Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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![**Problem 7: Finding the Unit Tangent Vector**
Objective: Find the unit tangent vector for the parameterized curve given by:
\[ \mathbf{r}(t) = \cos t \, \mathbf{i} + \sin t \, \mathbf{j} + \sin t \, \mathbf{k}, \quad 0 \le t < 2\pi \]
The goal of this problem is to determine the unit tangent vector for the specified parameterized curve. To do so, we will first differentiate \(\mathbf{r}(t)\) with respect to \(t\) to find the tangent vector, and then we will normalize this tangent vector to obtain the unit tangent vector.
**Steps:**
1. **Differentiate \(\mathbf{r}(t)\) with respect to \(t\)**:
\[ \frac{d\mathbf{r}(t)}{dt} = \frac{d}{dt} (\cos t \, \mathbf{i} + \sin t \, \mathbf{j} + \sin t \, \mathbf{k}) \]
2. **Compute the derivative**:
\[ \mathbf{r}'(t) = -\sin t \, \mathbf{i} + \cos t \, \mathbf{j} + \cos t \, \mathbf{k} \]
3. **Find the magnitude of \(\mathbf{r}'(t)\)**:
\[ \|\mathbf{r}'(t)\| = \sqrt{(-\sin t)^2 + (\cos t)^2 + (\cos t)^2} = \sqrt{\sin^2 t + \cos^2 t + \cos^2 t} = \sqrt{1 + \cos^2 t} \]
4. **Form the unit tangent vector \(\mathbf{T}(t)\)**:
\[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} = \frac{-\sin t \, \mathbf{i} + \cos t \, \mathbf{j} + \cos t \, \mathbf{k}}{\sqrt{1 + \cos^2 t}} \]
Thus, the unit tangent vector \(\mathbf{T}(t)\) for the given parameterized curve is:
\[ \mathbf{T}(t) = \frac](https://content.bartleby.com/qna-images/question/58097f1d-9f7f-481d-be94-f210527e0168/5428cc9b-9eee-4781-8c96-ee576bb3bdd5/8zznm2_processed.jpeg)
Transcribed Image Text:**Problem 7: Finding the Unit Tangent Vector**
Objective: Find the unit tangent vector for the parameterized curve given by:
\[ \mathbf{r}(t) = \cos t \, \mathbf{i} + \sin t \, \mathbf{j} + \sin t \, \mathbf{k}, \quad 0 \le t < 2\pi \]
The goal of this problem is to determine the unit tangent vector for the specified parameterized curve. To do so, we will first differentiate \(\mathbf{r}(t)\) with respect to \(t\) to find the tangent vector, and then we will normalize this tangent vector to obtain the unit tangent vector.
**Steps:**
1. **Differentiate \(\mathbf{r}(t)\) with respect to \(t\)**:
\[ \frac{d\mathbf{r}(t)}{dt} = \frac{d}{dt} (\cos t \, \mathbf{i} + \sin t \, \mathbf{j} + \sin t \, \mathbf{k}) \]
2. **Compute the derivative**:
\[ \mathbf{r}'(t) = -\sin t \, \mathbf{i} + \cos t \, \mathbf{j} + \cos t \, \mathbf{k} \]
3. **Find the magnitude of \(\mathbf{r}'(t)\)**:
\[ \|\mathbf{r}'(t)\| = \sqrt{(-\sin t)^2 + (\cos t)^2 + (\cos t)^2} = \sqrt{\sin^2 t + \cos^2 t + \cos^2 t} = \sqrt{1 + \cos^2 t} \]
4. **Form the unit tangent vector \(\mathbf{T}(t)\)**:
\[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} = \frac{-\sin t \, \mathbf{i} + \cos t \, \mathbf{j} + \cos t \, \mathbf{k}}{\sqrt{1 + \cos^2 t}} \]
Thus, the unit tangent vector \(\mathbf{T}(t)\) for the given parameterized curve is:
\[ \mathbf{T}(t) = \frac
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