r′(t)
  = ⟨8t+5,2,−3t√⟩


and r(1)
 = ⟨9,3,2⟩


r(t)=

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.

The query submitted by the user: r′(t) = ⟨8t+5,2,−3t√⟩ and r(1) = ⟨9,3,2⟩. Find r(t).

Learn how to find the vector function r(t) given its derivative r′(t) = ⟨8t+5, 2, -3√t⟩ and an initial condition r(1) = ⟨9, 3, 2⟩. This problem covers vector calculus concepts, including integration of vector components and the application of initial conditions to find the constants. Suitable for advanced high school or early college students.

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