/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 Eliminate the parameters to obta... [FREE SOLUTION] | 91Ó°ÊÓ

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Eliminate the parameters to obtain an equation in rectangular coordinates, and describe the surface. $$ \begin{array}{l}{x=2 u+v, y=u-v, z=3 v \text { for }-\infty

Short Answer

Expert verified
The surface is the plane \( z = x - y \).

Step by step solution

01

Understanding the Problem

We are given parametric equations with two parameters: \( x = 2u + v \), \( y = u - v \), and \( z = 3v \). Our goal is to eliminate the parameters \( u \) and \( v \) to find an equation for the surface in terms of \( x \), \( y \), and \( z \).
02

Express \( u \) and \( v \) in Terms of \( x \) and \( y \)

We solve the equations \( x = 2u + v \) and \( y = u - v \) for \( u \) and \( v \). From these equations, we can create the system: \( 2u + v = x \) and \( u - v = y \).
03

Solving for \( u \)

Add the equations \( 2u + v = x \) and \( u - v = y \) to eliminate \( v \):\[ (2u + v) + (u - v) = x + y \]This simplifies to \( 3u = x + y \), giving \( u = \frac{x + y}{3} \).
04

Solving for \( v \)

Subtract \( u - v = y \) from \( 2u + v = x \) to eliminate \( u \):\[ (2u + v) - (u - v) = x - y \]This simplifies to \( 3v = x - y \), giving \( v = \frac{x - y}{3} \).
05

Substitute \( u \) and \( v \) into \( z \)

We now have \( u = \frac{x + y}{3} \) and \( v = \frac{x - y}{3} \). Substitute \( v \) into the equation for \( z \):\[ z = 3v = 3\left(\frac{x - y}{3}\right) = x - y \].
06

Final Equation in Rectangular Coordinates

After substituting, we obtain the equation \( z = x - y \). This equation represents the surface in rectangular coordinates.
07

Describing the Surface

The equation \( z = x - y \) represents a plane in three-dimensional space. This plane is parallel to the line \( x = y \) in the \( xy \)-plane and has a slope of 1 with respect to both the \( x \)-axis and the \( y \)-axis in the \( z \)-direction.

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Key Concepts

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

Parametric Equations
Parametric equations are a powerful tool in mathematics, often used to describe mathematical objects like curves and surfaces. They work by expressing the coordinates of each point using one or more parameters instead of standard variables like \( x \), \( y \), and sometimes \( z \). By using parameters, we can describe complex forms and motions that are otherwise difficult to model directly in rectangular coordinates. In this case, we have parametric equations defined by:
  • \( x = 2u + v \)
  • \( y = u - v \)
  • \( z = 3v \)
Here, \( u \) and \( v \) are parameters. Parameters offer flexibility, allowing for the representation of a wider range of shapes. To convert parametric equations into rectangular form, parameters are eliminated by expressing them solely in terms of rectangular coordinates. In our exercise, this transformation is achieved through algebraic manipulation and results in the equation \( z = x - y \).
This process demonstrates how parametric equations can be rearranged to give insight into the nature of the object they describe.
Three-Dimensional Space
Three-dimensional space is the mathematical setting we often use to model the world around us. This space is defined by three perpendicular axes: the \( x \)-axis, \( y \)-axis, and \( z \)-axis. Together, they form a three-dimensional coordinate system, where every point can be described by a trio of numbers \((x, y, z)\). In three-dimensional space, objects such as volumes, surfaces, and planes are represented, each with unique properties.
The given parametric equations in our exercise describe a surface in this three-dimensional environment. Specifically, we use the parameters \( u \) and \( v \) to generate points \( (x, y, z) \) that belong to a surface. By eliminating the parameters, we transform the description of the surface into its direct form \( z = x - y \), a plane in three-dimensional space.
This plane, a flat two-dimensional object extending infinitely, showcases how transformations from parametric equations to rectangular coordinates reveal the essential nature of geometrical surfaces.
Coordinate System
A coordinate system is a framework used in geometry and physics to uniquely determine the position of points or other geometric elements. The most common systems in use are the Cartesian coordinate system for flat surfaces and the three-dimensional Cartesian coordinate system for volumetric spaces.
  • The Cartesian system relies on perpendicular axes intersecting at the origin \((0,0)\) in two dimensions, expanding to \((0,0,0)\) in three dimensions.
  • Each point in this space is identified by its distance along each axis, described by coordinates like \( (x, y) \) for two-dimensional space, or \( (x, y, z) \) for three-dimensional space.
The exercise engages a three-dimensional coordinate system where the formula \( z = x - y \) describes a plane. This shows how points along this plane have their position determined by the subtraction of their respective \( y \) and \( x \) coordinates in a rectangular format.
Coordinate systems allow for precise location and analysis of points and shapes, crucial in fields ranging from navigation to computer graphics.

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Most popular questions from this chapter

The tendency of a lamina to resist a change in rotational motion about an axis is measured by its moment of inertia about that axis. If a lamina occupies a region \(R\) of the \(x y\) -plane, and if its density function \(\delta(x, y)\) is continuous on \(R,\) then the moments of inertia about the \(x\) -axis, the \(y\) -axis, and the \(z\) -axis are denoted by \(I_{x}, I_{y},\) and \(I_{z},\) respectively, and are defined by $$\begin{aligned} I_{x} &=\iint_{R} y^{2} \delta(x, y) d A, \quad I_{y}=\iint_{R} x^{2} \delta(x, y) d A \\ I_{z} &=\iint_{R}\left(x^{2}+y^{2}\right) \delta(x, y) d A \end{aligned}$$ Use these definitions in Exercises 45 and 46 Consider the circular lamina that occupies the region described by the inequalities \(0 \leq x^{2}+y^{2} \leq a^{2} .\) Assuming that the lamina has constant density \(\delta,\) show that $$ I_{x}=I_{y}=\frac{\delta \pi a^{4}}{4}, \quad I_{z}=\frac{\delta \pi a^{4}}{2} $$

Let \(R\) be the rectangle bounded by the lines \(x=0, x=3,\) \(y=0,\) and \(y=2 .\) By inspection, find the centroid of \(R\) and use it to evaluate $$ \iint_{R} x d A \text { and } \iint_{R} y d A $$

The tendency of a solid to resist a change in rotational motion about an axis is measured by its moment of inertia about that axis. If the solid occupies a region \(G\) in an \(x y z\) -coordinate system, and if its density function \(\delta(x, y, z)\) is continuous on \(G,\) then the moments of inertia about the \(x\) -axis, the \(y\) -axis, and the \(z\) -axis are denoted by \(I_{x}, I_{y},\) and \(I_{z},\) respectively, and are defined by $$\begin{aligned} I_{x} &=\iiint_{G}\left(y^{2}+z^{2}\right) \delta(x, y, z) d V \\ I_{y} &=\iiint_{G}\left(x^{2}+z^{2}\right) \delta(x, y, z) d V \\ I_{z} &=\iiint_{G}\left(x^{2}+y^{2}\right) \delta(x, y, z) d V \end{aligned}$$ In these exercises, find the indicated moments of inertia of the solid, assuming that it has constant density \(\delta .\) \(I_{z}\) for the solid sphere \(x^{2}+y^{2}+z^{2} \leq a^{2}\)

Use an appropriate change of variables to find the area of the region in the first quadrant enclosed by the curves \(y=x\) \(y=2 x, x=y^{2}, x=4 y^{2}\).

Express the volume of the solid described as a double integral in polar coordinates. $$ \begin{array}{l}{\text { Below } z=1-x^{2}-y^{2}} \\ {\text { Inside of } x^{2}+y^{2}-x=0} \\ {\text { Above } z=0}\end{array} $$

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