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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![### Problem Statement
Find the location and velocity at \( t = 4 \) of a particle whose path satisfies \( \frac{d\mathbf{r}}{dt} = \langle 3t^{-1/2}, 5, 8t \rangle \) and \( \mathbf{r}(1) = \langle 4, 9, 2 \rangle \).
### Solution
To solve this problem, we need to:
1. **Compute the position vector \( \mathbf{r}(4) \).**
2. **Compute the velocity vector \( \mathbf{r}'(4) \).**
#### Step-by-Step Solution
**Integration of \( \frac{d\mathbf{r}}{dt} \):**
Given:
\[ \frac{d\mathbf{r}}{dt} = \langle 3t^{-1/2}, 5, 8t \rangle \]
Integrate each component with respect to \( t \):
\[ r(t) = \int \langle 3t^{-1/2}, 5, 8t \rangle dt \]
**First Component:**
\[ \int 3t^{-1/2} dt = 3 \int t^{-1/2} dt = 3 \left( \frac{t^{1/2}}{1/2} \right) = 6t^{1/2} \]
**Second Component:**
\[ \int 5 dt = 5t \]
**Third Component:**
\[ \int 8t dt = 8 \left( \frac{t^2}{2} \right) = 4t^2 \]
Thus:
\[ \mathbf{r}(t) = \langle 6t^{1/2}, 5t, 4t^2 \rangle + \mathbf{C} \]
where \( \mathbf{C} \) is the constant vector.
**Finding the Constant Vector \( \mathbf{C} \):**
Using the initial condition \( \mathbf{r}(1) = \langle 4, 9, 2 \rangle \):
\[ \mathbf{r}(1) = \langle 6(1)^{1/2}, 5(1), 4(1)^2 \rangle + \mathbf{C} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fee7991ba-658a-4880-acb7-04815ad6e6cd%2F39a36a73-ca96-448c-823e-7387cd9dac71%2Flf523mq.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Find the location and velocity at \( t = 4 \) of a particle whose path satisfies \( \frac{d\mathbf{r}}{dt} = \langle 3t^{-1/2}, 5, 8t \rangle \) and \( \mathbf{r}(1) = \langle 4, 9, 2 \rangle \).
### Solution
To solve this problem, we need to:
1. **Compute the position vector \( \mathbf{r}(4) \).**
2. **Compute the velocity vector \( \mathbf{r}'(4) \).**
#### Step-by-Step Solution
**Integration of \( \frac{d\mathbf{r}}{dt} \):**
Given:
\[ \frac{d\mathbf{r}}{dt} = \langle 3t^{-1/2}, 5, 8t \rangle \]
Integrate each component with respect to \( t \):
\[ r(t) = \int \langle 3t^{-1/2}, 5, 8t \rangle dt \]
**First Component:**
\[ \int 3t^{-1/2} dt = 3 \int t^{-1/2} dt = 3 \left( \frac{t^{1/2}}{1/2} \right) = 6t^{1/2} \]
**Second Component:**
\[ \int 5 dt = 5t \]
**Third Component:**
\[ \int 8t dt = 8 \left( \frac{t^2}{2} \right) = 4t^2 \]
Thus:
\[ \mathbf{r}(t) = \langle 6t^{1/2}, 5t, 4t^2 \rangle + \mathbf{C} \]
where \( \mathbf{C} \) is the constant vector.
**Finding the Constant Vector \( \mathbf{C} \):**
Using the initial condition \( \mathbf{r}(1) = \langle 4, 9, 2 \rangle \):
\[ \mathbf{r}(1) = \langle 6(1)^{1/2}, 5(1), 4(1)^2 \rangle + \mathbf{C} \]
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