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Chapter 10: Problem 6

Describe and sketch the surface. \(y=z^{2}\)

Short Answer

Expert verified
The surface is a parabolic cylinder extending along the x-axis.

Step by step solution

01

Identify the Equation Type

The given equation is \(y = z^2\). This is a quadratic equation in the yz-plane. It represents a parabola opening in the positive y-direction because it can be written in the form \(y = ax^2 + bx + c\) where \(x\) is replaced by \(z\), and \(a = 1, b = 0, c = 0\).
02

Recognize the Symmetry and Direction

The equation \(y = z^2\) is symmetrical about the z-axis. This means that for each positive and negative value of \(z\), \(y\) is the same, forming a U-shape along the yz-plane. This symmetry indicates that the 3D surface created by revolving this shape will also be symmetrical about the z-axis, creating a parabolic cylinder.
03

Visualize the Surface in Three Dimensions

In three dimensions, the equation \(y = z^2\) describes a surface where the cross-section in the yz-plane is a parabola. Since there is no restriction on \(x\), the parabola extends along the x-axis, forming a cylindrical shape. This results in a parabolic cylinder aligned parallel to the x-axis.
04

Sketch the Parabolic Cylinder

To sketch the surface described by \(y = z^2\), draw the parabola in the yz-plane, with the vertex at the origin (0, 0), curving upwards. Extend this parabolic shape infinitely along the x-axis in both directions, maintaining the same U-shaped cross-section at every x-coordinate. This forms a parabolic cylinder along the x-axis.

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Key Concepts

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

Quadratic Equation in 3D Geometry
A quadratic equation typically takes the form of \( y = ax^2 + bx + c \). In three-dimensional geometry, such equations are not limited to simple 2D curves. The equation \( y = z^2 \) is a specific type of quadratic equation, representing a parabola. In this case, \( a = 1 \), \( b = 0 \), and \( c = 0 \). It opens upwards, indicating that as \( z \) becomes larger or smaller, \( y \) correspondingly increases.
  • The parabola is centered at the origin (0, 0) in the yz-plane, making it a simple U-shape.
  • Given its quadratic nature, this equation generates a continuous and smooth curve.
  • One key characteristic of quadratic equations in 3D is their versatility in shaping various geometric surfaces.
For the equation \( y = z^2 \), the lack of terms involving \( x \) implies the figure can extend infinitely along the x-axis, ultimately forming a cylindrical surface in three dimensions.
Symmetry and Its Role in Parabolic Surfaces
Symmetry is a fundamental characteristic of many geometric forms. In the equation \( y = z^2 \), symmetry about the z-axis is evident. This means that the parabola created is perfectly mirrored on either side of the z-axis. Each positive value of \( z \) has a corresponding negative value, maintaining equal values of \( y \).
Such symmetry results in significant implications for the 3D surface.
  • When considering the whole surface in three dimensions, this symmetry means the geometric formation will appear identical from multiple perspectives perpendicular to the x-axis.
  • This bilateral symmetry provides easy visualization and consistency in mathematical representation.
  • Understanding this symmetrical nature aids in sketching and conceptualizing 3D shapes efficiently.
By examining symmetry, one can predict the outcome of rotating or extending basic parabolic patterns in space.
Visualizing the Parabolic Cylinder in Three Dimensions
In three-dimensional space, a parabolic equation like \( y = z^2 \) translates into a parabolic cylinder. Though 'cylinder' might evoke thoughts of smooth, rounded tubes, a parabolic cylinder is fundamentally different.
The surface formed extends the familiar 2D parabolic curve along a new dimension.
  • This shape features a consistent cross-section, which remains a parabola as you "move" along the x-axis.
  • The absence of an \( x \) variable in the equation allows free expansion along this axis, creating the cylindrical appearance.
  • Visualizing this involves considering a parabola standing in the yz-plane, then imagining it running infinitely parallel to the x-axis.
To sketch this surface, start by drawing a single parabola on the yz-plane, with the vertex at (0,0) and opening upwards. Then, replicate this shape consistently along the x-axis, picturing an endless corridor of parabolas side-by-side. This exercise reflects the elegant simplicity yet complexity of geometry transitioning from two to three dimensions.

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Most popular questions from this chapter

If \(\mathbf{r}(t) \neq \mathbf{0},$$ show that \)\frac{d}{d t}|\mathbf{r}(t)|=\frac{1}{|\mathbf{r}(t)|} \mathbf{r}(t) \cdot \mathbf{r}^{\prime}(t)$$

Two particles travel along the space curves $$\mathbf{r}_{1}(t)=\left\langle t, t^{2}, t^{3}\right\rangle \quad \mathbf{r}_{2}(t)=\langle 1+2 t, 1+6 t, 1+14 t\rangle$$ Do the particles collide? Do their paths intersect?

Let \(\mathbf{v}=5 \mathbf{j}\) and let \(\mathbf{u}\) be a vector with length 3 that starts at the origin and rotates in the \(x y\) -plane. Find the maximum and minimum values of the length of the vector \(\mathbf{u} \times \mathbf{v} .\) In what direction does \(\mathbf{u} \times \mathbf{v}\) point?

When \(a \ne 0\), there are two solutions to \\(ax^2 + bx + c = 0\\) and they are $$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$ $$ \int_{0}^{\pi / 2}\left(3 \sin ^{2} t \cos t \mathbf{i}+3 \sin t \cos ^{2} t \mathbf{j}+2 \sin t \cos t \mathbf{k}\right) d t $$

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