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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.

Find the tangential and normal components of the acceleration for the given trajectory r(t) = (13e^t sin t, 13e^t cos t, 13e^t).

Learn how to find the tangential and normal components of acceleration for the trajectory r(t) = (13e^t sin t, 13e^t cos t, 13e^t). The solution involves calculating velocity, acceleration, and breaking down the components of acceleration. Suitable for undergraduate students studying vector calculus and kinematics.

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