Answer :
To find [tex]\( r(t) \)[/tex] given the vector-valued function [tex]\( r'(t) = t^2 \mathbf{i} + e^t \mathbf{j} + 2t e^{2t} \mathbf{k} \)[/tex] and the initial condition [tex]\( r(0) = \mathbf{i} + \mathbf{j} + \mathbf{k} \)[/tex], we perform the following steps:
1. Integrate each component of [tex]\( r'(t) \)[/tex] with respect to [tex]\( t \)[/tex] to find [tex]\( r(t) \)[/tex]:
- For the [tex]\(\mathbf{i}\)[/tex] component:
[tex]\[ \int t^2 \, dt = \frac{t^3}{3} + C_1 \][/tex]
- For the [tex]\(\mathbf{j}\)[/tex] component:
[tex]\[ \int e^t \, dt = e^t + C_2 \][/tex]
- For the [tex]\(\mathbf{k}\)[/tex] component:
[tex]\[ \int 2t e^{2t} \, dt = \int (2t)e^{2t} \, dt \][/tex]
To solve this, we use integration by parts:
Let [tex]\( u = 2t \)[/tex] and [tex]\( dv = e^{2t} dt \)[/tex].
Then, [tex]\( du = 2 dt \)[/tex] and [tex]\( v = \frac{e^{2t}}{2} \)[/tex].
Applying integration by parts,
[tex]\[ \int 2t e^{2t} \, dt = t e^{2t} - \int e^{2t} \, dt = t e^{2t} - \frac{e^{2t}}{2} + C_3 = \left(2t - \frac{1}{2}\right) \frac{e^{2t}}{2} + C_3 \][/tex]
Simplifying,
[tex]\[ \int 2t e^{2t} \, dt = \left(2t - 1\right) \frac{e^{2t}}{2} + C_3 \][/tex]
2. Combine the integrated results with their respective constants:
Thus, we have:
[tex]\[ r(t) = \frac{t^3}{3} \mathbf{i} + e^t \mathbf{j} + \left(\left(2t - 1\right) \frac{e^{2t}}{2} + C_3\right) \mathbf{k} \][/tex]
3. Determine the constants using the initial condition [tex]\( r(0) = \mathbf{i} + \mathbf{j} + \mathbf{k} \)[/tex]:
At [tex]\( t = 0 \)[/tex]:
[tex]\[ r(0) = \frac{0^3}{3} \mathbf{i} + e^0 \mathbf{j} + \left(\left(2 \cdot 0 - 1\right) \frac{e^{2 \cdot 0}}{2} + C_3\right) \mathbf{k} = 0 \mathbf{i} + 1\mathbf{j} + \left(-\frac{1}{2} + C_3 \right)\mathbf{k} \][/tex]
This simplifies to:
[tex]\[ r(0) = \mathbf{j} + \left(-\frac{1}{2} + C_3 \right) \mathbf{k} \][/tex]
Given [tex]\( r(0) = \mathbf{i} + \mathbf{j} + \mathbf{k} \)[/tex], we equate components:
[tex]\[ 0 \mathbf{i} = \mathbf{i} \implies C_1 = 1 \][/tex]
[tex]\[ 1 \mathbf{j} = 1 \mathbf{j} \implies C_2 = 0 \][/tex]
[tex]\[ \left(-\frac{1}{2} + C_3 \right) \mathbf{k} = \mathbf{k} \implies -\frac{1}{2} + C_3 = 1 \implies C_3 = \frac{3}{2} \][/tex]
4. Substitute the constants [tex]\( C_1 \)[/tex], [tex]\( C_2 \)[/tex], and [tex]\( C_3 \)[/tex] back into the integrated components:
Finally, we get:
[tex]\[ r(t) = \left( \frac{t^3}{3} + 1 \right) \mathbf{i} + \left( e^t \right) \mathbf{j} + \left( \frac{(2t - 1)e^{2t}}{2} + \frac{3}{2} \right) \mathbf{k} \][/tex]
Thus, the solution is:
[tex]\[ r(t) = \frac{t^3}{3} + 1, \quad r_j(t) = e^t, \quad r_k(t) = \frac{(2t - 1)e^{2t}}{2} + \frac{3}{2} \][/tex]
So,
[tex]\[ r(t) = \left(\frac{t^3}{3} + 1 \right) \mathbf{i} + e^t \mathbf{j} + \left( \frac{(2t - 1)e^{2t}}{2} + \frac{3}{2} \right) \mathbf{k} \][/tex]
1. Integrate each component of [tex]\( r'(t) \)[/tex] with respect to [tex]\( t \)[/tex] to find [tex]\( r(t) \)[/tex]:
- For the [tex]\(\mathbf{i}\)[/tex] component:
[tex]\[ \int t^2 \, dt = \frac{t^3}{3} + C_1 \][/tex]
- For the [tex]\(\mathbf{j}\)[/tex] component:
[tex]\[ \int e^t \, dt = e^t + C_2 \][/tex]
- For the [tex]\(\mathbf{k}\)[/tex] component:
[tex]\[ \int 2t e^{2t} \, dt = \int (2t)e^{2t} \, dt \][/tex]
To solve this, we use integration by parts:
Let [tex]\( u = 2t \)[/tex] and [tex]\( dv = e^{2t} dt \)[/tex].
Then, [tex]\( du = 2 dt \)[/tex] and [tex]\( v = \frac{e^{2t}}{2} \)[/tex].
Applying integration by parts,
[tex]\[ \int 2t e^{2t} \, dt = t e^{2t} - \int e^{2t} \, dt = t e^{2t} - \frac{e^{2t}}{2} + C_3 = \left(2t - \frac{1}{2}\right) \frac{e^{2t}}{2} + C_3 \][/tex]
Simplifying,
[tex]\[ \int 2t e^{2t} \, dt = \left(2t - 1\right) \frac{e^{2t}}{2} + C_3 \][/tex]
2. Combine the integrated results with their respective constants:
Thus, we have:
[tex]\[ r(t) = \frac{t^3}{3} \mathbf{i} + e^t \mathbf{j} + \left(\left(2t - 1\right) \frac{e^{2t}}{2} + C_3\right) \mathbf{k} \][/tex]
3. Determine the constants using the initial condition [tex]\( r(0) = \mathbf{i} + \mathbf{j} + \mathbf{k} \)[/tex]:
At [tex]\( t = 0 \)[/tex]:
[tex]\[ r(0) = \frac{0^3}{3} \mathbf{i} + e^0 \mathbf{j} + \left(\left(2 \cdot 0 - 1\right) \frac{e^{2 \cdot 0}}{2} + C_3\right) \mathbf{k} = 0 \mathbf{i} + 1\mathbf{j} + \left(-\frac{1}{2} + C_3 \right)\mathbf{k} \][/tex]
This simplifies to:
[tex]\[ r(0) = \mathbf{j} + \left(-\frac{1}{2} + C_3 \right) \mathbf{k} \][/tex]
Given [tex]\( r(0) = \mathbf{i} + \mathbf{j} + \mathbf{k} \)[/tex], we equate components:
[tex]\[ 0 \mathbf{i} = \mathbf{i} \implies C_1 = 1 \][/tex]
[tex]\[ 1 \mathbf{j} = 1 \mathbf{j} \implies C_2 = 0 \][/tex]
[tex]\[ \left(-\frac{1}{2} + C_3 \right) \mathbf{k} = \mathbf{k} \implies -\frac{1}{2} + C_3 = 1 \implies C_3 = \frac{3}{2} \][/tex]
4. Substitute the constants [tex]\( C_1 \)[/tex], [tex]\( C_2 \)[/tex], and [tex]\( C_3 \)[/tex] back into the integrated components:
Finally, we get:
[tex]\[ r(t) = \left( \frac{t^3}{3} + 1 \right) \mathbf{i} + \left( e^t \right) \mathbf{j} + \left( \frac{(2t - 1)e^{2t}}{2} + \frac{3}{2} \right) \mathbf{k} \][/tex]
Thus, the solution is:
[tex]\[ r(t) = \frac{t^3}{3} + 1, \quad r_j(t) = e^t, \quad r_k(t) = \frac{(2t - 1)e^{2t}}{2} + \frac{3}{2} \][/tex]
So,
[tex]\[ r(t) = \left(\frac{t^3}{3} + 1 \right) \mathbf{i} + e^t \mathbf{j} + \left( \frac{(2t - 1)e^{2t}}{2} + \frac{3}{2} \right) \mathbf{k} \][/tex]
