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To solve the vector differential equation [tex]\(\mathbf{r}'(t) = \cos(7t)\mathbf{i} + \sin(7t)\mathbf{j} + 7\mathbf{k}\)[/tex] with the initial condition [tex]\(\mathbf{r}(0) = \mathbf{i} + \mathbf{k}\)[/tex], we follow these steps:
1. Component Form: Decompose the differential equation into its component vectors:
[tex]\[ \mathbf{r}'(t) = \cos(7t)\mathbf{i} + \sin(7t)\mathbf{j} + 7\mathbf{k} \][/tex]
means:
[tex]\[ \begin{cases} x'(t) = \cos(7t) \\ y'(t) = \sin(7t) \\ z'(t) = 7 \end{cases} \][/tex]
2. Integration:
- Integrate [tex]\(x'(t)\)[/tex]:
[tex]\[ x(t) = \int \cos(7t) \, dt = \frac{1}{7} \sin(7t) + C_1 \][/tex]
- Integrate [tex]\(y'(t)\)[/tex]:
[tex]\[ y(t) = \int \sin(7t) \, dt = -\frac{1}{7} \cos(7t) + C_2 \][/tex]
- Integrate [tex]\(z'(t)\)[/tex]:
[tex]\[ z(t) = \int 7 \, dt = 7t + C_3 \][/tex]
3. Initial Condition:
Use the initial condition [tex]\(\mathbf{r}(0) = \mathbf{i} + \mathbf{k}\)[/tex] to find the constants [tex]\(C_1\)[/tex], [tex]\(C_2\)[/tex], and [tex]\(C_3\)[/tex].
- For [tex]\(x(t)\)[/tex]:
[tex]\[ x(0) = 1 \quad \text{(from \(\mathbf{i}\))} \\ \frac{1}{7} \sin(0) + C_1 = 1 \implies C_1 = 1 \][/tex]
- For [tex]\(y(t)\)[/tex]:
[tex]\[ y(0) = 0 \quad \text{(there is no \(\mathbf{j}\) component initially)} \\ -\frac{1}{7} \cos(0) + C_2 = 0 \implies -\frac{1}{7} + C_2 = 0 \implies C_2 = \frac{1}{7} \][/tex]
- For [tex]\(z(t)\)[/tex]:
[tex]\[ z(0) = 1 \quad \text{(from \(\mathbf{k}\))} \\ 7 \cdot 0 + C_3 = 1 \implies C_3 = 1 \][/tex]
4. Solution:
Substitute the constants back into the integrated functions:
[tex]\[ x(t) = \frac{1}{7} \sin(7t) + 1 \][/tex]
[tex]\[ y(t) = -\frac{1}{7} \cos(7t) + \frac{1}{7} \][/tex]
[tex]\[ z(t) = 7t + 1 \][/tex]
5. Vector Form:
Combine the component functions into the vector [tex]\(\mathbf{r}(t)\)[/tex]:
[tex]\[ \mathbf{r}(t) = \left( \frac{1}{7} \sin(7t) + 1 \right) \mathbf{i} + \left( -\frac{1}{7} \cos(7t) + \frac{1}{7} \right) \mathbf{j} + \left( 7t + 1 \right) \mathbf{k} \][/tex]
So, the solution to the vector differential equation [tex]\(\mathbf{r}'(t) = \cos(7t)\mathbf{i} + \sin(7t)\mathbf{j} + 7\mathbf{k}\)[/tex] with the initial condition [tex]\(\mathbf{r}(0) = \mathbf{i} + \mathbf{k}\)[/tex] is:
[tex]\[ \mathbf{r}(t) = \left[ \frac{1}{7} \sin(7t) + 1 \right] \mathbf{i} + \left[ -\frac{1}{7} \cos(7t) + \frac{1}{7} \right] \mathbf{j} + \left[ 7t + 1 \right] \mathbf{k} \][/tex]
1. Component Form: Decompose the differential equation into its component vectors:
[tex]\[ \mathbf{r}'(t) = \cos(7t)\mathbf{i} + \sin(7t)\mathbf{j} + 7\mathbf{k} \][/tex]
means:
[tex]\[ \begin{cases} x'(t) = \cos(7t) \\ y'(t) = \sin(7t) \\ z'(t) = 7 \end{cases} \][/tex]
2. Integration:
- Integrate [tex]\(x'(t)\)[/tex]:
[tex]\[ x(t) = \int \cos(7t) \, dt = \frac{1}{7} \sin(7t) + C_1 \][/tex]
- Integrate [tex]\(y'(t)\)[/tex]:
[tex]\[ y(t) = \int \sin(7t) \, dt = -\frac{1}{7} \cos(7t) + C_2 \][/tex]
- Integrate [tex]\(z'(t)\)[/tex]:
[tex]\[ z(t) = \int 7 \, dt = 7t + C_3 \][/tex]
3. Initial Condition:
Use the initial condition [tex]\(\mathbf{r}(0) = \mathbf{i} + \mathbf{k}\)[/tex] to find the constants [tex]\(C_1\)[/tex], [tex]\(C_2\)[/tex], and [tex]\(C_3\)[/tex].
- For [tex]\(x(t)\)[/tex]:
[tex]\[ x(0) = 1 \quad \text{(from \(\mathbf{i}\))} \\ \frac{1}{7} \sin(0) + C_1 = 1 \implies C_1 = 1 \][/tex]
- For [tex]\(y(t)\)[/tex]:
[tex]\[ y(0) = 0 \quad \text{(there is no \(\mathbf{j}\) component initially)} \\ -\frac{1}{7} \cos(0) + C_2 = 0 \implies -\frac{1}{7} + C_2 = 0 \implies C_2 = \frac{1}{7} \][/tex]
- For [tex]\(z(t)\)[/tex]:
[tex]\[ z(0) = 1 \quad \text{(from \(\mathbf{k}\))} \\ 7 \cdot 0 + C_3 = 1 \implies C_3 = 1 \][/tex]
4. Solution:
Substitute the constants back into the integrated functions:
[tex]\[ x(t) = \frac{1}{7} \sin(7t) + 1 \][/tex]
[tex]\[ y(t) = -\frac{1}{7} \cos(7t) + \frac{1}{7} \][/tex]
[tex]\[ z(t) = 7t + 1 \][/tex]
5. Vector Form:
Combine the component functions into the vector [tex]\(\mathbf{r}(t)\)[/tex]:
[tex]\[ \mathbf{r}(t) = \left( \frac{1}{7} \sin(7t) + 1 \right) \mathbf{i} + \left( -\frac{1}{7} \cos(7t) + \frac{1}{7} \right) \mathbf{j} + \left( 7t + 1 \right) \mathbf{k} \][/tex]
So, the solution to the vector differential equation [tex]\(\mathbf{r}'(t) = \cos(7t)\mathbf{i} + \sin(7t)\mathbf{j} + 7\mathbf{k}\)[/tex] with the initial condition [tex]\(\mathbf{r}(0) = \mathbf{i} + \mathbf{k}\)[/tex] is:
[tex]\[ \mathbf{r}(t) = \left[ \frac{1}{7} \sin(7t) + 1 \right] \mathbf{i} + \left[ -\frac{1}{7} \cos(7t) + \frac{1}{7} \right] \mathbf{j} + \left[ 7t + 1 \right] \mathbf{k} \][/tex]
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