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Chapter 11: Problem 14

Sketch the surfaces in Exercises \(13-44\) $$z=y^{2}-1$$

Short Answer

Expert verified
The surface is a parabolic cylinder with axis parallel to the \(x\)-axis, shifting the parabola \(z = y^2 - 1\) along \(x\).

Step by step solution

01

Understand the Equation

The given equation is \(z = y^2 - 1\). This is a function in three-dimensional space where \(z\) is expressed in terms of \(y\). The equation suggests for every \(y\), \(z\) is translated vertically by -1. This is a quadratic function in terms of \(y\).
02

Identify the Shape

Since the equation is \(z = y^2 - 1\), it's a parabolic surface where for each constant \(x\), the curve is a parabola opening in the positive \(z\) direction. The minimum point of the parabola is at \(z = -1\) when \(y = 0\). This is a parabola facing upwards shifted down by 1 unit along the \(z\)-axis.
03

Describe the Sketch

Visualize the surface: Imagine a two-dimensional parabola, \(z = y^2 - 1\), with its vertex at \( (0,0,-1) \), and now extend this two-dimensional shape infinitely along the \(x\)-axis. This forms a parabolic cylinder with its axis parallel to the \(x\)-axis.
04

Draw the Sketch

To sketch the graph:- Draw the \(y\)-\(z\) plane, focusing on \(z = y^2 - 1\), a parabola with its vertex at \((0,-1)\).- Extend this parabola infinitely along the \(x\)-axis, creating a three-dimensional parabolic cylinder.- Ensure to show the symmetry about the \(z\)-axis.

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Key Concepts

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

Three-Dimensional Space
In math, we often deal with two-dimensional graphs, like lines or circles. However, in three-dimensional space, or 3D space, things get a bit more complex but also way more interesting.
Three-dimensional space lets us graph functions like our equation, \(z = y^2 - 1\), by introducing an extra axis. Think of 3D space as a real-world scenario, like height, length, and width.
  • In this case, we have three axes: \(x\), \(y\), and \(z\).
  • The \(x\)-axis generally runs left to right.
  • The \(y\)-axis runs back and forth.
  • The \(z\)-axis goes up and down.
In our problem, the equation \(z = y^2 - 1\) fits into this space by forming a shape extending along the \(x\)-axis, affecting how the graph looks in a 3D world. As you move in the \(x\)-direction, the graph doesn't change in terms of \(z\) and \(y\), which helps create the parabolic cylinder shape.
Quadratic Functions
Quadratic functions are a vital part of mathematics, often forming parabolas. A typical quadratic function has the form \(y = ax^2 + bx + c\).
In our exercise, the quadratic part is \(z = y^2 - 1\), which can be understood as:
  • \(y^2\) causing the curve to open upwards since it's positive.
  • The \(-1\) shifts the entire curve downward by one unit.
This type of graph looks like a bowl or a U-shaped curve.
The interesting twist in our problem is that this curvature doesn't relate to \(x\) at all, making it constant; hence, for each \(x\)-value, \(z\) is determined solely by \(y\).
This simplifies its plotting in three-dimensional space, focusing on translating the 2D quadratic idea into a 3D graph.
Parabolic Cylinder
The concept of a parabolic cylinder is fascinating because it combines parabolas and cylinders in 3D space.
Visualize a parabolic cylinder by starting with a simple 2D parabola, like \(z = y^2 - 1\), on the \(y\)-\(z\) plane. Now, imagine stretching this parabola along the \(x\)-axis:
  • This creates a shape that never bends or curves in the \(x\)-direction.
  • For any fixed \(x\), you'll find the same parabola.
  • Its main axis is parallel to the \(x\)-axis, which is constant and why it's called a *cylinder*.
So, unlike a circular cylinder that has a round profile, a parabolic cylinder has a parabola's profile when viewed along the \(y\)-\(z\) plane.
This shape happens to model certain physical phenomena and can appear in architecture or physics, illustrating how math can bridge into real-life applications.

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Most popular questions from this chapter

Let \(A B C D\) be a general, not necessarily planar, quadrilateral in space. Show that the two segments joining the midpoints of opposite sides of \(A B C D\) bisect each other. (Hint: Show that the segments have the same midpoint.)

Find a formula for the distance from the point \(P(x, y, z)\) to the a. \(x\) -axis. b. \(y\) -axis. c. \(z\) -axis.

Sketch the surfaces in Exercises \(13-44\) $$x^{2}-y^{2}=z$$

Sketch the surfaces in Exercises \(13-44\) $$y^{2}-x^{2}=z$$

Find the acute angles between the planes in Exercises \(49-52\) to the nearest hundredth of a radian. $$2 x+2 y+2 z=3, \quad 2 x-2 y-z=5$$
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