To find the normal component of acceleration of a particle at $t = 4$ given the position vector:

$\mathbf{r}(t) = \langle 3\cos(2t), 5\sin(3t), \sin(0.3t) + \cos(t + 5) \rangle,$

we first need to determine the velocity and acceleration vectors and then compute the normal component of acceleration.

Step 1: Find the Velocity Vector $\mathbf{v}(t)$

The velocity vector $\mathbf{v}(t)$ is the first derivative of the position vector $\mathbf{r}(t)$ with respect to $t$:

$\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \langle -6\sin(2t), 15\cos(3t), 0.3\cos(0.3t) - \sin(t + 5) \rangle.$

Step 2: Find the Acceleration Vector $\mathbf{a}(t)$

The acceleration vector $\mathbf{a}(t)$ is the first derivative of the velocity vector $\mathbf{v}(t)$ with respect to $t$:

$\mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt} = \langle -12\cos(2t), -45\sin(3t), -0.09\sin(0.3t) - \cos(t + 5) \rangle.$

Step 3: Find the Magnitude of the Velocity $|\mathbf{v}(t)|$

The magnitude of the velocity vector $|\mathbf{v}(t)|$ is given by:

$|\mathbf{v}(t)| = \sqrt{(-6\sin(2t))^2 + (15\cos(3t))^2 + (0.3\cos(0.3t) - \sin(t + 5))^2}.$

Step 4: Find the Tangential and Normal Components of Acceleration

The normal component of acceleration $\mathbf{a}_N$ is found using:

$a_N = \frac{|\mathbf{v}(t) \times \mathbf{a}(t)|}{|\mathbf{v}(t)|},$

where $\times$ denotes the cross product.

Step 5: Evaluate at $t = 4$

Let's calculate the normal component of acceleration at $t = 4$. We will compute the cross product, magnitudes, and substitute $t = 4$.The normal component of acceleration at $t = 4$ is approximately:

$a_N \approx 11.90$

Would you like more details, or do you have any questions?

Here are five related questions to explore further:

1. How is the tangential component of acceleration computed?
2. What is the physical interpretation of the normal component of acceleration?
3. How do the velocity and acceleration vectors relate to the path of a particle?
4. What role does curvature play in determining the normal acceleration?
5. Can you find the radius of curvature at a given point on the path?

Tip: The normal component of acceleration is responsible for changing the direction of the velocity vector, while the tangential component affects the speed of the particle.