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Chapter 2: Problem 304

For the following exercises, sketch and describe the cylindrical surface of the given equation. $$ x^{2}+y^{2}=9 $$

Short Answer

Expert verified
The equation \( x^2 + y^2 = 9 \) describes a cylinder with a radius of 3, centered on the \( z \)-axis and extending infinitely along it.

Step by step solution

01

Identify the Type of Equation

The given equation is \( x^2 + y^2 = 9 \). This is the equation of a circle in a two-dimensional plane with its center at the origin \((0, 0)\) and a radius of 3. This is because the general equation for a circle is \( x^2 + y^2 = r^2 \), where \( r \) is the radius. In this case, \( r^2 = 9 \), so \( r = 3 \).
02

Recognize the Cylindrical Nature

Since the equation involves only the variables \( x \) and \( y \), and not \( z \), the cylinder extends infinitely along the \( z \)-axis. This is a classic definition of a cylindrical surface where the cross-section is a circle and the height (or depth in the case of \( z \)) is infinite.
03

Sketch the Circle and Form the Cylinder

To sketch the cylindrical surface, begin by drawing the circle \( x^2 + y^2 = 9 \) in the \( xy \)-plane. This circle has a radius of 3, centered at the origin. Now, extend this circle infinitely in both the positive and negative \( z \)-directions to form a cylindrical surface.
04

Describe the Cylindrical Surface

The cylindrical surface is circular in the \( xy \)-plane with a radius of 3, centered at the origin. The surface is open and extends infinitely in both directions along the \( z \)-axis. It forms a 'tube' shape with the circular cross-section as its base.

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Key Concepts

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

Equation of a Circle
The equation of a circle is fundamental in geometry and can be expressed as \( x^2 + y^2 = r^2 \). Here, **\( x \)** and **\( y \)** represent the coordinates on a plane, and **\( r \)** is the radius of the circle. In the given equation \( x^2 + y^2 = 9 \), the circle is
  • centered at the origin, \((0, 0)\)
  • has a radius of 3 because \( r^2 = 9 \) implies \( r = 3 \).
This circle exists solely within a two-dimensional plane consisting of the x and y axes. An understanding of this equation helps with visualizing the shape and size of the circle before extending it into three dimensions.
3D Geometry
Moving from a 2D to 3D perspective opens up a whole new understanding of shapes and forms. In 3D geometry, shapes like circles become more complex objects such as spheres or cylinders. The key to understanding 3D shapes like a cylinder comes from visualizing them as extensions of their 2D counterparts. For our circle,
  • it lies on the x-y plane
  • it is uninfluenced by the z dimension initially.
When we consider the z-axis, we add depth, making it crucial to understand how these 2D shapes project into 3D space. In 3D, the equation \( x^2 + y^2 = 9 \) describes a set of circles stacked along the z-axis, without being affected by the z-coordinates, showcasing the beauty of symmetry in spatial figures.
Cylinder Visualization
Visualizing a cylinder in 3D helps make the abstract concept of a cylindrical surface more concrete. The cylindrical surface described by our equation \( x^2 + y^2 = 9 \) can be visualized as follows:
  • Begin with a circle of radius 3 on the xy-plane.
  • This circle acts as a cross-section of the cylinder.
  • Extend this circle along the z-axis in both directions.
The result is an infinite cylinder, stretching through space, and forming a tube-like structure. The beauty of this cylinder lies in its infinite nature along the z-axis, creating an endless surface that wraps around an invisible core, perfectly symmetrical along its length.

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Most popular questions from this chapter

For the following exercises, point \(P\) and vector \(\mathbf{n}\) are given. a. Find the scalar equation of the plane that passes through \(P\) and has normal vector \(\mathbf{n} .\) b. Find the general form of the equation of the plane that passes through \(P\) and has normal vector \(\mathbf{n} .\) \(P(3,2,2), \quad \mathbf{n}=2 \mathbf{i}+3 \mathbf{j}-\mathbf{k}\)

For the following exercises, points \(P, Q,\) and \(R\) are given. a. Find the general equation of the plane passing through \(P, Q,\) and \(R .\) b. Write the vector equation \(\mathbf{n} \cdot \overrightarrow{P S}=0\) of the plane at a, where \(S(x, y, z)\) is an arbitrary point of the plane. c. Find parametric equations of the line passing through the origin that is perpendicular to the plane passing through \(P, Q,\) and \(R\) \(P(-2,1,4), Q(3,1,3), and R(-2,1,0)\)

\([T]\) The force vector \(F\) acting on a proton with anelectric charge of \(1.6 \times 10^{-19} \mathrm{C}\) moving in a magnetic field \(B\) where the velocity vector \(V\) is given by\(\mathbf{F}=1.6 \times 10^{-19}(\mathbf{v} \times \mathbf{B})\) (here, v is expressed in meters per second, \(B\) in \(T\), and \(F\) in \(N\)). If the magnitude of force \(F\) acting on a proton is\(5.9 \times 10^{-17} \mathrm{~N}\) and the proton is moving at the speed of 300 m/sec in magnetic field \(B\) of magnitude \(2.4 T\), find the angle between velocity vector \(V\) of the proton and magnetic field \(B\). Express the answer in degrees rounded to the nearest integer.

Find the real number \(\alpha\) such that the line of parametric equations \(\quad x=t, y=2-t, z=3+t\) , \(t \in \mathbb{R} \quad\) is parallel to the plane of equation \(a x+5 y+z-10=0\)

Consider point \(C(-3,2,4)\) and the plane of equation \(2 x+4 y-3 z=8\) . a. Find the radius of the sphere with center \(C\) tangent to the given plane. b. Find point \(P\) of tangency.
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