### Problem Statement Find the speed over the path \( \mathbf{r}(t) = \langle 2t + 4, 8t + 2, 3t - 7 \rangle \) at \( t = 8 \). (Use symbolic notation and fractions where needed.) ### Attempted Solution \[ \mathbf{v}(8) = \langle 2, 8, 3 \rangle \] **Result: Incorrect** ### Explanation To find the speed of the particle at \( t = 8 \) for the path \( \mathbf{r}(t) \), we need to follow these steps: 1. **Differentiate** each component of \( \mathbf{r}(t) \) with respect to \( t \) to find the velocity vector, \( \mathbf{v}(t) = \mathbf{r}'(t) \): - For \( x(t) = 2t + 4 \): \[ \frac{{dx}}{{dt}} = 2 \] - For \( y(t) = 8t + 2 \): \[ \frac{{dy}}{{dt}} = 8 \] - For \( z(t) = 3t - 7 \): \[ \frac{{dz}}{{dt}} = 3 \] Therefore, the velocity vector is: \[ \mathbf{v}(t) = \langle 2, 8, 3 \rangle \] 2. **Evaluate the velocity** vector at \( t = 8 \): \[ \mathbf{v}(8) = \langle 2, 8, 3 \rangle \] 3. **Calculate the speed** as the magnitude of the velocity vector: \[ \text{Speed} = \| \mathbf{v}(8) \| = \sqrt{2^2 + 8^2 + 3^2} \] Simplify the calculation: \[ \sqrt{2^2 + 8^2 + 3^2} = \sqrt{4 + 64 + 9} = \sqrt{77} \] Therefore, the speed of the particle at \( t = 8 \) is \( \sqrt{
### Problem Statement Find the speed over the path \( \mathbf{r}(t) = \langle 2t + 4, 8t + 2, 3t - 7 \rangle \) at \( t = 8 \). (Use symbolic notation and fractions where needed.) ### Attempted Solution \[ \mathbf{v}(8) = \langle 2, 8, 3 \rangle \] **Result: Incorrect** ### Explanation To find the speed of the particle at \( t = 8 \) for the path \( \mathbf{r}(t) \), we need to follow these steps: 1. **Differentiate** each component of \( \mathbf{r}(t) \) with respect to \( t \) to find the velocity vector, \( \mathbf{v}(t) = \mathbf{r}'(t) \): - For \( x(t) = 2t + 4 \): \[ \frac{{dx}}{{dt}} = 2 \] - For \( y(t) = 8t + 2 \): \[ \frac{{dy}}{{dt}} = 8 \] - For \( z(t) = 3t - 7 \): \[ \frac{{dz}}{{dt}} = 3 \] Therefore, the velocity vector is: \[ \mathbf{v}(t) = \langle 2, 8, 3 \rangle \] 2. **Evaluate the velocity** vector at \( t = 8 \): \[ \mathbf{v}(8) = \langle 2, 8, 3 \rangle \] 3. **Calculate the speed** as the magnitude of the velocity vector: \[ \text{Speed} = \| \mathbf{v}(8) \| = \sqrt{2^2 + 8^2 + 3^2} \] Simplify the calculation: \[ \sqrt{2^2 + 8^2 + 3^2} = \sqrt{4 + 64 + 9} = \sqrt{77} \] Therefore, the speed of the particle at \( t = 8 \) is \( \sqrt{
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I really need help, it is incorrect
![### Problem Statement
Find the speed over the path \( \mathbf{r}(t) = \langle 2t + 4, 8t + 2, 3t - 7 \rangle \) at \( t = 8 \).
(Use symbolic notation and fractions where needed.)
### Attempted Solution
\[ \mathbf{v}(8) = \langle 2, 8, 3 \rangle \]
**Result: Incorrect**
### Explanation
To find the speed of the particle at \( t = 8 \) for the path \( \mathbf{r}(t) \), we need to follow these steps:
1. **Differentiate** each component of \( \mathbf{r}(t) \) with respect to \( t \) to find the velocity vector, \( \mathbf{v}(t) = \mathbf{r}'(t) \):
- For \( x(t) = 2t + 4 \):
\[
\frac{{dx}}{{dt}} = 2
\]
- For \( y(t) = 8t + 2 \):
\[
\frac{{dy}}{{dt}} = 8
\]
- For \( z(t) = 3t - 7 \):
\[
\frac{{dz}}{{dt}} = 3
\]
Therefore, the velocity vector is:
\[
\mathbf{v}(t) = \langle 2, 8, 3 \rangle
\]
2. **Evaluate the velocity** vector at \( t = 8 \):
\[
\mathbf{v}(8) = \langle 2, 8, 3 \rangle
\]
3. **Calculate the speed** as the magnitude of the velocity vector:
\[
\text{Speed} = \| \mathbf{v}(8) \| = \sqrt{2^2 + 8^2 + 3^2}
\]
Simplify the calculation:
\[
\sqrt{2^2 + 8^2 + 3^2} = \sqrt{4 + 64 + 9} = \sqrt{77}
\]
Therefore, the speed of the particle at \( t = 8 \) is \( \sqrt{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe815a6c0-dcfd-4fc5-aa4e-03e2f1f5f601%2Fcdab5b66-82dc-4cfb-93cd-81c79b8db6d7%2Fg7yjtz_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Find the speed over the path \( \mathbf{r}(t) = \langle 2t + 4, 8t + 2, 3t - 7 \rangle \) at \( t = 8 \).
(Use symbolic notation and fractions where needed.)
### Attempted Solution
\[ \mathbf{v}(8) = \langle 2, 8, 3 \rangle \]
**Result: Incorrect**
### Explanation
To find the speed of the particle at \( t = 8 \) for the path \( \mathbf{r}(t) \), we need to follow these steps:
1. **Differentiate** each component of \( \mathbf{r}(t) \) with respect to \( t \) to find the velocity vector, \( \mathbf{v}(t) = \mathbf{r}'(t) \):
- For \( x(t) = 2t + 4 \):
\[
\frac{{dx}}{{dt}} = 2
\]
- For \( y(t) = 8t + 2 \):
\[
\frac{{dy}}{{dt}} = 8
\]
- For \( z(t) = 3t - 7 \):
\[
\frac{{dz}}{{dt}} = 3
\]
Therefore, the velocity vector is:
\[
\mathbf{v}(t) = \langle 2, 8, 3 \rangle
\]
2. **Evaluate the velocity** vector at \( t = 8 \):
\[
\mathbf{v}(8) = \langle 2, 8, 3 \rangle
\]
3. **Calculate the speed** as the magnitude of the velocity vector:
\[
\text{Speed} = \| \mathbf{v}(8) \| = \sqrt{2^2 + 8^2 + 3^2}
\]
Simplify the calculation:
\[
\sqrt{2^2 + 8^2 + 3^2} = \sqrt{4 + 64 + 9} = \sqrt{77}
\]
Therefore, the speed of the particle at \( t = 8 \) is \( \sqrt{
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