To find the position vector $r(t)$, we will need to integrate the given acceleration vector twice.

Step 1: Find the velocity vector $v(t)$

The acceleration vector $a(t) = \langle t, e^t, \sin(2t) \rangle$ is the derivative of the velocity vector $v(t)$. To find $v(t)$, we integrate each component of $a(t)$.

$v(t) = \int a(t) \, dt = \int \langle t, e^t, \sin(2t) \rangle \, dt$

First component:

$\int t \, dt = \frac{t^2}{2} + C_1$

Second component:

$\int e^t \, dt = e^t + C_2$

Third component:

$\int \sin(2t) \, dt = -\frac{1}{2} \cos(2t) + C_3$

Thus, the velocity vector is: $v(t) = \left\langle \frac{t^2}{2} + C_1, e^t + C_2, -\frac{1}{2} \cos(2t) + C_3 \right\rangle$

Now, we apply the initial condition $v(0) = \langle 1, 1, 1 \rangle$ to determine the constants $C_1, C_2, C_3$.

At $t = 0$: $v(0) = \left\langle \frac{0^2}{2} + C_1, e^0 + C_2, -\frac{1}{2} \cos(0) + C_3 \right\rangle = \langle 1, 1, 1 \rangle$ $\Rightarrow \langle C_1, 1 + C_2, -\frac{1}{2} + C_3 \rangle = \langle 1, 1, 1 \rangle$

From this, we get: $C_1 = 1, \quad C_2 = 0, \quad C_3 = \frac{3}{2}$

Thus, the velocity vector is: $v(t) = \left\langle \frac{t^2}{2} + 1, e^t, -\frac{1}{2} \cos(2t) + \frac{3}{2} \right\rangle$

Step 2: Find the position vector $r(t)$

To find the position vector $r(t)$, we integrate the velocity vector:

$r(t) = \int v(t) \, dt = \int \left\langle \frac{t^2}{2} + 1, e^t, -\frac{1}{2} \cos(2t) + \frac{3}{2} \right\rangle \, dt$

First component:

$\int \left( \frac{t^2}{2} + 1 \right) \, dt = \frac{t^3}{6} + t + D_1$

Second component:

$\int e^t \, dt = e^t + D_2$

Third component:

$\int \left( -\frac{1}{2} \cos(2t) + \frac{3}{2} \right) \, dt = -\frac{1}{4} \sin(2t) + \frac{3}{2}t + D_3$

Thus, the position vector is: $r(t) = \left\langle \frac{t^3}{6} + t + D_1, e^t + D_2, -\frac{1}{4} \sin(2t) + \frac{3}{2} t + D_3 \right\rangle$

Now, we apply the initial condition $r(0) = \langle 0, 0, 0 \rangle$ to determine the constants $D_1, D_2, D_3$.

At $t = 0$: $r(0) = \left\langle \frac{0^3}{6} + 0 + D_1, e^0 + D_2, -\frac{1}{4} \sin(0) + \frac{3}{2} \cdot 0 + D_3 \right\rangle = \langle 0, 0, 0 \rangle$ $\Rightarrow \langle D_1, 1 + D_2, D_3 \rangle = \langle 0, 0, 0 \rangle$

From this, we get: $D_1 = 0, \quad D_2 = -1, \quad D_3 = 0$

Thus, the position vector is: $r(t) = \left\langle \frac{t^3}{6} + t, e^t - 1, -\frac{1}{4} \sin(2t) + \frac{3}{2} t \right\rangle$

Final Answer:

The formula for the position vector is: $r(t) = \left\langle \frac{t^3}{6} + t, e^t - 1, -\frac{1}{4} \sin(2t) + \frac{3}{2} t \right\rangle$

Would you like more details on any of these steps or further explanation?

Related Questions:

1. How do you compute the velocity from an acceleration vector in general?
2. Can this process be extended to higher-dimensional problems?
3. How do initial conditions affect the constants of integration in vector calculus?
4. What is the physical interpretation of integrating acceleration to find velocity and position?
5. How would the result change if different initial conditions were given for velocity and position?

Tip:

Always apply the initial conditions after integrating to find the unknown constants—this ensures the solution matches the given conditions.