Find a vector function that represents the curve of intersection between the paraboloid 2 z = 4x² + y² and the parabolic cylinder x = y² -1 . Then draw all three in GeoGebra (:
Find a vector function that represents the curve of intersection between the paraboloid 2 z = 4x² + y² and the parabolic cylinder x = y² -1 . Then draw all three in GeoGebra (:
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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![### Finding the Curve of Intersection Between Surfaces
**Problem Statement:**
Find a vector function that represents the curve of intersection between the paraboloid:
\[ z = 4x^2 + y^2 \]
and the parabolic cylinder:
\[ x = y^2 - 1 \]
Then draw all three in GeoGebra (paraboloid, parabolic cylinder, and their intersection curve).
**Solution:**
1. **Identifying the Intersection Curve:**
To find the curve of intersection, we need to express both surfaces in one set of parametric equations.
2. **Substitute \( x \) from the Parabolic Cylinder into the Paraboloid Equation:**
From the parabolic cylinder equation:
\[ x = y^2 - 1 \]
Now, substitute \( x \) in the paraboloid equation:
\[ z = 4(y^2 - 1)^2 + y^2 \]
3. **Simplify:**
Simplify the equation to find \( z \) in terms of \( y \):
\[ z = 4(y^2 - 1)^2 + y^2 \]
\[ z = 4(y^4 - 2y^2 + 1) + y^2 \]
\[ z = 4y^4 - 8y^2 + 4 + y^2 \]
\[ z = 4y^4 - 7y^2 + 4 \]
4. **Forming the Vector Function:**
The vector function representing the curve of intersection can then be written as:
\[ \mathbf{r}(t) = \left( t^2 - 1, t, 4t^4 - 7t^2 + 4 \right) \]
Here, we replaced the parameter \( y \) with \( t \).
5. **Graphing in GeoGebra:**
- **Graph the Paraboloid:** Input the equation \( z = 4x^2 + y^2 \) into GeoGebra.
- **Graph the Parabolic Cylinder:** Input the equation \( x = y^2 - 1 \).
- **Graph the Intersection Curve:** Use the parametric equations from the vector function \( \mathbf{r}(t) = \left( t^2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc303f822-5e20-4d30-8920-178170554c96%2F33eb2946-dea0-4462-8e83-902622fb212a%2Ffe7olh_processed.png&w=3840&q=75)
Transcribed Image Text:### Finding the Curve of Intersection Between Surfaces
**Problem Statement:**
Find a vector function that represents the curve of intersection between the paraboloid:
\[ z = 4x^2 + y^2 \]
and the parabolic cylinder:
\[ x = y^2 - 1 \]
Then draw all three in GeoGebra (paraboloid, parabolic cylinder, and their intersection curve).
**Solution:**
1. **Identifying the Intersection Curve:**
To find the curve of intersection, we need to express both surfaces in one set of parametric equations.
2. **Substitute \( x \) from the Parabolic Cylinder into the Paraboloid Equation:**
From the parabolic cylinder equation:
\[ x = y^2 - 1 \]
Now, substitute \( x \) in the paraboloid equation:
\[ z = 4(y^2 - 1)^2 + y^2 \]
3. **Simplify:**
Simplify the equation to find \( z \) in terms of \( y \):
\[ z = 4(y^2 - 1)^2 + y^2 \]
\[ z = 4(y^4 - 2y^2 + 1) + y^2 \]
\[ z = 4y^4 - 8y^2 + 4 + y^2 \]
\[ z = 4y^4 - 7y^2 + 4 \]
4. **Forming the Vector Function:**
The vector function representing the curve of intersection can then be written as:
\[ \mathbf{r}(t) = \left( t^2 - 1, t, 4t^4 - 7t^2 + 4 \right) \]
Here, we replaced the parameter \( y \) with \( t \).
5. **Graphing in GeoGebra:**
- **Graph the Paraboloid:** Input the equation \( z = 4x^2 + y^2 \) into GeoGebra.
- **Graph the Parabolic Cylinder:** Input the equation \( x = y^2 - 1 \).
- **Graph the Intersection Curve:** Use the parametric equations from the vector function \( \mathbf{r}(t) = \left( t^2
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