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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Question
3.2.7
![**Problem Statement**
Find the unit tangent vector to the curve defined by
\[
\vec{r}(t) = \langle 5t, -5t, \sqrt{1 - t^2} \rangle
\]
at \( t = 0 \).
**Solution:**
To find the unit tangent vector, we need to determine the derivative of the vector function \(\vec{r}(t)\) and then evaluate it at \( t = 0 \). Then, we will normalize this vector to obtain the unit tangent vector.
1. **Differentiate \(\vec{r}(t)\):**
\[
\vec{r}'(t) = \left\langle \frac{d}{dt}(5t), \frac{d}{dt}(-5t), \frac{d}{dt}\left(\sqrt{1 - t^2}\right) \right\rangle
\]
\[
= \langle 5, -5, \frac{-t}{\sqrt{1-t^2}} \rangle
\]
2. **Evaluate \(\vec{r}'(t)\) at \( t = 0 \):**
\[
\vec{r}'(0) = \langle 5, -5, 0 \rangle
\]
3. **Find the magnitude of \(\vec{r}'(0)\):**
\[
\|\vec{r}'(0)\| = \sqrt{5^2 + (-5)^2 + 0^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}
\]
4. **Normalize \(\vec{r}'(0)\) to find the unit tangent vector \(\vec{T}(0)\):**
\[
\vec{T}(0) = \frac{\vec{r}'(0)}{\|\vec{r}'(0)\|} = \frac{\langle 5, -5, 0 \rangle}{5\sqrt{2}} = \left\langle \frac{5}{5\sqrt{2}}, \frac{-5}{5\sqrt{2}}, 0 \right\rangle
\]
\[
= \left\langle \frac{1}{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6b463452-b960-4b00-bce0-3a69d9f467e2%2F6e7ac726-71b9-45ff-a143-54945e182347%2Fdf623q_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement**
Find the unit tangent vector to the curve defined by
\[
\vec{r}(t) = \langle 5t, -5t, \sqrt{1 - t^2} \rangle
\]
at \( t = 0 \).
**Solution:**
To find the unit tangent vector, we need to determine the derivative of the vector function \(\vec{r}(t)\) and then evaluate it at \( t = 0 \). Then, we will normalize this vector to obtain the unit tangent vector.
1. **Differentiate \(\vec{r}(t)\):**
\[
\vec{r}'(t) = \left\langle \frac{d}{dt}(5t), \frac{d}{dt}(-5t), \frac{d}{dt}\left(\sqrt{1 - t^2}\right) \right\rangle
\]
\[
= \langle 5, -5, \frac{-t}{\sqrt{1-t^2}} \rangle
\]
2. **Evaluate \(\vec{r}'(t)\) at \( t = 0 \):**
\[
\vec{r}'(0) = \langle 5, -5, 0 \rangle
\]
3. **Find the magnitude of \(\vec{r}'(0)\):**
\[
\|\vec{r}'(0)\| = \sqrt{5^2 + (-5)^2 + 0^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}
\]
4. **Normalize \(\vec{r}'(0)\) to find the unit tangent vector \(\vec{T}(0)\):**
\[
\vec{T}(0) = \frac{\vec{r}'(0)}{\|\vec{r}'(0)\|} = \frac{\langle 5, -5, 0 \rangle}{5\sqrt{2}} = \left\langle \frac{5}{5\sqrt{2}}, \frac{-5}{5\sqrt{2}}, 0 \right\rangle
\]
\[
= \left\langle \frac{1}{
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