To find the position vector $\mathbf{r}(t)$, we start from the acceleration vector $\mathbf{a}(t) = \langle 5t, 3\sin(t), \cos(3t) \rangle$, then integrate it step by step to find the velocity $\mathbf{v}(t)$ and position $\mathbf{r}(t)$.

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Step 1: Find Velocity Vector $\mathbf{v}(t)$

The velocity vector is obtained by integrating the acceleration vector with respect to $t$: $\mathbf{v}(t) = \int \mathbf{a}(t) \, dt = \int \langle 5t, 3\sin(t), \cos(3t) \rangle \, dt$

Integrate each component:

1. $\int 5t \, dt = \frac{5t^2}{2} + C_1$
2. $\int 3\sin(t) \, dt = -3\cos(t) + C_2$
3. $\int \cos(3t) \, dt = \frac{\sin(3t)}{3} + C_3$

So: $\mathbf{v}(t) = \left\langle \frac{5t^2}{2} + C_1, -3\cos(t) + C_2, \frac{\sin(3t)}{3} + C_3 \right\rangle$

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Step 2: Solve for Constants in $\mathbf{v}(t)$

Using the initial velocity $\mathbf{v}(0) = \langle 0, 1, -3 \rangle$, substitute $t = 0$: $\mathbf{v}(0) = \left\langle \frac{5(0)^2}{2} + C_1, -3\cos(0) + C_2, \frac{\sin(3(0))}{3} + C_3 \right\rangle = \langle 0, 1, -3 \rangle$ From each component:

1. $0 = C_1$
2. $1 = -3(1) + C_2 \implies C_2 = 4$
3. $-3 = C_3$

Thus: $\mathbf{v}(t) = \left\langle \frac{5t^2}{2}, -3\cos(t) + 4, \frac{\sin(3t)}{3} - 3 \right\rangle$

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Step 3: Find Position Vector $\mathbf{r}(t)$

The position vector is obtained by integrating the velocity vector $\mathbf{v}(t)$: $\mathbf{r}(t) = \int \mathbf{v}(t) \, dt = \int \left\langle \frac{5t^2}{2}, -3\cos(t) + 4, \frac{\sin(3t)}{3} - 3 \right\rangle \, dt$

Integrate each component:

1. $\int \frac{5t^2}{2} \, dt = \frac{5t^3}{6} + D_1$
2. $\int (-3\cos(t) + 4) \, dt = -3\sin(t) + 4t + D_2$
3. $\int \left(\frac{\sin(3t)}{3} - 3\right) \, dt = -\frac{\cos(3t)}{9} - 3t + D_3$

So: $\mathbf{r}(t) = \left\langle \frac{5t^3}{6} + D_1, -3\sin(t) + 4t + D_2, -\frac{\cos(3t)}{9} - 3t + D_3 \right\rangle$

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Step 4: Solve for Constants in $\mathbf{r}(t)$

Using the initial position $\mathbf{r}(0) = \langle -2, -5, -5 \rangle$, substitute $t = 0$: $\mathbf{r}(0) = \left\langle \frac{5(0)^3}{6} + D_1, -3\sin(0) + 4(0) + D_2, -\frac{\cos(3(0))}{9} - 3(0) + D_3 \right\rangle = \langle -2, -5, -5 \rangle$ From each component:

1. $-2 = D_1$
2. $-5 = D_2$
3. $-5 = -\frac{1}{9} + D_3 \implies D_3 = -\frac{44}{9}$

Thus: $\mathbf{r}(t) = \left\langle \frac{5t^3}{6} - 2, -3\sin(t) + 4t - 5, -\frac{\cos(3t)}{9} - 3t - \frac{44}{9} \right\rangle$

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Final Answer:

$\mathbf{r}(t) = \left\langle \frac{5t^3}{6} - 2, -3\sin(t) + 4t - 5, -\frac{\cos(3t)}{9} - 3t - \frac{44}{9} \right\rangle$

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Do you want a step-by-step explanation of any part or have further questions?

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5 Related Questions:

1. How is the integration of vector components performed?
2. What are the physical interpretations of position, velocity, and acceleration vectors?
3. Can this method be applied for higher-dimensional vector functions?
4. How do initial conditions influence the constants of integration?
5. What are common mistakes to avoid when integrating trigonometric functions?

Tip: Always double-check constants of integration using the initial conditions provided.