To determine the function [tex]\( r(t) \)[/tex] given that [tex]\( r'(t) = 3 t^2 \mathbf{i} + 7 t^6 \mathbf{j} + \sqrt{t} \mathbf{k} \)[/tex] and the initial condition [tex]\( r(1) = \mathbf{i} + \mathbf{j} \)[/tex], follow these steps:
1. Integrate each component of [tex]\( r'(t) \)[/tex] to find [tex]\( r(t) \)[/tex]:
- For the [tex]\( \mathbf{i} \)[/tex] component:
[tex]\[
\int 3 t^2 \, dt = t^3 + C_1
\][/tex]
- For the [tex]\( \mathbf{j} \)[/tex] component:
[tex]\[
\int 7 t^6 \, dt = t^7 + C_2
\][/tex]
- For the [tex]\( \mathbf{k} \)[/tex] component:
[tex]\[
\int \sqrt{t} \, dt = \int t^{1/2} \, dt = \frac{2}{3} t^{3/2} + C_3
\][/tex]
Therefore, the general form of [tex]\( r(t) \)[/tex] before applying the initial conditions is:
[tex]\[
r(t) = (t^3 + C_1) \mathbf{i} + (t^7 + C_2) \mathbf{j} + \left( \frac{2}{3} t^{3/2} + C_3 \right) \mathbf{k}
\][/tex]
2. Apply the initial condition [tex]\( r(1) = \mathbf{i} + \mathbf{j} \)[/tex]:
[tex]\[
r(1) = (\mathbf{i}, \mathbf{j}, \mathbf{k}) = (1, 1, 0)
\][/tex]
Substitute [tex]\( t = 1 \)[/tex] into [tex]\( r(t) \)[/tex] and set it equal to the initial condition [tex]\( (1, 1, 0) \)[/tex]:
- For the [tex]\(\mathbf{i}\)[/tex] component:
[tex]\[
1^3 + C_1 = 1 \implies 1 + C_1 = 1 \implies C_1 = 0
\][/tex]
- For the [tex]\(\mathbf{j}\)[/tex] component:
[tex]\[
1^7 + C_2 = 1 \implies 1 + C_2 = 1 \implies C_2 = 0
\][/tex]
- For the [tex]\(\mathbf{k}\)[/tex] component:
[tex]\[
\frac{2}{3} \cdot 1^{3/2} + C_3 = 0 \implies \frac{2}{3} + C_3 = 0 \implies C_3 = -\frac{2}{3}
\][/tex]
3. Substitute these constants back into [tex]\( r(t) \)[/tex]:
Incorporate [tex]\( C_1 = 0 \)[/tex], [tex]\( C_2 = 0 \)[/tex], and [tex]\( C_3 = -\frac{2}{3} \)[/tex] into the expression for [tex]\( r(t) \)[/tex]:
[tex]\[
r(t) = t^3 \mathbf{i} + t^7 \mathbf{j} + \left( \frac{2}{3} t^{3/2} - \frac{2}{3} \right) \mathbf{k}
\][/tex]
Thus, the function [tex]\( r(t) \)[/tex] is:
[tex]\[
r(t) = t^3 \mathbf{i} + t^7 \mathbf{j} + \left( \frac{2}{3} t^{3/2} - \frac{2}{3} \right) \mathbf{k}
\][/tex]
