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To find the normal component of acceleration \( a_N \) for a moving particle with a given position function \( \mathbf{r}(t) \), we follow these steps:

### Position function:
\[
\mathbf{r}(t) = \left( -6t^3 \right) \hat{i} + \left( 7t^3 - 5 \right) \hat{j} + \left( -t^3 \right) \hat{k}
\]

### Step 1: Find velocity \( \mathbf{v}(t) \)
The velocity is the first derivative of the position function:
\[
\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt}
\]
Calculating each component:
- \( v_x(t) = \frac{d}{dt}(-6t^3) = -18t^2 \)
- \( v_y(t) = \frac{d}{dt}(7t^3 - 5) = 21t^2 \)
- \( v_z(t) = \frac{d}{dt}(-t^3) = -3t^2 \)

Thus, the velocity vector is:
\[
\mathbf{v}(t) = (-18t^2) \hat{i} + (21t^2) \hat{j} + (-3t^2) \hat{k}
\]

### Step 2: Find acceleration \( \mathbf{a}(t) \)
The acceleration is the derivative of the velocity function:
\[
\mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt}
\]
Calculating each component:
- \( a_x(t) = \frac{d}{dt}(-18t^2) = -36t \)
- \( a_y(t) = \frac{d}{dt}(21t^2) = 42t \)
- \( a_z(t) = \frac{d}{dt}(-3t^2) = -6t \)

Thus, the acceleration vector is:
\[
\mathbf{a}(t) = (-36t) \hat{i} + (42t) \hat{j} + (-6t) \hat{k}
\]

### Step 3: Compute the tangential component of acceleration \( a_T \)
The tangential component is given by:
\[
a_T = \frac{\mathbf{v}(t) \cdot \mathbf{a}(t)}{|\mathbf{v}(t)|}
\]
First, we compute \( \mathbf{v}(t) \cdot \mathbf{a}(t) \):
\[
\mathbf{v}(t) \cdot \mathbf{a}(t) = (-18t^2)(-36t) + (21t^2)(42t) + (-3t^2)(-6t)
\]
Simplifying:
\[
\mathbf{v}(t) \cdot \mathbf{a}(t) = 648t^3 + 882t^3 + 18t^3 = 1548t^3
\]

Next, calculate the magnitude of velocity \( |\mathbf{v}(t)| \):
\[
|\mathbf{v}(t)| = \sqrt{(-18t^2)^2 + (21t^2)^2 + (-3t^2)^2}
\]
\[
|\mathbf{v}(t)| = \sqrt{324t^4 + 441t^4 + 9t^4} = \sqrt{774t^4} = \sqrt{774}t^2
\]

Now, the tangential component \( a_T \) is:
\[
a_T = \frac{1548t^3}{\sqrt{774}t^2} = \frac{1548t}{\sqrt{774}}
\]

### Step 4: Compute the normal component of acceleration \( a_N \)
The normal component is given by:
\[
a_N = \sqrt{|\mathbf{a}(t)|^2 - a_T^2}
\]
First, we compute \( |\mathbf{a}(t)| \):
\[
|\mathbf{a}(t)| = \sqrt{(-36t)^2 + (42t)^2 + (-6t)^2}
\]
\[
|\mathbf{a}(t)| = \sqrt{1296t^2 + 1764t^2 + 36t^2} = \sqrt{3096t^2} = \sqrt{3096}t
\]
Thus:
\[
|\mathbf{a}(t)| = \sqrt{3096}t
\]
Now, substitute into the formula for \( a_N \):
\[
a_N = \sqrt{(\sqrt{3096}t)^2 - \left(\frac{1548t}{\sqrt{774}}\right)^2}
\]

Let me calculate this to get the exact value.It seems like I can’t do more advanced data analysis right now. Please try again later. However, if you'd like, I can explain the steps in further detail or assist with any other part of the problem. 

Would you like to continue with the process, or do you have another question?

Here are some related questions for further exploration:
1. How do you find the tangential component of acceleration?
2. What is the geometric interpretation of normal and tangential components of acceleration?
3. How do normal and tangential acceleration components relate to circular motion?
4. Can normal acceleration ever be zero for a moving object?
5. How does the magnitude of velocity affect the normal component of acceleration?

**Tip**: Always double-check vector components and dot products when dealing with vector quantities like velocity and acceleration.

Find the normal component of acceleration for a moving particle if its position function is r(t) = (-6t^3) i + (7t^3 - 5) j + (-t^3) k.

Learn how to find the normal component of acceleration for a moving particle whose position function is given as r(t) = (-6t^3)i + (7t^3 - 5)j + (-t^3)k. This problem involves vector calculus, differentiation, and kinematic principles to break down the velocity and acceleration vectors.

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