),\) where \(z\) is the height of the water surface above (or below) the \(x y\) -plane, which is the level of undisturbed water. a. Graph the height function using the window $$[-5,5] \times[-5,5] \times[-15,15]$$ b. For what values of \(x\) and \(y\) is \(z\) defined? c. What are the maximum and minimum values of the water height? d. Give a vector in the \(x y\) -plane that is orthogonal to the level curves of the crests and troughs of the wave (which is parallel to the direction of wave propagation).

Potential functions Potential functions arise frequently in physics and engineering. A potential function has the property that a field of interest (for example, an electric field, a gravitational field, or a velocity field) is the gradient of the potential (or sometimes the negative of the gradient of the potential). (Potential functions are considered in depth in Chapter \(17 .)\) The gravitational potential associated with two objects of mass \(M\) and \(m\) is \(\varphi=-G M m / r,\) where \(G\) is the gravitational constant. If one of the objects is at the origin and the other object is at \(P(x, y, z),\) then \(r^{2}=x^{2}+y^{2}+z^{2}\) is the square of the distance between the objects. The gravitational field at \(P\) is given by \(\mathbf{F}=-\nabla \varphi,\) where \(\nabla \varphi\) is the gradient in three dimensions. Show that the force has a magnitude \(|\mathbf{F}|=G M m / r^{2}\) Explain why this relationship is called an inverse square law.

Use Lagrange multipliers to find these values. \(f(x, y, z)=(x y z)^{1 / 2}\) subject to \(x+y+z=1\) with \(x \geq 0\) \(y \geq 0, z \geq 0\)
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