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To solve for \(\vec{r}'(t)\), given \(\vec{r}(t) = \langle -2t^2 + 5, -e^{-2t}, -4\sin(2t) \rangle\), we need to differentiate each component of the vector with respect to \(t\).

Let's differentiate each component:

1. First component: \(-2t^2 + 5\)
   \[
   \frac{d}{dt}(-2t^2 + 5) = -4t
   \]

2. Second component: \(-e^{-2t}\)
   \[
   \frac{d}{dt}(-e^{-2t}) = (-1) \cdot \frac{d}{dt}(e^{-2t}) = -(-2)e^{-2t} = 2e^{-2t}
   \]

3. Third component: \(-4\sin(2t)\)
   \[
   \frac{d}{dt}(-4\sin(2t)) = -4 \cdot \frac{d}{dt}(\sin(2t)) = -4 \cdot 2\cos(2t) = -8\cos(2t)
   \]

Thus, \(\vec{r}'(t)\) is:
\[
\vec{r}'(t) = \langle -4t, 2e^{-2t}, -8\cos(2t) \rangle
\]

Would you like further clarification or details? 

Here are 5 related questions to explore this topic further:
1. How do you find the second derivative \(\vec{r}''(t)\) for this problem?
2. What physical interpretations can be derived from the first and second derivatives of a position vector function?
3. How does the chain rule apply in differentiating the exponential and trigonometric functions here?
4. What changes would occur if the coefficients of each function were altered?
5. Can you generalize the process of differentiating a vector-valued function with multiple components?

**Tip:** When differentiating functions like \(e^{kt}\) or \(\sin(kt)\), always remember to apply the chain rule for the constant \(k\).

Find \(\vec{r}'(t)\), given \(\vec{r}(t) = \langle -2t^2 + 5, -e^{-2t}, -4\sin(2t) \rangle\).

Learn how to find the derivative of the vector-valued function \(\vec{r}(t) = \langle -2t^2 + 5, -e^{-2t}, -4\sin(2t) \rangle\). This problem involves key concepts from calculus, including the chain rule and differentiation of exponential and trigonometric functions.

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