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: