A particle moves so that r(t) = (3 cos t + 3t sin t)i + (3 sint – 3t cos t)j+ (2t² )k. Find the distance traveled by the particle from t = 0 to t = 67.
A particle moves so that r(t) = (3 cos t + 3t sin t)i + (3 sint – 3t cos t)j+ (2t² )k. Find the distance traveled by the particle from t = 0 to t = 67.
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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![**Problem Description:**
A particle moves such that its position vector \( \mathbf{r}(t) \) is given by:
\[
\mathbf{r}(t) = (3 \cos t + 3t \sin t) \mathbf{i} + (3 \sin t - 3t \cos t) \mathbf{j} + (2t^2) \mathbf{k}
\]
Calculate the distance traveled by the particle from \( t = 0 \) to \( t = 6\pi \).
**Answer Choices:**
- \( \circ \) 3
- \( \circ \) \(\sqrt{324\pi^2 + 5184\pi^4}\)
- \( \circ \) \(\sqrt{324\pi^2 + 5184\pi^4 + 9}\)
- \( \circ \) \(3\mathbf{i} - 18\pi \mathbf{j} + 72\pi^2 \mathbf{k}\)
- \( \circ \) \(90\pi^2\)
**Detailed Explanation:**
To solve this problem, you'll need to compute the integral of the magnitude of the derivative of the position vector, \(|\mathbf{v}(t)|\), where \(\mathbf{v}(t) = \frac{d\mathbf{r}}{dt}\). Apply the distance formula for a vector function to find the total distance traveled over the given interval.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F82704860-94cd-423d-b2ec-83bd4f56cfa0%2F74933b20-4679-4447-941f-a9bb28535d9a%2Fjaozsee_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Description:**
A particle moves such that its position vector \( \mathbf{r}(t) \) is given by:
\[
\mathbf{r}(t) = (3 \cos t + 3t \sin t) \mathbf{i} + (3 \sin t - 3t \cos t) \mathbf{j} + (2t^2) \mathbf{k}
\]
Calculate the distance traveled by the particle from \( t = 0 \) to \( t = 6\pi \).
**Answer Choices:**
- \( \circ \) 3
- \( \circ \) \(\sqrt{324\pi^2 + 5184\pi^4}\)
- \( \circ \) \(\sqrt{324\pi^2 + 5184\pi^4 + 9}\)
- \( \circ \) \(3\mathbf{i} - 18\pi \mathbf{j} + 72\pi^2 \mathbf{k}\)
- \( \circ \) \(90\pi^2\)
**Detailed Explanation:**
To solve this problem, you'll need to compute the integral of the magnitude of the derivative of the position vector, \(|\mathbf{v}(t)|\), where \(\mathbf{v}(t) = \frac{d\mathbf{r}}{dt}\). Apply the distance formula for a vector function to find the total distance traveled over the given interval.
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