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The problem asks for the derivative of the vector function \( \mathbf{r}(t) \), defined as:

\[
\mathbf{r}(t) = t\mathbf{a} \times (\mathbf{b} + t\mathbf{c}),
\]

where:
- \( \mathbf{a} = \langle 2, -1, 2 \rangle \),
- \( \mathbf{b} = \langle -2, 3, -5 \rangle \),
- \( \mathbf{c} = \langle -2, 5, 4 \rangle \).

The cross product in this function involves the time-dependent term \( t\mathbf{a} \) and the sum \( (\mathbf{b} + t\mathbf{c}) \).

### Step-by-step approach:

1. **Rewrite the vector function**: 
   
   \[
   \mathbf{r}(t) = t\mathbf{a} \times (\mathbf{b} + t\mathbf{c}) = t\mathbf{a} \times \mathbf{b} + t^2 \mathbf{a} \times \mathbf{c}.
   \]

   This expands into two parts:
   - \( t\mathbf{a} \times \mathbf{b} \),
   - \( t^2 \mathbf{a} \times \mathbf{c} \).

2. **Calculate the cross products**:
   
   - \( \mathbf{a} \times \mathbf{b} = \langle 2, -1, 2 \rangle \times \langle -2, 3, -5 \rangle \),
   - \( \mathbf{a} \times \mathbf{c} = \langle 2, -1, 2 \rangle \times \langle -2, 5, 4 \rangle \).

   Using the determinant method for the cross product:

   \[
   \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 2 \\ -2 & 3 & -5 \end{vmatrix} = \hat{i}( -1 \cdot (-5) - 2 \cdot 3 ) - \hat{j}( 2 \cdot (-5) - 2 \cdot (-2) ) + \hat{k}( 2 \cdot 3 - (-1) \cdot (-2) ),
   \]
   
   Simplifying this:

   \[
   \mathbf{a} \times \mathbf{b} = \hat{i}(5 - 6) - \hat{j}(-10 + 4) + \hat{k}(6 - 2) = \langle -1, 6, 4 \rangle.
   \]

   Similarly, for \( \mathbf{a} \times \mathbf{c} \):

   \[
   \mathbf{a} \times \mathbf{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 2 \\ -2 & 5 & 4 \end{vmatrix} = \hat{i}( -1 \cdot 4 - 2 \cdot 5 ) - \hat{j}( 2 \cdot 4 - 2 \cdot (-2) ) + \hat{k}( 2 \cdot 5 - (-1) \cdot (-2) ),
   \]
   
   Simplifying this:

   \[
   \mathbf{a} \times \mathbf{c} = \hat{i}(-4 - 10) - \hat{j}(8 + 4) + \hat{k}(10 - 2) = \langle -14, -12, 8 \rangle.
   \]

3. **Substitute into the original expression**:

   \[
   \mathbf{r}(t) = t \langle -1, 6, 4 \rangle + t^2 \langle -14, -12, 8 \rangle.
   \]

   Simplifying:

   \[
   \mathbf{r}(t) = \langle -t, 6t, 4t \rangle + \langle -14t^2, -12t^2, 8t^2 \rangle.
   \]

   Adding the components:

   \[
   \mathbf{r}(t) = \langle -t - 14t^2, 6t - 12t^2, 4t + 8t^2 \rangle.
   \]

4. **Differentiate with respect to \( t \)**:

   \[
   \mathbf{r}'(t) = \frac{d}{dt} \langle -t - 14t^2, 6t - 12t^2, 4t + 8t^2 \rangle.
   \]

   Differentiating component-wise:

   \[
   \mathbf{r}'(t) = \langle -1 - 28t, 6 - 24t, 4 + 16t \rangle.
   \]

Thus, the derivative is:

\[
\mathbf{r}'(t) = \langle -1 - 28t, 6 - 24t, 4 + 16t \rangle,
\]

which matches the given solution.

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Would you like a further breakdown of any step?

Here are 5 related questions to explore:

1. What is the geometric meaning of the cross product in vector calculus?
2. How can we apply this method to find the derivative of a more complex vector function?
3. Why do we use the product rule for vector-valued functions?
4. What are some applications of cross products in physics?
5. Can we use this method to analyze higher-dimensional vector functions?

**Tip:** When calculating cross products, always use the determinant method to keep track of signs and components systematically.

Find the derivative of the vector function r(t) = t a × (b + tc), where a = <2, -1, 2>, b = <-2, 3, -5>, and c = <-2, 5, 4>.

Explore the step-by-step solution for finding the derivative of a vector function involving cross products and time-dependent terms. Learn about vector calculus, cross products, and the differentiation of vector functions using the product rule. Suitable for undergraduate students in Calculus II or III.

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