Find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. A r(t) = (9 cos³t)j + (9 sin³t)k, Osts
Find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. A r(t) = (9 cos³t)j + (9 sin³t)k, Osts
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![To find the curve's unit tangent vector and the length of the indicated portion of the curve, consider the following vector function:
\[
\mathbf{r}(t) = \left(9 \cos^3 t \right) \mathbf{i} + \left(9 \sin^3 t \right) \mathbf{j}, \quad 0 \leq t \leq \frac{\pi}{3}
\]
**Explanation:**
- The vector function \(\mathbf{r}(t)\) describes a 2D parametric curve in terms of the parameter \(t\), with its components given by \(x(t) = 9 \cos^3 t\) and \(y(t) = 9 \sin^3 t\).
**Steps to solve:**
1. **Find the Derivative**:
- Differentiate \(\mathbf{r}(t)\) with respect to \(t\) to get \(\mathbf{r}'(t)\), the tangent vector.
2. **Unit Tangent Vector**:
- Normalize \(\mathbf{r}'(t)\) to find the unit tangent vector \(\mathbf{T}(t)\).
3. **Arc Length**:
- Compute the arc length of the curve from \(t = 0\) to \(t = \frac{\pi}{3}\) using the formula:
\[
L = \int_{0}^{\frac{\pi}{3}} \left\| \mathbf{r}'(t) \right\| \, dt
\]
These steps will determine the unit tangent vector and the length of the specified segment of the curve.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F671be518-f9e6-4df9-9304-33791fefeccb%2Ff73dd599-0231-4c48-aaeb-554747150ee6%2F1715kk8_processed.png&w=3840&q=75)
Transcribed Image Text:To find the curve's unit tangent vector and the length of the indicated portion of the curve, consider the following vector function:
\[
\mathbf{r}(t) = \left(9 \cos^3 t \right) \mathbf{i} + \left(9 \sin^3 t \right) \mathbf{j}, \quad 0 \leq t \leq \frac{\pi}{3}
\]
**Explanation:**
- The vector function \(\mathbf{r}(t)\) describes a 2D parametric curve in terms of the parameter \(t\), with its components given by \(x(t) = 9 \cos^3 t\) and \(y(t) = 9 \sin^3 t\).
**Steps to solve:**
1. **Find the Derivative**:
- Differentiate \(\mathbf{r}(t)\) with respect to \(t\) to get \(\mathbf{r}'(t)\), the tangent vector.
2. **Unit Tangent Vector**:
- Normalize \(\mathbf{r}'(t)\) to find the unit tangent vector \(\mathbf{T}(t)\).
3. **Arc Length**:
- Compute the arc length of the curve from \(t = 0\) to \(t = \frac{\pi}{3}\) using the formula:
\[
L = \int_{0}^{\frac{\pi}{3}} \left\| \mathbf{r}'(t) \right\| \, dt
\]
These steps will determine the unit tangent vector and the length of the specified segment of the curve.
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