We are given the derivative $r'(t)$ and the value of $r(t)$ at $t = 1$, and we are tasked with finding the original function $r(t)$.

Step-by-Step Solution

1. Integrate each component of $r'(t)$:

   The derivative of the vector function is given by: $r'(t) = \langle 8t + 5, 2, -3t^{1/2} \rangle$

   To find $r(t)$, we integrate each component with respect to $t$.

   • First component: $\int (8t + 5) \, dt = 4t^2 + 5t + C_1$

   • Second component: $\int 2 \, dt = 2t + C_2$

   • Third component: $\int -3t^{1/2} \, dt = -2t^{3/2} + C_3$

   So, the general form of $r(t)$ is: $r(t) = \langle 4t^2 + 5t + C_1, 2t + C_2, -2t^{3/2} + C_3 \rangle$

2. Use the initial condition $r(1) = \langle 9, 3, 2 \rangle$ to solve for $C_1, C_2, C_3$:

   We plug $t = 1$ into the general form of $r(t)$ and set it equal to $\langle 9, 3, 2 \rangle$: $\langle 4(1)^2 + 5(1) + C_1, 2(1) + C_2, -2(1)^{3/2} + C_3 \rangle = \langle 9, 3, 2 \rangle$ This gives the following system of equations: $4 + 5 + C_1 = 9 \quad \Rightarrow \quad C_1 = 0$ $2 + C_2 = 3 \quad \Rightarrow \quad C_2 = 1$ $-2 + C_3 = 2 \quad \Rightarrow \quad C_3 = 4$

3. Write the final form of $r(t)$:

   Substituting $C_1 = 0$, $C_2 = 1$, and $C_3 = 4$ into the general form of $r(t)$, we get: $r(t) = \langle 4t^2 + 5t, 2t + 1, -2t^{3/2} + 4 \rangle$

Thus, the vector function $r(t)$ is: $r(t) = \langle 4t^2 + 5t, 2t + 1, -2t^{3/2} + 4 \rangle$

Would you like more details or further clarification?

Here are 5 related questions:

1. How do you find the integral of a vector function?
2. What is the process of solving a system of equations with initial conditions?
3. How does the integration of $t^{1/2}$ work in more general cases?
4. What is the geometrical interpretation of the function $r(t)$?
5. Can you solve a similar problem for different initial conditions?

Tip: Always check your work by differentiating the final solution to ensure it matches the given derivative.