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The problem provided involves finding the velocity of a particle whose position is modeled by the function:

\[
f(t) = a \left( \sin(t^6) \right)^{10}
\]

To find the velocity function, we need to take the derivative of the position function \( f(t) \) with respect to time \( t \). The velocity \( v(t) \) is the derivative \( f'(t) \).

Let’s differentiate the function step-by-step:

1. **Function:** \( f(t) = a \left( \sin(t^6) \right)^{10} \)
   - Since \( a \) is a constant, we can ignore it during differentiation and multiply it later.

2. **Applying the Chain Rule:** 
   - We have an outer function \( \left( \sin(t^6) \right)^{10} \). To differentiate, we first apply the power rule to the outer exponent and then differentiate the inner functions.

   Let’s calculate:

   \[
   f'(t) = a \cdot 10 \left( \sin(t^6) \right)^9 \cdot \frac{d}{dt} \left( \sin(t^6) \right)
   \]

3. **Differentiate \( \sin(t^6) \):**
   - The derivative of \( \sin(t^6) \) requires the chain rule again.
   \[
   \frac{d}{dt} \sin(t^6) = \cos(t^6) \cdot \frac{d}{dt} t^6
   \]
   - Now differentiate \( t^6 \):
   \[
   \frac{d}{dt} t^6 = 6t^5
   \]
   - So, \( \frac{d}{dt} \sin(t^6) = \cos(t^6) \cdot 6t^5 \).

4. **Substitute back:**
   - Now, we substitute this derivative into our earlier expression:
   \[
   f'(t) = a \cdot 10 \left( \sin(t^6) \right)^9 \cdot \cos(t^6) \cdot 6t^5
   \]

5. **Simplify:**
   \[
   f'(t) = 60a \cdot t^5 \left( \sin(t^6) \right)^9 \cos(t^6)
   \]

Thus, the velocity function is:

\[
v(t) = 60a \cdot t^5 \left( \sin(t^6) \right)^9 \cos(t^6)
\]

Would you like further clarification on any part of the process or additional details?

Here are 5 related questions you might find helpful:
1. What is the acceleration function for this particle?
2. How does the velocity behave as \( t \to 0 \)?
3. What would the velocity be if \( a = 1 \)?
4. How does the power of the sine function influence the velocity?
5. What happens to the velocity if we change the exponent on \( t \) inside the sine function?

**Tip:** When differentiating complex functions, always break them down step-by-step using the chain rule and power rule carefully!

The function f(t) = a(sin(t^6))^10 models the position of an oscillating particle that has recently been discovered. Assume that a is a constant. What is the function that represents the velocity of this particle?

Discover how to find the velocity function for the position f(t) = a(sin(t^6))^10 using calculus techniques. This problem involves differentiation with trigonometric functions, applying the chain rule and power rule. Suitable for advanced high school calculus students.

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