Question

Computing the binormal vector and torsion Use the definitions to compute the unit binormal vector and torsion of the following curves. $$\mathbf{r}(t)=\langle t, \cosh t,-\sinh t\rangle$$

Short answer

Question: Find the unit binormal vector and torsion of the curve $$\mathbf{r}(t) = \langle t, \cosh t, -\sinh t \rangle.$$ Answer: The unit binormal vector is given by $$\mathbf{B}(t) = \frac{\langle -1, 1, 1 \rangle}{(1 + \sinh^2 t + \cosh^2 t)}$$ and the torsion of the curve is $$\tau(t) = \frac{-1}{\sqrt{1+\sinh^2 t +\cosh^2 t}}$$.

Step-by-step solution

Find the first and second derivatives of the curve

Calculate the first derivative and second derivative of the curve with respect to the parameter \(t\): $$\mathbf{r'}(t) = \frac{d}{dt}\langle t, \cosh t, -\sinh t\rangle = \langle 1, \sinh t, -\cosh t \rangle$$ $$\mathbf{r''}(t) = \frac{d}{dt}\langle 1, \sinh t, -\cosh t \rangle = \langle 0, \cosh t, \sinh t \rangle$$

Find the unit tangent vector

The unit tangent vector is defined as the normalized first derivative of the curve: $$\mathbf{T}(t) = \frac{\mathbf{r'}(t)}{|\mathbf{r'}(t)|}$$ First, we need to find the magnitude of \(\mathbf{r'}(t)\): $$|\mathbf{r'}(t)| = \sqrt{1^2 + (\sinh t)^2 + (-\cosh t)^2} = \sqrt{1 + \sinh^2 t + \cosh^2 t}$$ Now, we can compute the unit tangent vector: $$\mathbf{T}(t) = \frac{\langle 1, \sinh t, -\cosh t \rangle}{\sqrt{1 + \sinh^2 t + \cosh^2 t}}$$

Calculate the normal vector

To find the normal vector, take the derivative of the unit tangent vector with respect to \(t\) and normalize it: $$\mathbf{N}(t) = \frac{\mathbf{T'}(t)}{|\mathbf{T'}(t)|} = \frac{d}{dt}\left(\frac{\langle 1, \sinh t, -\cosh t \rangle}{\sqrt{1 + \sinh^2 t + \cosh^2 t}}\right)$$ After some calculations, we find: $$\mathbf{N}(t) = \frac{\langle 0, \cosh t, \sinh t \rangle}{\sqrt{1 + \sinh^2 t + \cosh^2 t}}$$

Calculate the unit binormal vector

The unit binormal vector is given by the cross product of the unit tangent vector and the normal vector: $$\mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t)$$ Calculating the cross product, we get: $$\mathbf{B}(t) = \frac{\langle -1, 1, 1 \rangle}{(1 + \sinh^2 t + \cosh^2 t)}$$

Compute the torsion

The torsion of the curve is defined as: $$\tau(t) = -\left(\frac{d\mathbf{B}(t)}{dt} \cdot \mathbf{N}(t)\right)$$ Computing the derivative of the binormal vector, we get: $$\frac{d\mathbf{B}(t)}{dt} = \frac{\langle 0, -\sinh t + \cosh t, -\sinh t - \cosh t \rangle}{(1+\sinh^2 t +\cosh^2 t)}$$ Now, we can find the torsion by taking the dot product of the derivative of the binormal vector with the normal vector: $$\tau(t) = -\left( \frac{\langle 0, -\sinh t + \cosh t, -\sinh t - \cosh t \rangle \cdot \langle 0, \cosh t, \sinh t \rangle }{(1+\sinh^2 t +\cosh^2 t)(\sqrt{1 + \sinh^2 t + \cosh^2 t})}\right)= \frac{-1}{\sqrt{1+\sinh^2 t +\cosh^2 t}}$$ And that's the final answer: The unit binormal vector is: $$\mathbf{B}(t) = \frac{\langle -1, 1, 1 \rangle}{(1 + \sinh^2 t + \cosh^2 t)}$$ And the torsion of the curve is: $$\tau(t) = \frac{-1}{\sqrt{1+\sinh^2 t +\cosh^2 t}}$$

Key concepts

Torsion

Torsion is a key concept in the study of the geometry of curves. It describes how sharply a curve twists in three-dimensional space. If you think of a curve as a piece of wire, torsion indicates how the wire rotates about its path, rather than bending left or right. This property is particularly relevant in physics and engineering, where understanding the torsion can inform the stability and stress of structures like bridges or coils.

Mathematically, torsion is calculated using derivatives of vectors that define the curve. Specifically, the torsion \( \tau(t) \) is given by:

• The derivative of the binormal vector \( \mathbf{B}(t) \)
• Taking the dot product with the normal vector \( \mathbf{N}(t) \)
• Multiplying the result by -1.

An important point to remember is that if the torsion is zero, the curve lies in a plane, indicating no twisting about its path.

Unit Tangent Vector

The unit tangent vector \( \mathbf{T}(t) \) is a crucial component when examining curves in space. As the name suggests, it represents the direction of the curve at any given point and does so with a magnitude (or length) of one. This property makes it particularly useful in analyzing the path of a curve as it provides a consistent direction without scaling issues.

To calculate the unit tangent vector, follow these steps:

• Firstly, find the derivative of the curve, which provides the velocity vector of the curve.
• Next, compute the magnitude of this velocity vector.
• Finally, normalize the derivative by dividing it by its magnitude.

For the given curve, the unit tangent vector is derived from \( \mathbf{r'}(t) \) and gives insights into the curve's behavior and direction as it progresses through space.

Normal Vector

The normal vector \( \mathbf{N}(t) \) is perpendicular to the unit tangent vector and plays a crucial role in understanding the curvature of a curve. It helps in identifying how sharply the curve bends at any point. The normal vector indicates the direction in which the curve is turning and is always orthogonal to the tangent vector.

The process of calculating the normal vector involves:

• Finding the derivative of the unit tangent vector \( \mathbf{T}'(t) \).
• Calculating the magnitude of this derivative.
• Normalizing it to create a unit vector.

This method ensures that the normal vector maintains the necessary properties of orthogonality and unit length, making it essential for understanding curve behavior and its geometric features.

Cross Product

In the context of three-dimensional vectors, the cross product is an operation that takes two vectors and returns a third vector that is perpendicular to the plane formed by the initial two vectors. This concept is fundamental in calculations involving orientations and rotations.

The cross product is symbolized by \( \times \) and has several important properties:

• It produces a vector that is orthogonal to both input vectors.
• The magnitude of the resulting vector corresponds to the area of the parallelogram spanned by the original vectors.
• The direction of the cross product follows the right-hand rule, a convenient method to determine the orientation of the resulting vector.

In the binormal vector context, the cross product is used to create a vector that is orthogonal to both the unit tangent and the normal vector, forming the basis for the Frenet-Serret frame, which is vital for describing the local behavior of curves.