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The problem asks to decompose the acceleration vector into its tangential and normal components for a particle moving on a path defined by the vector-valued function:

\[
\mathbf{r}(t) = (\sin(7t))\hat{i} + (1)\hat{j} + (\cos(7t))\hat{k}
\]

To solve this, follow these steps:

### Step 1: Find the velocity vector
The velocity vector \(\mathbf{v}(t)\) is the derivative of the position vector \(\mathbf{r}(t)\) with respect to time \(t\):

\[
\mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t)
\]

Since \(\mathbf{r}(t) = \sin(7t)\hat{i} + 1\hat{j} + \cos(7t)\hat{k}\):

\[
\mathbf{v}(t) = \frac{d}{dt} [\sin(7t)\hat{i} + 1\hat{j} + \cos(7t)\hat{k}]
\]
\[
\mathbf{v}(t) = 7\cos(7t)\hat{i} - 7\sin(7t)\hat{k}
\]

### Step 2: Find the acceleration vector
The acceleration vector \(\mathbf{a}(t)\) is the derivative of the velocity vector \(\mathbf{v}(t)\):

\[
\mathbf{a}(t) = \frac{d}{dt} \mathbf{v}(t)
\]
\[
\mathbf{a}(t) = \frac{d}{dt} [7\cos(7t)\hat{i} - 7\sin(7t)\hat{k}]
\]
\[
\mathbf{a}(t) = -49\sin(7t)\hat{i} - 49\cos(7t)\hat{k}
\]

### Step 3: Tangential and Normal components
The acceleration vector can be decomposed into tangential and normal components:
\[
\mathbf{a}(t) = a_T(t)\hat{T}(t) + a_N(t)\hat{N}(t)
\]
where:
- \(a_T(t) = \frac{d}{dt}|\mathbf{v}(t)|\) (tangential acceleration),
- \(a_N(t) = \frac{|\mathbf{v}(t) \times \mathbf{a}(t)|}{|\mathbf{v}(t)|}\) (normal acceleration).

#### Tangential component
First, find \(|\mathbf{v}(t)|\), the magnitude of the velocity vector:
\[
|\mathbf{v}(t)| = \sqrt{(7\cos(7t))^2 + (-7\sin(7t))^2} = \sqrt{49\cos^2(7t) + 49\sin^2(7t)} = \sqrt{49} = 7
\]
The tangential component \(a_T(t)\) is the derivative of the velocity magnitude:
\[
a_T(t) = \frac{d}{dt}(7) = 0
\]

#### Normal component
The normal component \(a_N(t)\) requires calculating the magnitude of the cross product \(\mathbf{v}(t) \times \mathbf{a}(t)\). First, compute the cross product:
\[
\mathbf{v}(t) \times \mathbf{a}(t) = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
7\cos(7t) & 0 & -7\sin(7t) \\
-49\sin(7t) & 0 & -49\cos(7t)
\end{vmatrix}
\]
\[
= \hat{i}(0 - 0) - \hat{j}(7\cos(7t)(-49\cos(7t)) - (-7\sin(7t))(-49\sin(7t))) + \hat{k}(0 - 0)
\]
\[
= \hat{j}(-343(\cos^2(7t) + \sin^2(7t))) = -343\hat{j}
\]

The magnitude of the cross product is:
\[
|\mathbf{v}(t) \times \mathbf{a}(t)| = |-343\hat{j}| = 343
\]

Finally, the normal component is:
\[
a_N(t) = \frac{343}{7} = 49
\]

### Final Answer
Thus, the acceleration vector is:
\[
\mathbf{a}(t) = 0\hat{T}(t) + 49\hat{N}(t)
\]

Would you like further clarification or details?

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**Related questions:**
1. How do you compute the curvature of a particle's path?
2. What is the difference between tangential and normal acceleration?
3. How do you find the unit tangent vector for a curve?
4. Can you provide a geometrical interpretation of acceleration components?
5. How does the cross product relate to normal acceleration?

**Tip:** When calculating cross products, always remember to use the determinant formula for vectors in 3D.

Decompose the acceleration vector into its components in the tangential and normal directions for a particle moving along a path given by the vector-valued function r(t) = (sin(7t))i + (1)j + (cos(7t))k.

Learn how to decompose the acceleration vector into its tangential and normal components for a particle moving along the path r(t) = (sin(7t))i + (1)j + (cos(7t))k. Detailed solution includes velocity, acceleration, and vector calculus.

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