Given 7(t) = (et, et, vZet), for t > 0. a) Determine whether 7 (t) uses arc length as the parameter. %3| b) If not, re-parameterize 7 (t) with respect to arc length.
Given 7(t) = (et, et, vZet), for t > 0. a) Determine whether 7 (t) uses arc length as the parameter. %3| b) If not, re-parameterize 7 (t) with respect to arc length.
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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Vector Arithmetic
Vectors are those objects which have a magnitude along with the direction. In vector arithmetic, we will see how arithmetic operators like addition and multiplication are used on any two vectors. Arithmetic in basic means dealing with numbers. Here, magnitude means the length or the size of an object. The notation used is the arrow over the head of the vector indicating its direction.
Vector Calculus
Vector calculus is an important branch of mathematics and it relates two important branches of mathematics namely vector and calculus.
Question
![### Problem Statement
Given the vector function \(\vec{r}(t) = \langle e^t, \, e^t, \, \sqrt{2}e^t \rangle\), for \(t \geq 0\):
a) Determine whether \(\vec{r}(t)\) uses arc length as the parameter.
b) If not, re-parameterize \(\vec{r}(t)\) with respect to arc length.
### Explanation and Solution Approach
**a) Determine whether \(\vec{r}(t)\) uses arc length as the parameter:**
To determine whether the given vector function \(\vec{r}(t)\) uses arc length \(s\) as the parameter, we need to check if the magnitude of the derivative of \(\vec{r}(t)\) with respect to \(t\) is 1 at all points. This means we need to calculate \(\left\|\frac{d\vec{r}}{dt}\right\|\).
Let's find \(\vec{r}'(t)\):
\[\vec{r}(t) = \langle e^t, \, e^t, \, \sqrt{2}e^t \rangle\]
\[\frac{d\vec{r}}{dt} = \langle \frac{d}{dt}(e^t), \, \frac{d}{dt}(e^t), \, \frac{d}{dt}(\sqrt{2}e^t) \rangle\]
\[\vec{r}'(t) = \langle e^t, \, e^t, \, \sqrt{2}e^t \rangle\]
Now, calculate the magnitude of \(\vec{r}'(t)\):
\[\left\|\vec{r}'(t)\right\| = \sqrt{(e^t)^2 + (e^t)^2 + (\sqrt{2}e^t)^2}\]
\[= \sqrt{e^{2t} + e^{2t} + 2e^{2t}}\]
\[= \sqrt{4e^{2t}}\]
\[= 2e^t\]
For \(\vec{r}(t)\) to use arc length as the parameter, \(\left\|\vec{r}'(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0beceb5e-99bc-4360-aaf5-87226e0ffc77%2Fe8923b7c-cf58-49d6-a22e-460d8ebd22be%2Fwu8q8pk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Given the vector function \(\vec{r}(t) = \langle e^t, \, e^t, \, \sqrt{2}e^t \rangle\), for \(t \geq 0\):
a) Determine whether \(\vec{r}(t)\) uses arc length as the parameter.
b) If not, re-parameterize \(\vec{r}(t)\) with respect to arc length.
### Explanation and Solution Approach
**a) Determine whether \(\vec{r}(t)\) uses arc length as the parameter:**
To determine whether the given vector function \(\vec{r}(t)\) uses arc length \(s\) as the parameter, we need to check if the magnitude of the derivative of \(\vec{r}(t)\) with respect to \(t\) is 1 at all points. This means we need to calculate \(\left\|\frac{d\vec{r}}{dt}\right\|\).
Let's find \(\vec{r}'(t)\):
\[\vec{r}(t) = \langle e^t, \, e^t, \, \sqrt{2}e^t \rangle\]
\[\frac{d\vec{r}}{dt} = \langle \frac{d}{dt}(e^t), \, \frac{d}{dt}(e^t), \, \frac{d}{dt}(\sqrt{2}e^t) \rangle\]
\[\vec{r}'(t) = \langle e^t, \, e^t, \, \sqrt{2}e^t \rangle\]
Now, calculate the magnitude of \(\vec{r}'(t)\):
\[\left\|\vec{r}'(t)\right\| = \sqrt{(e^t)^2 + (e^t)^2 + (\sqrt{2}e^t)^2}\]
\[= \sqrt{e^{2t} + e^{2t} + 2e^{2t}}\]
\[= \sqrt{4e^{2t}}\]
\[= 2e^t\]
For \(\vec{r}(t)\) to use arc length as the parameter, \(\left\|\vec{r}'(
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