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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![### Vector Direction Calculation
**Question:**
What is the direction of the vector with an initial point of (3, -5) and a terminal point of (8, 6)?
**Answer Options:**
- **A. 11.31°**
- **B. 24.44°**
- **C. 65.56°**
- **D. 114.44°**
To solve such questions, we calculate the direction (or angle) of the vector using the tangent function. Specifically, we use the formula for the angle θ formed with the positive x-axis:
\[\theta = \tan^{-1}\left(\frac{y_2 - y_1}{x_2 - x_1}\right)\]
Where:
- \((x_1, y_1)\) are the coordinates of the initial point
- \((x_2, y_2)\) are the coordinates of the terminal point
For this problem:
- Initial point (x_1, y_1) = (3, -5)
- Terminal point (x_2, y_2) = (8, 6)
First, calculate the difference in coordinates:
- \(\Delta x = x_2 - x_1 = 8 - 3 = 5\)
- \(\Delta y = y_2 - y_1 = 6 + 5 = 11\) (Note: adding 5 because y_1 is negative)
Next, apply these to the tangent formula:
\[\theta = \tan^{-1}\left(\frac{11}{5}\right)\]
Using a calculator to find the arctangent value:
\[\theta ≈ 65.56°\]
Thus, the correct answer is:
- **C. 65.56°**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6d5b7d63-40af-4a8b-b18d-6fc41e951270%2F32aaaa69-e7d4-4dae-92dd-3dc1b7149d1e%2F4ofjnc7_processed.png&w=3840&q=75)
Transcribed Image Text:### Vector Direction Calculation
**Question:**
What is the direction of the vector with an initial point of (3, -5) and a terminal point of (8, 6)?
**Answer Options:**
- **A. 11.31°**
- **B. 24.44°**
- **C. 65.56°**
- **D. 114.44°**
To solve such questions, we calculate the direction (or angle) of the vector using the tangent function. Specifically, we use the formula for the angle θ formed with the positive x-axis:
\[\theta = \tan^{-1}\left(\frac{y_2 - y_1}{x_2 - x_1}\right)\]
Where:
- \((x_1, y_1)\) are the coordinates of the initial point
- \((x_2, y_2)\) are the coordinates of the terminal point
For this problem:
- Initial point (x_1, y_1) = (3, -5)
- Terminal point (x_2, y_2) = (8, 6)
First, calculate the difference in coordinates:
- \(\Delta x = x_2 - x_1 = 8 - 3 = 5\)
- \(\Delta y = y_2 - y_1 = 6 + 5 = 11\) (Note: adding 5 because y_1 is negative)
Next, apply these to the tangent formula:
\[\theta = \tan^{-1}\left(\frac{11}{5}\right)\]
Using a calculator to find the arctangent value:
\[\theta ≈ 65.56°\]
Thus, the correct answer is:
- **C. 65.56°**
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