Solve: D₁(r(t) = u(t)) X r(t) = 6ti - 4t³j + 5t²k u(t) = -2i+ 5t²j - 6t³k
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
Please solve the problem provided in the photo
![### Problem Statement
**Solve: \( D_t (\mathbf{r}(t) \times \mathbf{u}(t)) \)**
Given the vectors:
\[ \mathbf{r}(t) = 6t\mathbf{i} - 4t^3\mathbf{j} + 5t^2\mathbf{k} \]
\[ \mathbf{u}(t) = -2\mathbf{i} + 5t^2\mathbf{j} - 6t^3\mathbf{k} \]
### Explanation
We are tasked with finding the derivative of the cross product of two time-dependent vectors, \(\mathbf{r}(t)\) and \(\mathbf{u}(t)\).
#### Steps to Solve
1. **Calculate the Cross Product:**
\[
\mathbf{r}(t) \times \mathbf{u}(t)
\]
2. **Find the Derivative with Respect to \( t \):**
\[
D_t (\mathbf{r}(t) \times \mathbf{u}(t))
\]
### Detailed Steps
1. **Calculate \(\mathbf{r}(t) \times \mathbf{u}(t)\):**
Using the determinant form for the cross product of two vectors \(\mathbf{r}(t)\) and \(\mathbf{u}(t)\):
\[
\mathbf{r}(t) \times \mathbf{u}(t) =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
6t & -4t^3 & 5t^2 \\
-2 & 5t^2 & -6t^3
\end{vmatrix}
\]
Expanding this determinant:
\[
\mathbf{r}(t) \times \mathbf{u}(t) = \mathbf{i} \left( (-4t^3)(-6t^3) - (5t^2)(5t^2) \right) - \mathbf{j} \left( (6t)(-6t^3) - (5t^2)(-2) \right) + \mathbf{k} \left( (6t)(5t^2) - (-4t^3)(-2) \right)
\]
Simplify each component:
\[
= \mathbf{i}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1fc33cc1-e822-44c8-a303-0b5fd2ff9f49%2F07807e53-dcff-4950-966f-46178b7d4338%2Fek5wty_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
**Solve: \( D_t (\mathbf{r}(t) \times \mathbf{u}(t)) \)**
Given the vectors:
\[ \mathbf{r}(t) = 6t\mathbf{i} - 4t^3\mathbf{j} + 5t^2\mathbf{k} \]
\[ \mathbf{u}(t) = -2\mathbf{i} + 5t^2\mathbf{j} - 6t^3\mathbf{k} \]
### Explanation
We are tasked with finding the derivative of the cross product of two time-dependent vectors, \(\mathbf{r}(t)\) and \(\mathbf{u}(t)\).
#### Steps to Solve
1. **Calculate the Cross Product:**
\[
\mathbf{r}(t) \times \mathbf{u}(t)
\]
2. **Find the Derivative with Respect to \( t \):**
\[
D_t (\mathbf{r}(t) \times \mathbf{u}(t))
\]
### Detailed Steps
1. **Calculate \(\mathbf{r}(t) \times \mathbf{u}(t)\):**
Using the determinant form for the cross product of two vectors \(\mathbf{r}(t)\) and \(\mathbf{u}(t)\):
\[
\mathbf{r}(t) \times \mathbf{u}(t) =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
6t & -4t^3 & 5t^2 \\
-2 & 5t^2 & -6t^3
\end{vmatrix}
\]
Expanding this determinant:
\[
\mathbf{r}(t) \times \mathbf{u}(t) = \mathbf{i} \left( (-4t^3)(-6t^3) - (5t^2)(5t^2) \right) - \mathbf{j} \left( (6t)(-6t^3) - (5t^2)(-2) \right) + \mathbf{k} \left( (6t)(5t^2) - (-4t^3)(-2) \right)
\]
Simplify each component:
\[
= \mathbf{i}
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