Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Hello, can you please answer the following question. Please show all your work. Thank you.
![**Problem Statement:**
Eliminate the parameter. Graph the curve and indicate orientation. Show points on graph.
\[ x = \sqrt{t + 1}, \quad y = \sqrt{t - 1} \]
**Solution:**
1. **Eliminate the parameter \( t \):**
Express \( t \) in terms of \( x \) and \( y \).
From \( x = \sqrt{t + 1} \), we have:
\[ t = x^2 - 1 \]
From \( y = \sqrt{t - 1} \), we have:
\[ t = y^2 + 1 \]
Setting the two expressions for \( t \) equal gives:
\[ x^2 - 1 = y^2 + 1 \]
Simplifying, we get the relation:
\[ x^2 - y^2 = 2 \]
2. **Graph the curve:**
The equation \( x^2 - y^2 = 2 \) represents a hyperbola.
3. **Indicate orientation and points:**
To indicate orientation, solve for specific values of \( t \) to find corresponding \( (x, y) \) points.
- For \( t = 2 \):
- \( x = \sqrt{2 + 1} = \sqrt{3} \)
- \( y = \sqrt{2 - 1} = 1 \)
- Point: \( (\sqrt{3}, 1) \)
- For \( t = 3 \):
- \( x = \sqrt{3 + 1} = 2 \)
- \( y = \sqrt{3 - 1} = \sqrt{2} \)
- Point: \( (2, \sqrt{2}) \)
The orientation of the curve is determined by increasing values of \( t \), moving from \( (\sqrt{3}, 1) \) to \( (2, \sqrt{2}) \).
**Notes for Educators:**
- When graphing, ensure students understand the transformation from parameterized equations to an expression involving only \( x \) and \( y \).
- Highlight the process and logic used to eliminate parameters and find corresponding points to establish orientation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0ea55b65-6a30-46ce-96b8-2b732a599bb3%2Fdfe8dabd-9aa2-4a9b-b1c9-e33f8f2f7c06%2Fwvpitt_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Eliminate the parameter. Graph the curve and indicate orientation. Show points on graph.
\[ x = \sqrt{t + 1}, \quad y = \sqrt{t - 1} \]
**Solution:**
1. **Eliminate the parameter \( t \):**
Express \( t \) in terms of \( x \) and \( y \).
From \( x = \sqrt{t + 1} \), we have:
\[ t = x^2 - 1 \]
From \( y = \sqrt{t - 1} \), we have:
\[ t = y^2 + 1 \]
Setting the two expressions for \( t \) equal gives:
\[ x^2 - 1 = y^2 + 1 \]
Simplifying, we get the relation:
\[ x^2 - y^2 = 2 \]
2. **Graph the curve:**
The equation \( x^2 - y^2 = 2 \) represents a hyperbola.
3. **Indicate orientation and points:**
To indicate orientation, solve for specific values of \( t \) to find corresponding \( (x, y) \) points.
- For \( t = 2 \):
- \( x = \sqrt{2 + 1} = \sqrt{3} \)
- \( y = \sqrt{2 - 1} = 1 \)
- Point: \( (\sqrt{3}, 1) \)
- For \( t = 3 \):
- \( x = \sqrt{3 + 1} = 2 \)
- \( y = \sqrt{3 - 1} = \sqrt{2} \)
- Point: \( (2, \sqrt{2}) \)
The orientation of the curve is determined by increasing values of \( t \), moving from \( (\sqrt{3}, 1) \) to \( (2, \sqrt{2}) \).
**Notes for Educators:**
- When graphing, ensure students understand the transformation from parameterized equations to an expression involving only \( x \) and \( y \).
- Highlight the process and logic used to eliminate parameters and find corresponding points to establish orientation.
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