Find the unit tangent vector for the parameterized curve: r(t) = 2ti + 2cos tj - 2 sint k
Find the unit tangent vector for the parameterized curve: r(t) = 2ti + 2cos tj - 2 sint k
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Unit Tangent Vector for a Parameterized Curve
To find the unit tangent vector for the parameterized curve, follow these steps:
Given the vector function:
\[ \mathbf{r}(t) = 2t\mathbf{i} + 2\cos t \mathbf{j} - 2\sin t \mathbf{k} \]
1. **Compute the Derivative of \(\mathbf{r}(t)\)**
Find \(\mathbf{r}'(t)\), which is the derivative of \(\mathbf{r}(t)\) with respect to \(t\):
\[
\mathbf{r}'(t) = \frac{d}{dt} (2t\mathbf{i} + 2\cos t \mathbf{j} - 2\sin t \mathbf{k})
\]
This yields:
\[
\mathbf{r}'(t) = 2\mathbf{i} - 2\sin t \mathbf{j} - 2\cos t \mathbf{k}
\]
2. **Compute the Magnitude of \(\mathbf{r}'(t)\)**
Find the magnitude \(|\mathbf{r}'(t)|\):
\[
|\mathbf{r}'(t)| = \sqrt{(2)^2 + (-2\sin t)^2 + (-2\cos t)^2}
\]
Simplify inside the square root:
\[
|\mathbf{r}'(t)| = \sqrt{4 + 4\sin^2 t + 4\cos^2 t}
\]
Using the Pythagorean identity \(\sin^2 t + \cos^2 t = 1\):
\[
|\mathbf{r}'(t)| = \sqrt{4 + 4(1)} = \sqrt{8} = 2\sqrt{2}
\]
3. **Compute the Unit Tangent Vector**
The unit tangent vector \(\mathbf{T}(t)\) is given by:
\[
\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}
\]
Substitute \(\mathbf{r}'(t)\) and \(|\mathbf{r}'(t)|\):
\[
\mathbf](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F58097f1d-9f7f-481d-be94-f210527e0168%2F3e761c52-c59f-4191-8e38-d2d3fce7a412%2Fi684cts_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Unit Tangent Vector for a Parameterized Curve
To find the unit tangent vector for the parameterized curve, follow these steps:
Given the vector function:
\[ \mathbf{r}(t) = 2t\mathbf{i} + 2\cos t \mathbf{j} - 2\sin t \mathbf{k} \]
1. **Compute the Derivative of \(\mathbf{r}(t)\)**
Find \(\mathbf{r}'(t)\), which is the derivative of \(\mathbf{r}(t)\) with respect to \(t\):
\[
\mathbf{r}'(t) = \frac{d}{dt} (2t\mathbf{i} + 2\cos t \mathbf{j} - 2\sin t \mathbf{k})
\]
This yields:
\[
\mathbf{r}'(t) = 2\mathbf{i} - 2\sin t \mathbf{j} - 2\cos t \mathbf{k}
\]
2. **Compute the Magnitude of \(\mathbf{r}'(t)\)**
Find the magnitude \(|\mathbf{r}'(t)|\):
\[
|\mathbf{r}'(t)| = \sqrt{(2)^2 + (-2\sin t)^2 + (-2\cos t)^2}
\]
Simplify inside the square root:
\[
|\mathbf{r}'(t)| = \sqrt{4 + 4\sin^2 t + 4\cos^2 t}
\]
Using the Pythagorean identity \(\sin^2 t + \cos^2 t = 1\):
\[
|\mathbf{r}'(t)| = \sqrt{4 + 4(1)} = \sqrt{8} = 2\sqrt{2}
\]
3. **Compute the Unit Tangent Vector**
The unit tangent vector \(\mathbf{T}(t)\) is given by:
\[
\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}
\]
Substitute \(\mathbf{r}'(t)\) and \(|\mathbf{r}'(t)|\):
\[
\mathbf
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