To solve the problem of finding the tangential and normal components of the acceleration for the given trajectory $\mathbf{r}(t)$, we need to follow these steps:

Given vector function:

$\mathbf{r}(t) = \left( 13e^t \sin t, 13e^t \cos t, 13e^t \right)$

Steps:

1. Find the velocity vector $\mathbf{v}(t)$: This is the first derivative of $\mathbf{r}(t)$ with respect to $t$: $\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt}$

2. Find the acceleration vector $\mathbf{a}(t)$: This is the second derivative of $\mathbf{r}(t)$: $\mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt}$

3. Find the speed: The speed is the magnitude of the velocity vector: $v(t) = \|\mathbf{v}(t)\|$

4. Find the tangential component of acceleration: The tangential component $a_T$ is the rate of change of speed: $a_T = \frac{dv(t)}{dt}$

5. Find the normal component of acceleration: The normal component $a_N$ is given by: $a_N = \sqrt{\|\mathbf{a}(t)\|^2 - a_T^2}$

Let me compute these derivatives and magnitudes to get the tangential and normal components of the acceleration.

I'll start by calculating the derivatives.### Solution Breakdown:

1. Velocity Vector $\mathbf{v}(t)$: $\mathbf{v}(t) = \left( 13e^t(\sin t + \cos t), -13e^t(\sin t - \cos t), 13e^t \right)$

2. Acceleration Vector $\mathbf{a}(t)$: $\mathbf{a}(t) = \left( 26e^t \cos t, -26e^t \sin t, 13e^t \right)$

3. Speed $v(t) = \|\mathbf{v}(t)\|$: $v(t) = \sqrt{676e^{2t} + 169e^{2t}} = 13e^t \sqrt{6}$

4. Tangential Component of Acceleration $a_T$: $a_T = \frac{dv(t)}{dt} = 13e^t \sqrt{6}$

5. Normal Component of Acceleration $a_N$: $a_N = 13e^t$

Thus, the tangential component of acceleration is $13e^t \sqrt{6}$, and the normal component of acceleration is $13e^t$.

Would you like any additional details or explanations on this process?

Related Questions:

1. What is the significance of tangential and normal components of acceleration in physical systems?
2. How do the tangential and normal components relate to the curvature of a trajectory?
3. Can this method be applied to a different parametric equation of motion?
4. What role does the exponential factor $e^t$ play in the acceleration of the object?
5. How would these components change if the trajectory involved a more complex function?

Tip:

The tangential component of acceleration always represents the change in speed along the curve, while the normal component indicates how sharply the trajectory is changing direction.