Find the speed: r(t) = 9ti + 6tj + 4tk
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section: Chapter Questions
Problem 39RE
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Question
Find the speed and refer back to the photo provided
![**Finding Speed from a Vector Function**
**Problem Statement:**
Given the vector function \( \mathbf{r}(t) = 9t\mathbf{i} + 6t\mathbf{j} + 4t\mathbf{k} \), find the speed.
**Solution:**
To find the speed, we need to calculate the magnitude of the velocity vector \( \mathbf{v}(t) \). The velocity vector is the derivative of the position vector function \( \mathbf{r}(t) \).
1. Given:
\[
\mathbf{r}(t) = 9t\mathbf{i} + 6t\mathbf{j} + 4t\mathbf{k}
\]
2. Find the derivative of \( \mathbf{r}(t) \) with respect to \( t \) to obtain \( \mathbf{v}(t) \):
\[
\mathbf{v}(t) = \frac{d}{dt}(9t\mathbf{i} + 6t\mathbf{j} + 4t\mathbf{k}) = 9\mathbf{i} + 6\mathbf{j} + 4\mathbf{k}
\]
3. The speed is the magnitude of the velocity vector \( \mathbf{v}(t) \):
\[
\text{Speed} = \|\mathbf{v}(t)\| = \sqrt{(9)^2 + (6)^2 + (4)^2}
\]
4. Calculate the magnitude:
\[
\text{Speed} = \sqrt{81 + 36 + 16} = \sqrt{133}
\]
Therefore, the speed is \( \sqrt{133} \) units per time unit.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1fc33cc1-e822-44c8-a303-0b5fd2ff9f49%2F7426eba5-82e1-4d8d-9d04-9a7b83820c65%2F85b1r57_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Finding Speed from a Vector Function**
**Problem Statement:**
Given the vector function \( \mathbf{r}(t) = 9t\mathbf{i} + 6t\mathbf{j} + 4t\mathbf{k} \), find the speed.
**Solution:**
To find the speed, we need to calculate the magnitude of the velocity vector \( \mathbf{v}(t) \). The velocity vector is the derivative of the position vector function \( \mathbf{r}(t) \).
1. Given:
\[
\mathbf{r}(t) = 9t\mathbf{i} + 6t\mathbf{j} + 4t\mathbf{k}
\]
2. Find the derivative of \( \mathbf{r}(t) \) with respect to \( t \) to obtain \( \mathbf{v}(t) \):
\[
\mathbf{v}(t) = \frac{d}{dt}(9t\mathbf{i} + 6t\mathbf{j} + 4t\mathbf{k}) = 9\mathbf{i} + 6\mathbf{j} + 4\mathbf{k}
\]
3. The speed is the magnitude of the velocity vector \( \mathbf{v}(t) \):
\[
\text{Speed} = \|\mathbf{v}(t)\| = \sqrt{(9)^2 + (6)^2 + (4)^2}
\]
4. Calculate the magnitude:
\[
\text{Speed} = \sqrt{81 + 36 + 16} = \sqrt{133}
\]
Therefore, the speed is \( \sqrt{133} \) units per time unit.
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