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
![**Vector Calculus Problem: Velocity and Acceleration**
Consider the curve given by the vector function \( \mathbf{r}(t) = (t^2 + 1)\mathbf{i} + t\mathbf{j} + (2t^2 + 3t)\mathbf{k} \).
### Problems:
1. **Velocity**: Determine the velocity vector \( \mathbf{v}(t) \) of the curve.
2. **Acceleration**: Determine the acceleration vector \( \mathbf{a}(t) \) of the curve.
3. **Constancy**: Assess whether these vectors are constant.
### Solution:
#### Step 1: Find Velocity
- **Velocity, \( \mathbf{v}(t) \)**, is the first derivative of the position vector \( \mathbf{r}(t) \).
Given:
\[ \mathbf{r}(t) = (t^2 + 1)\mathbf{i} + t\mathbf{j} + (2t^2 + 3t)\mathbf{k} \]
Differentiating component-wise:
- For the \(\mathbf{i}\) component: \(\frac{d}{dt}(t^2 + 1) = 2t\)
- For the \(\mathbf{j}\) component: \(\frac{d}{dt}(t) = 1\)
- For the \(\mathbf{k}\) component: \(\frac{d}{dt}(2t^2 + 3t) = 4t + 3\)
Thus:
\[ \mathbf{v}(t) = 2t\mathbf{i} + 1\mathbf{j} + (4t + 3)\mathbf{k} \]
#### Step 2: Find Acceleration
- **Acceleration, \( \mathbf{a}(t) \)**, is the derivative of the velocity vector \( \mathbf{v}(t) \).
Given:
\[ \mathbf{v}(t) = 2t\mathbf{i} + 1\mathbf{j} + (4t + 3)\mathbf{k} \]
Differentiating component-wise:
- For the \(\mathbf{i}\) component: \(\frac{d}{dt}(2t) = 2\)
- For the \(\mathbf{j}\) component: \(\frac{d}{dt}(1) = 0\)
- For the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F934ff6bb-8e29-4028-baf5-67e0d726d963%2Fcdc5a612-b52f-4d3a-8e42-b91ba3a5947b%2Fwf8f0kq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Vector Calculus Problem: Velocity and Acceleration**
Consider the curve given by the vector function \( \mathbf{r}(t) = (t^2 + 1)\mathbf{i} + t\mathbf{j} + (2t^2 + 3t)\mathbf{k} \).
### Problems:
1. **Velocity**: Determine the velocity vector \( \mathbf{v}(t) \) of the curve.
2. **Acceleration**: Determine the acceleration vector \( \mathbf{a}(t) \) of the curve.
3. **Constancy**: Assess whether these vectors are constant.
### Solution:
#### Step 1: Find Velocity
- **Velocity, \( \mathbf{v}(t) \)**, is the first derivative of the position vector \( \mathbf{r}(t) \).
Given:
\[ \mathbf{r}(t) = (t^2 + 1)\mathbf{i} + t\mathbf{j} + (2t^2 + 3t)\mathbf{k} \]
Differentiating component-wise:
- For the \(\mathbf{i}\) component: \(\frac{d}{dt}(t^2 + 1) = 2t\)
- For the \(\mathbf{j}\) component: \(\frac{d}{dt}(t) = 1\)
- For the \(\mathbf{k}\) component: \(\frac{d}{dt}(2t^2 + 3t) = 4t + 3\)
Thus:
\[ \mathbf{v}(t) = 2t\mathbf{i} + 1\mathbf{j} + (4t + 3)\mathbf{k} \]
#### Step 2: Find Acceleration
- **Acceleration, \( \mathbf{a}(t) \)**, is the derivative of the velocity vector \( \mathbf{v}(t) \).
Given:
\[ \mathbf{v}(t) = 2t\mathbf{i} + 1\mathbf{j} + (4t + 3)\mathbf{k} \]
Differentiating component-wise:
- For the \(\mathbf{i}\) component: \(\frac{d}{dt}(2t) = 2\)
- For the \(\mathbf{j}\) component: \(\frac{d}{dt}(1) = 0\)
- For the
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