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The task involves finding the **unit binormal vector** \( \mathbf{B}(t) \) and **torsion** \( \tau(t) \) for the given parameterized curve \( \mathbf{r}(t) = (5\cos{t}, 5\sin{t}, 14t) \).

We are provided with the following:

1. The **unit tangent vector** \( \mathbf{T}(t) \):
   \[
   \mathbf{T}(t) = \frac{(-5\sin t, 5\cos t, 14)}{\sqrt{221}}
   \]

2. The **principal unit normal vector** \( \mathbf{N}(t) \):
   \[
   \mathbf{N}(t) = (-\cos{t}, -\sin{t}, 0)
   \]

### Step 1: Finding the Unit Binormal Vector \( \mathbf{B}(t) \)

The unit binormal vector is defined as:
\[
\mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t)
\]
where \( \times \) represents the cross product of the vectors \( \mathbf{T}(t) \) and \( \mathbf{N}(t) \).

### Step 2: Computing Torsion \( \tau(t) \)

The torsion is given by:
\[
\tau(t) = -\frac{\mathbf{dB}(t)}{\mathbf{ds}(t)} \cdot \mathbf{N}(t)
\]
where \( \mathbf{dB}(t) \) is the derivative of the binormal vector with respect to \( t \), and \( \mathbf{ds}(t) \) is the differential arc length.

Let me start with calculating the cross product of \( \mathbf{T}(t) \times \mathbf{N}(t) \). I will compute that now.The unit binormal vector \( \mathbf{B}(t) \) at \( t = 0 \) is:

\[
\mathbf{B}(t) = (0, -0.9417, 0.3363)
\]

### Next steps:
1. Compute the derivative of the binormal vector \( \mathbf{dB}(t) \).
2. Calculate the torsion \( \tau(t) \) using the formula.

I will now proceed with the torsion calculation by differentiating the binormal vector.The derivative of the binormal vector \( \mathbf{dB}(t) \) at \( t = 0 \) is approximately:

\[
\mathbf{dB}(t) = (0.9417, 4.71 \times 10^{-6}, 0)
\]

### Step 3: Torsion Calculation

To compute the torsion \( \tau(t) \), we use the formula:

\[
\tau(t) = -\frac{\mathbf{dB}(t) \cdot \mathbf{N}(t)}{\mathbf{ds}(t)}
\]

Since \( \mathbf{ds}(t) = 1 \) in this context (unit speed curve), the torsion simplifies to:

\[
\tau(t) = -\mathbf{dB}(t) \cdot \mathbf{N}(t)
\]

I will now calculate the torsion at \( t = 0 \).The torsion \( \tau(t) \) at \( t = 0 \) is approximately:

\[
\tau(0) = 0.9417
\]

### Summary:
- The **unit binormal vector** at \( t = 0 \) is \( \mathbf{B}(0) = (0, -0.9417, 0.3363) \).
- The **torsion** at \( t = 0 \) is \( \tau(0) = 0.9417 \).

Would you like a detailed breakdown of any part of the process or further clarification?

Here are some related questions you might want to explore:
1. How is torsion related to curvature in a 3D curve?
2. What are the geometric interpretations of the unit tangent, normal, and binormal vectors?
3. How do you compute curvature from a parameterized curve?
4. How does torsion affect the shape of the curve in space?
5. Can you apply these concepts to a real-world problem, such as motion along a space curve?

**Tip:** The torsion describes how much a curve twists out of the plane formed by the tangent and normal vectors. A zero torsion implies the curve lies in a plane.

Find the unit binormal vector and torsion for the parameterized curve r(t) = (5cos(t), 5sin(t), 14t).

Learn how to compute the unit binormal vector and torsion for the parameterized curve r(t) = (5cos(t), 5sin(t), 14t). This problem covers advanced concepts in vector calculus such as the unit tangent vector, normal vector, binormal vector, and torsion. Suitable for advanced high school students.

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