Reparametrize the curve with respect to arc length measured from the point wheret = 0 in the direction of increasing t. r(t) = (- 2+3t)i +(-1- 2t)j+ 4k 35 - 1 V13 a. r(s) = (-2 + b. r(s)=(-T13 35 25 V13 35 25 C. r(s) = (-2+ - 1 V13 d. r(s) = (-2 + 13 e. r(s) = (-2+V13s, – 1-/135,4) O c O a O d O b
Reparametrize the curve with respect to arc length measured from the point wheret = 0 in the direction of increasing t. r(t) = (- 2+3t)i +(-1- 2t)j+ 4k 35 - 1 V13 a. r(s) = (-2 + b. r(s)=(-T13 35 25 V13 35 25 C. r(s) = (-2+ - 1 V13 d. r(s) = (-2 + 13 e. r(s) = (-2+V13s, – 1-/135,4) O c O a O d O b
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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4. as fast as possible
![**Problem Statement:**
Reparametrize the curve with respect to arc length measured from the point where \( t = 0 \) in the direction of increasing \( t \).
**Given Function:**
\[ \mathbf{r}(t) = (-2 + 3t)\mathbf{i} + (-1 - 2t)\mathbf{j} + 4\mathbf{k} \]
**Options:**
a. \[ \mathbf{r}(s) = \left\langle -2 + \frac{3s}{\sqrt{13}}, -1 - \frac{2s}{\sqrt{13}}, 0 \right\rangle \]
b. \[ \mathbf{r}(s) = \left\langle -\frac{3s}{\sqrt{13}}, -\frac{2s}{\sqrt{13}}, 0 \right\rangle \]
c. \[ \mathbf{r}(s) = \left\langle -2 + \frac{3s}{\sqrt{13}}, -1 - \frac{2s}{\sqrt{13}}, 4 \right\rangle \]
d. \[ \mathbf{r}(s) = \left\langle -2 + \frac{s}{\sqrt{13}}, -1 - \frac{s}{\sqrt{13}}, 4 \right\rangle \]
e. \[ \mathbf{r}(s) = \langle -2 + \sqrt{13}s, -1 - \sqrt{13}s, 4 \rangle \]
**Answer Choices:**
- o c
- o a
- o d
- o b
- o e](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa72e8965-bd79-4464-a740-a72095a0be2c%2F5ba5907f-0b43-4768-9ae4-09a45d64ab5f%2Fq3zky6l_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Reparametrize the curve with respect to arc length measured from the point where \( t = 0 \) in the direction of increasing \( t \).
**Given Function:**
\[ \mathbf{r}(t) = (-2 + 3t)\mathbf{i} + (-1 - 2t)\mathbf{j} + 4\mathbf{k} \]
**Options:**
a. \[ \mathbf{r}(s) = \left\langle -2 + \frac{3s}{\sqrt{13}}, -1 - \frac{2s}{\sqrt{13}}, 0 \right\rangle \]
b. \[ \mathbf{r}(s) = \left\langle -\frac{3s}{\sqrt{13}}, -\frac{2s}{\sqrt{13}}, 0 \right\rangle \]
c. \[ \mathbf{r}(s) = \left\langle -2 + \frac{3s}{\sqrt{13}}, -1 - \frac{2s}{\sqrt{13}}, 4 \right\rangle \]
d. \[ \mathbf{r}(s) = \left\langle -2 + \frac{s}{\sqrt{13}}, -1 - \frac{s}{\sqrt{13}}, 4 \right\rangle \]
e. \[ \mathbf{r}(s) = \langle -2 + \sqrt{13}s, -1 - \sqrt{13}s, 4 \rangle \]
**Answer Choices:**
- o c
- o a
- o d
- o b
- o e
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