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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![**Finding Polar Coordinates: An Educational Tutorial**
To understand how to find polar coordinates \((r, \theta)\) for a given point in the Cartesian coordinate system, we will look at an example step-by-step.
Here is a problem statement along with possible answers:
---
**Problem:**
Find polar coordinates \((r, \theta)\) of the point \((-2\sqrt{2}, 2\sqrt{2})\), where \(r < 0\) and \(0 \leq \theta \leq 2\pi\).
**Options:**
a. \[\left(-2\sqrt{2}, \frac{5\pi}{3}\right)\]
b. \[\left(2\sqrt{2}, \frac{2\pi}{3}\right)\]
c. \[\left(-2, \frac{3\pi}{4}\right)\]
d. \[\left(4, \frac{3\pi}{4}\right)\]
e. \[\left(-4, \frac{7\pi}{4}\right)\]
---
Let's solve this step-by-step.
### Step 1: Cartesian to Polar Conversion
The given Cartesian coordinates are \((-2\sqrt{2}, 2\sqrt{2})\). To convert this to polar coordinates \((r, \theta)\), we use the formulas:
\[ r = \sqrt{x^2 + y^2} \]
\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \]
### Step 2: Calculating \(r\)
Given \(x = -2\sqrt{2}\) and \(y = 2\sqrt{2}\):
\[ r = \sqrt{(-2\sqrt{2})^2 + (2\sqrt{2})^2} \]
\[ r = \sqrt{8 + 8} \]
\[ r = \sqrt{16} \]
\[ r = 4 \]
Since \(r < 0\) as per the problem, we have:
\[ r = -4 \]
### Step 3: Calculating \(\theta\)
\[ \theta = \tan^{-1}\left(\frac{2\sqrt{2}}{-2\sqrt{2}}\right) = \tan^{-1}(-1) \]
The value of \(\theta\) where \(\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0d8d0c97-e66d-424a-9c52-d4c7d8464ad8%2Fc6fe2651-c2c5-43c5-953f-66a67fadb790%2F6ixoxpv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Finding Polar Coordinates: An Educational Tutorial**
To understand how to find polar coordinates \((r, \theta)\) for a given point in the Cartesian coordinate system, we will look at an example step-by-step.
Here is a problem statement along with possible answers:
---
**Problem:**
Find polar coordinates \((r, \theta)\) of the point \((-2\sqrt{2}, 2\sqrt{2})\), where \(r < 0\) and \(0 \leq \theta \leq 2\pi\).
**Options:**
a. \[\left(-2\sqrt{2}, \frac{5\pi}{3}\right)\]
b. \[\left(2\sqrt{2}, \frac{2\pi}{3}\right)\]
c. \[\left(-2, \frac{3\pi}{4}\right)\]
d. \[\left(4, \frac{3\pi}{4}\right)\]
e. \[\left(-4, \frac{7\pi}{4}\right)\]
---
Let's solve this step-by-step.
### Step 1: Cartesian to Polar Conversion
The given Cartesian coordinates are \((-2\sqrt{2}, 2\sqrt{2})\). To convert this to polar coordinates \((r, \theta)\), we use the formulas:
\[ r = \sqrt{x^2 + y^2} \]
\[ \theta = \tan^{-1}\left(\frac{y}{x}\right) \]
### Step 2: Calculating \(r\)
Given \(x = -2\sqrt{2}\) and \(y = 2\sqrt{2}\):
\[ r = \sqrt{(-2\sqrt{2})^2 + (2\sqrt{2})^2} \]
\[ r = \sqrt{8 + 8} \]
\[ r = \sqrt{16} \]
\[ r = 4 \]
Since \(r < 0\) as per the problem, we have:
\[ r = -4 \]
### Step 3: Calculating \(\theta\)
\[ \theta = \tan^{-1}\left(\frac{2\sqrt{2}}{-2\sqrt{2}}\right) = \tan^{-1}(-1) \]
The value of \(\theta\) where \(\
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