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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Question
Find the points of the line where the tangent is vertical
![## Problem Statement
Find the points on the line where the two curves meet.
### Given Parametric Equations:
\[ x(t) = t^3 - 3t + 2 \]
\[ y(t) = t^3 - 3t^2 + 2 \]
### Instructions:
1. **Identify the Intersection Points:**
- Determine the values of \( t \) where the two curves intersect.
- Substitute these values back into the parametric equations to find the coordinates of the intersection points.
2. **Graphical Representation:**
- If applicable, graph the parametric equations to visualize the points of intersection.
### Explanation:
To locate the intersection points, you need to solve the equations simultaneously for the parameter \( t \). This will involve setting the \( x(t) \) equation equal to the \( y(t) \) equation and solving for \( t \).
#### Step-by-Step Solution:
1. **Set the two equations equal to each other:**
\[ t^3 - 3t + 2 = t^3 - 3t^2 + 2 \]
2. **Simplify the equation:**
\[ -3t + 2 = -3t^2 + 2 \]
\[ -3t^2 + 3t = 0 \]
\[ 3t(t - 1) = 0 \]
3. **Solve for \( t \):**
\[ t = 0 \] or \[ t = 1 \]
4. **Substitute \( t \) back into the parametric equations to find the coordinates:**
- For \( t = 0 \):
\[ x(0) = 0^3 - 3(0) + 2 = 2 \]
\[ y(0) = 0^3 - 3(0)^2 + 2 = 2 \]
Intersection point: \((2, 2)\)
- For \( t = 1 \):
\[ x(1) = 1^3 - 3(1) + 2 = 0 \]
\[ y(1) = 1^3 - 3(1)^2 + 2 = 0 \]
Intersection point: \((0, 0)\)
### Conclusion:
The points of intersection are](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffe12ea73-12ee-4ced-9931-af14c1db1366%2F3da3bcfc-3e88-4ca8-8357-81410610d418%2Fkguavrr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Problem Statement
Find the points on the line where the two curves meet.
### Given Parametric Equations:
\[ x(t) = t^3 - 3t + 2 \]
\[ y(t) = t^3 - 3t^2 + 2 \]
### Instructions:
1. **Identify the Intersection Points:**
- Determine the values of \( t \) where the two curves intersect.
- Substitute these values back into the parametric equations to find the coordinates of the intersection points.
2. **Graphical Representation:**
- If applicable, graph the parametric equations to visualize the points of intersection.
### Explanation:
To locate the intersection points, you need to solve the equations simultaneously for the parameter \( t \). This will involve setting the \( x(t) \) equation equal to the \( y(t) \) equation and solving for \( t \).
#### Step-by-Step Solution:
1. **Set the two equations equal to each other:**
\[ t^3 - 3t + 2 = t^3 - 3t^2 + 2 \]
2. **Simplify the equation:**
\[ -3t + 2 = -3t^2 + 2 \]
\[ -3t^2 + 3t = 0 \]
\[ 3t(t - 1) = 0 \]
3. **Solve for \( t \):**
\[ t = 0 \] or \[ t = 1 \]
4. **Substitute \( t \) back into the parametric equations to find the coordinates:**
- For \( t = 0 \):
\[ x(0) = 0^3 - 3(0) + 2 = 2 \]
\[ y(0) = 0^3 - 3(0)^2 + 2 = 2 \]
Intersection point: \((2, 2)\)
- For \( t = 1 \):
\[ x(1) = 1^3 - 3(1) + 2 = 0 \]
\[ y(1) = 1^3 - 3(1)^2 + 2 = 0 \]
Intersection point: \((0, 0)\)
### Conclusion:
The points of intersection are
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