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Use what you learned about surfaces in Section 12.1 to sketch a graph of the following functions. In each case identify the surface and state the domain and range of the function. $$G(x, y)=-\sqrt{1+x^{2}+y^{2}}$$

Short Answer

Expert verified
Short Answer: The surface of the given function, $$G(x, y)=-\sqrt{1+x^{2}+y^{2}}$$, is a downward-opening paraboloid centered at the origin (0,0,0). The domain of this function is all real values of x and y, represented as $$Domain(G) = \{ (x, y) | x, y \in \mathbb{R} \}$$. The range consists of all values between minus infinity and zero, expressed as $$Range(G)= \{ G(x, y)| G(x, y)\in(-\infty,0] \}$$.

Step by step solution

01

Identify the surface

Notice that $$G(x, y)$$ involves a square root. The expression inside the square root is $$1+x^2+y^2$$, and it is greater than or equal to 1. So after taking the square root, the function will have negative values, and they will be larger when x and y are small. This indicates that the surface described by the function is opening downwards.
02

Find the domain

The domain of the function is the set of all possible values of x and y for which the function G(x, y) is defined. Since the square root function is defined for non-negative values, we have to consider the following: $$1+x^2+y^2 \geq 0$$ This inequality holds for all real values of x and y, meaning the domain of G(x, y) is: $$Domain(G) = \{ (x, y) | x, y \in \mathbb{R} \}$$
03

Find the range

The range of the function is the set of all possible values of G(x, y). Since G(x, y) is always negative and the value of the square root approaches zero as the expression inside the square root becomes larger, we can deduce that the range of G(x, y) is: $$Range(G)= \{ G(x, y)| G(x, y)\in(-\infty,0] \}$$
04

Sketch the graph

To help sketch the graph of G(x,y), think of some key features. At every point on the x-y plane where G(x,y)=0, the expression inside the square root is exactly 1. This means that the surface intersects the x-y plane when $$x^2+y^2=0$$, which is a single point, the origin (0,0). Additionally, the surface opens downwards since G(x,y) is always negative. Taking all these observations into account, we can visualize the surface as a downward-opening paraboloid, centered at the origin. From the above analysis, we have identified the surface, domain, and range. The surface is a downward-opening paraboloid centered at the origin (0,0,0), domain is all real values of x and y, and the range consists of all values between minus infinity and zero.

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Key Concepts

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

Surfaces
In multivariable calculus, surfaces are two-dimensional shapes that reside in a three-dimensional space. A function of two variables, such as \( G(x, y) \), defines a surface by assigning a single z-value to every point on the xy-plane. In this exercise, the function \( G(x, y) = -\sqrt{1 + x^2 + y^2} \) describes a surface. Here, the presence of the square root indicates a relationship between x, y, and the z-value. Understanding what a surface is, paves the way for tackling problems involving functions of more than one variable. Surfaces can be complex, but they serve as a fundamental stepping-stone in understanding multidimensional geometry.
Domain and Range
The domain and range are vital concepts in understanding functions, especially in multiple dimensions. For a function \( G(x, y) = -\sqrt{1 + x^2 + y^2} \), the domain refers to all possible pairs of \((x, y)\) values that we can input into the function. Here, the domain is all real values of x and y, because the expression \(1 + x^2 + y^2 \geq 0\) is satisfied for all real numbers. This occurs as any real number, when squared, is non-negative. On the other hand, the range of the function is the set of output values, \( G(x, y) \), that result from plugging in all numbers from the domain. Since the function involves a negative square root, all output values are negative or zero, hence the range is \((-\infty, 0] \). This indicates that at no point is \( G(x, y) \) positive.
Sketching Graphs
Sketching graphs in multivariable calculus involves visualizing the surface defined by a function. For the function \( G(x, y) = -\sqrt{1 + x^2 + y^2} \), the graph depicts a three-dimensional surface. To sketch it, we need to understand key features. At the origin, \((0,0)\), the graph has its peak because \( G(x, y) = 0 \) there. This is where \( x^2 + y^2 = 0 \), matching the condition when the expression inside the square root equals one. After determining critical points, we note that because \( G(x, y) \) is a quadratic form inside the square root, the surface opens downwards as x and y increase. The resulting sketch would illustrate a sombrero-shaped surface tilted downwards, centered at the origin in three-dimensional space.
Paraboloid
The graph of \( G(x, y) = -\sqrt{1 + x^2 + y^2} \) represents a specific type of surface called a paraboloid. In this case, it's a downward-opening paraboloid. Paraboloids can be thought of as three-dimensional parabolas, resembling a bowl or dish shape either pointing upwards or downwards. This function forms a downward paraboloid because all values of \( G(x, y) \) are non-positive. The center of this paraboloid is at the origin \((0, 0, 0)\), where it achieves its maximum possible value, which is zero. As x and y move away from the center, the value of \( G(x, y) \) decreases, creating the downward shape. Understanding paraboloids is crucial for visualizing and solving problems in multivariable calculus, as they frequently appear in contexts involving quadratic surfaces.

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Most popular questions from this chapter

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 14 .) 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.

Given positive numbers \(x_{1}, \ldots, x_{n},\) prove that the geometric mean \(\left(x_{1} x_{2} \cdots x_{n}\right)^{1 / n}\) is no greater than the arithmetic mean \(\left(x_{1}+\cdots+x_{n}\right) / n\) in the following cases. a. Find the maximum value of \(x y z,\) subject to \(x+y+z=k\) where \(k\) is a real number and \(x>0, y>0\), and \(z>0 .\) Use the result to prove that $$(x y z)^{1 / 3} \leq \frac{x+y+z}{3}.$$ b. Generalize part (a) and show that $$\left(x_{1} x_{2} \cdots x_{n}\right)^{1 / n} \leq \frac{x_{1}+\cdots+x_{n}}{n}.$$

Show that the plane \(a x+b y+c z=d\) and the line \(\mathbf{r}(t)=\mathbf{r}_{0}+\mathbf{v} t,\) not in the plane, have no points of intersection if and only if \(\mathbf{v} \cdot\langle a, b, c\rangle=0 .\) Give a geometric explanation of this result.

Two resistors in an electrical circuit with resistance \(R_{1}\) and \(R_{2}\) wired in parallel with a constant voltage give an effective resistance of \(R,\) where \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\). a. Find \(\frac{\partial R}{\partial R_{1}}\) and \(\frac{\partial R}{\partial R_{2}}\) by solving for \(R\) and differentiating. b. Find \(\frac{\partial R}{\partial R_{1}}\) and \(\frac{\partial R}{\partial R_{2}}\) by differentiating implicitly. c. Describe how an increase in \(R_{1}\) with \(R_{2}\) constant affects \(R\). d. Describe how a decrease in \(R_{2}\) with \(R_{1}\) constant affects \(R\).

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