4. Consider the equation 16y" - 4 + 2² = 16. By first deseribing all of its traces, sketch the surface in R'.

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Chapter2: Second-order Linear Odes
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**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 tackle this problem, you'll need to analyze the traces of the given equation in three-dimensional space. The equation defines a surface which we can explore by understanding its cross-sections (traces) in the planes \(xy\), \(xz\), and \(yz\).

Here's how to break it down:

1. **Trace in the \(xy\)-plane:**
   - Set \(z = 0\) in the equation: \(16y^2 - 4x^2 = 16\).
   - This can be rewritten as a hyperbola: \(\frac{y^2}{1} - \frac{x^2}{4} = 1\).

2. **Trace in the \(xz\)-plane:**
   - Set \(y = 0\) in the equation: \(-4x^2 + z^2 = 16\).
   - This can be rewritten as another hyperbola: \(-\frac{x^2}{4} + \frac{z^2}{16} = 1\).

3. **Trace in the \(yz\)-plane:**
   - Set \(x = 0\) in the equation: \(16y^2 + z^2 = 16\).
   - This can be rewritten as an ellipse: \(\frac{y^2}{1} + \frac{z^2}{16} = 1\).

**Surface Description:**
The surface described by this equation is a hyperboloid of one sheet. It stems from the fact there are hyperbolic traces in two planes and an elliptic trace in the other plane.

By plotting these traces and understanding their intersections, you can sketch the hyperboloid in \(\mathbb{R}^3\), showing how these geometric shapes fit together in three dimensions.
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 tackle this problem, you'll need to analyze the traces of the given equation in three-dimensional space. The equation defines a surface which we can explore by understanding its cross-sections (traces) in the planes \(xy\), \(xz\), and \(yz\). Here's how to break it down: 1. **Trace in the \(xy\)-plane:** - Set \(z = 0\) in the equation: \(16y^2 - 4x^2 = 16\). - This can be rewritten as a hyperbola: \(\frac{y^2}{1} - \frac{x^2}{4} = 1\). 2. **Trace in the \(xz\)-plane:** - Set \(y = 0\) in the equation: \(-4x^2 + z^2 = 16\). - This can be rewritten as another hyperbola: \(-\frac{x^2}{4} + \frac{z^2}{16} = 1\). 3. **Trace in the \(yz\)-plane:** - Set \(x = 0\) in the equation: \(16y^2 + z^2 = 16\). - This can be rewritten as an ellipse: \(\frac{y^2}{1} + \frac{z^2}{16} = 1\). **Surface Description:** The surface described by this equation is a hyperboloid of one sheet. It stems from the fact there are hyperbolic traces in two planes and an elliptic trace in the other plane. By plotting these traces and understanding their intersections, you can sketch the hyperboloid in \(\mathbb{R}^3\), showing how these geometric shapes fit together in three dimensions.
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