Consider the equation 16y – 4² + 2² = 16. By first describing all of its traces, sketch the surface in R’.
Consider the equation 16y – 4² + 2² = 16. By first describing all of its traces, sketch the surface in R’.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I had asked this question before and I got computer generated graph which was no help at all. I am looking for a detailed descriptin of all traces and a final sketch of the surface. Is this a one sheet hyperbola or two? Compter graph shows it as two sheet hyperbola which does not make sense to me. Please help.
![**Problem 4:**
Consider the equation \(16y^2 - 4x^2 + z^2 = 16\). By first describing all of its traces, sketch the surface in \(\mathbb{R}^3\).
To analyze this equation, we need to explore its traces, which are obtained by setting each of the variables \(x\), \(y\), and \(z\) to constants and examining the resulting two-dimensional curves. Let’s break this down:
1. **\(xy\)-plane (Set \(z = 0\)):**
The equation simplifies to \(16y^2 - 4x^2 = 16\), which can be rewritten as:
\[
\frac{y^2}{1} - \frac{x^2}{4} = 1
\]
This is the equation of a hyperbola centered at the origin with vertices along the \(y\)-axis.
2. **\(xz\)-plane (Set \(y = 0\)):**
The equation becomes \(-4x^2 + z^2 = 16\), or:
\[
\frac{z^2}{16} - \frac{x^2}{4} = 1
\]
This is another hyperbola, where the vertices lie along the \(z\)-axis.
3. **\(yz\)-plane (Set \(x = 0\)):**
Simplifying gives \(16y^2 + z^2 = 16\), which is:
\[
\frac{y^2}{1} + \frac{z^2}{16} = 1
\]
This is an ellipse centered at the origin.
**Surface in \(\mathbb{R}^3\):**
Given the traces, this surface is a hyperboloid of one sheet. The behavior of these slices shows that the surface extends infinitely and is symmetric with respect to the origin. The axes of symmetry correspond to the \(x\), \(y\), and \(z\) axes.
To graph this, visualize a hyperboloid that narrows towards the \(z\)-axis, with elliptical cross-sections parallel to the \(yz\)-plane and hyperbolic cross-sections parallel to the \(xy\) and \(xz\) planes.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd93ae1f0-5948-4075-83b0-747f4fbdf146%2F62532576-2372-4d27-9e1a-7e767e122a8a%2Frfv5zkf_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 4:**
Consider the equation \(16y^2 - 4x^2 + z^2 = 16\). By first describing all of its traces, sketch the surface in \(\mathbb{R}^3\).
To analyze this equation, we need to explore its traces, which are obtained by setting each of the variables \(x\), \(y\), and \(z\) to constants and examining the resulting two-dimensional curves. Let’s break this down:
1. **\(xy\)-plane (Set \(z = 0\)):**
The equation simplifies to \(16y^2 - 4x^2 = 16\), which can be rewritten as:
\[
\frac{y^2}{1} - \frac{x^2}{4} = 1
\]
This is the equation of a hyperbola centered at the origin with vertices along the \(y\)-axis.
2. **\(xz\)-plane (Set \(y = 0\)):**
The equation becomes \(-4x^2 + z^2 = 16\), or:
\[
\frac{z^2}{16} - \frac{x^2}{4} = 1
\]
This is another hyperbola, where the vertices lie along the \(z\)-axis.
3. **\(yz\)-plane (Set \(x = 0\)):**
Simplifying gives \(16y^2 + z^2 = 16\), which is:
\[
\frac{y^2}{1} + \frac{z^2}{16} = 1
\]
This is an ellipse centered at the origin.
**Surface in \(\mathbb{R}^3\):**
Given the traces, this surface is a hyperboloid of one sheet. The behavior of these slices shows that the surface extends infinitely and is symmetric with respect to the origin. The axes of symmetry correspond to the \(x\), \(y\), and \(z\) axes.
To graph this, visualize a hyperboloid that narrows towards the \(z\)-axis, with elliptical cross-sections parallel to the \(yz\)-plane and hyperbolic cross-sections parallel to the \(xy\) and \(xz\) planes.
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