Final Answer:
The steps outlined above provide a comprehensive method for finding the curvature of the path described by the vector function \( \mathbf{r}(t) = e^t \cos(t) \mathbf{i} + e^t \sin(t) \mathbf{j} + e^t \mathbf{k} \). The process involves calculating the velocity vector, acceleration vector, speed, curvature, unit tangent vector, unit normal vector, and tangential and normal components of acceleration.
Step-by-step explanation:
To find the curvature of the path defined by the vector function \( \mathbf{r}(t) = e^t \cos(t) \mathbf{i} + e^t \sin(t) \mathbf{j} + e^t \mathbf{k} \), we can follow these steps:
1. Find the Velocity and Acceleration Vectors:
\[ \mathbf{r}(t) = e^t \cos(t) \mathbf{i} + e^t \sin(t) \mathbf{j} + e^t \mathbf{k} \]
The velocity vector \( \mathbf{v}(t) \) is the derivative of \( \mathbf{r}(t) \) with respect to \( t \):
\[ \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} \]
\[ \mathbf{v}(t) = -e^t \cos(t) \mathbf{i} + e^t \sin(t) \mathbf{j} + e^t \mathbf{k} \]
The acceleration vector \( \mathbf{a}(t) \) is the derivative of \( \mathbf{v}(t) \):
\[ \mathbf{a}(t) = \frac{d\mathbf{v}}{dt} \]
\[ \mathbf{a}(t) = -2e^t \cos(t) \mathbf{i} - e^t \sin(t) \mathbf{j} + e^t \mathbf{k} \]
2. Find the Speed (\( \lVert \mathbf{v} \rVert \)):
\[ \lVert \mathbf{v} \rVert = \sqrt{(-e^t \cos(t))^2 + (e^t \sin(t))^2 + (e^t)^2} \]
3. Find the Curvature (\( \kappa \)):
\[ \kappa = \frac{\lVert \mathbf{v} \times \mathbf{a} \rVert}{\lVert \mathbf{v} \rVert^3} \]
4. Find the Unit Tangent Vector (\( \mathbf{T} \)):
\[ \mathbf{T}(t) = \frac{\mathbf{v}(t)}{\lVert \mathbf{v}(t) \rVert} \]
5. Find the Unit Normal Vector (\( \mathbf{N} \)):
\[ \mathbf{N}(t) = \frac{\mathbf{v}(t) \times \mathbf{a}(t)}{\lVert \mathbf{v}(t) \times \mathbf{a}(t) \rVert} \]
6. Find the Tangential Component of Acceleration (\( a_T \)):
\[ a_T = \mathbf{a} \cdot \mathbf{T} \]
7. Find the Normal Component of Acceleration (\( a_N \)):
\[ a_N = \mathbf{a} \cdot \mathbf{N} \]