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
![**Problem:**
Find the exact length of the helix \(\vec{r}(t) = \left\langle 9 \cos \left( \frac{4t}{9} \right), \, 9 \sin \left( \frac{4t}{9} \right), \, 3t \right\rangle\) for \(-1 \le t \le 5\).
[Insert interactive graphing tool here, if available]
**Explanation:**
To find the length of the helix \( \vec{r}(t) \), we will use the arc length formula for parametric equations. The length \( L \) of the curve given by \( \vec{r}(t) \) from \( t = a \) to \( t = b \) is calculated using the formula:
\[ L = \int_a^b \left\| \vec{r}'(t) \right\| \, dt, \]
where \( \vec{r}'(t) \) is the derivative of \( \vec{r}(t) \) with respect to \( t \).
First, we need to find the derivative \( \vec{r}'(t) \):
\[ \vec{r}(t) = \left\langle 9 \cos \left( \frac{4t}{9} \right), \, 9 \sin \left( \frac{4t}{9} \right), \, 3t \right\rangle. \]
Taking the derivative component-wise, we get:
\[ \vec{r}'(t) = \left\langle \frac{d}{dt} \left( 9 \cos \left( \frac{4t}{9} \right) \right), \, \frac{d}{dt} \left( 9 \sin \left( \frac{4t}{9} \right) \right), \, \frac{d}{dt} (3t) \right\rangle. \]
Calculate each derivative:
1. \( \frac{d}{dt} \left( 9 \cos \left( \frac{4t}{9} \right) \right) = 9 \cdot \left( -\sin \left( \frac{4t}{9} \right) \cdot \frac{4}{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F16864198-3c29-44d0-a974-0ddc23a60123%2Fef3b0db5-de27-4400-b209-0a1a3ff6c138%2Fri3st7k_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem:**
Find the exact length of the helix \(\vec{r}(t) = \left\langle 9 \cos \left( \frac{4t}{9} \right), \, 9 \sin \left( \frac{4t}{9} \right), \, 3t \right\rangle\) for \(-1 \le t \le 5\).
[Insert interactive graphing tool here, if available]
**Explanation:**
To find the length of the helix \( \vec{r}(t) \), we will use the arc length formula for parametric equations. The length \( L \) of the curve given by \( \vec{r}(t) \) from \( t = a \) to \( t = b \) is calculated using the formula:
\[ L = \int_a^b \left\| \vec{r}'(t) \right\| \, dt, \]
where \( \vec{r}'(t) \) is the derivative of \( \vec{r}(t) \) with respect to \( t \).
First, we need to find the derivative \( \vec{r}'(t) \):
\[ \vec{r}(t) = \left\langle 9 \cos \left( \frac{4t}{9} \right), \, 9 \sin \left( \frac{4t}{9} \right), \, 3t \right\rangle. \]
Taking the derivative component-wise, we get:
\[ \vec{r}'(t) = \left\langle \frac{d}{dt} \left( 9 \cos \left( \frac{4t}{9} \right) \right), \, \frac{d}{dt} \left( 9 \sin \left( \frac{4t}{9} \right) \right), \, \frac{d}{dt} (3t) \right\rangle. \]
Calculate each derivative:
1. \( \frac{d}{dt} \left( 9 \cos \left( \frac{4t}{9} \right) \right) = 9 \cdot \left( -\sin \left( \frac{4t}{9} \right) \cdot \frac{4}{
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