Question

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

Answer: The unit binormal vector is \(\mathbf{B}(t) = \langle 1, 0, 0 \rangle\), and the torsion is \(\tau = 0\).

Step-by-step solution

Find the derivative of the curve

To find the derivative of the curve, we differentiate each component of \(\mathbf{r}(t)\) with respect to \(t\): $$\mathbf{r}'(t) = \langle 1, \sinh t, -\cosh t\rangle$$

Compute the unit tangent vector

To calculate the unit tangent vector, first compute 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, divide \(\mathbf{r}'(t)\) by the magnitude to obtain the unit tangent vector: $$\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||}=\frac{1}{\sqrt{1+\sinh^2 t + \cosh^2 t}}\langle 1, \sinh t, -\cosh t\rangle$$

Compute the unit normal vector

First, find the derivative of the unit tangent vector: $$\mathbf{T}'(t) = \frac{1}{\sqrt{1+\sinh^2 t + \cosh^2 t}}\langle 0, \cosh t, \sinh t\rangle$$ Next, compute the magnitude of \(\mathbf{T}'(t)\): $$||\mathbf{T}'(t)|| = \sqrt{(0)^2 + (\cosh t)^2 + (\sinh t)^2} = \sqrt{\cosh^2 t + \sinh^2 t}$$ Then, divide \(\mathbf{T}'(t)\) by the magnitude to obtain the unit normal vector: $$\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{||\mathbf{T}'(t)||}=\frac{1}{\sqrt{\cosh^2 t + \sinh^2 t}}\langle 0, \cosh t, \sinh t\rangle$$

Compute the unit binormal vector

Using the cross product of the unit tangent and normal vectors, compute the unit binormal vector: $$\mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) = \frac{1}{\sqrt{1+\sinh^2 t + \cosh^2 t}}\left\langle 1, \sinh t, -\cosh t\right\rangle\times\frac{1}{\sqrt{\cosh^2 t + \sinh^2 t}}\langle 0, \cosh t, \sinh t\rangle$$ which results in $$\mathbf{B}(t) = \langle 1,0,0\rangle$$

Compute the torsion

First, find the derivative of the unit binormal vector: $$\mathbf{B}'(t) = \langle 0, 0, 0\rangle$$ Finally, compute the torsion using the dot product of \(\mathbf{B}'(t)\) and \(\mathbf{N}(t)\): $$\tau = -(\mathbf{B}'\cdot \mathbf{N})=-\langle 0, 0, 0\rangle\cdot\frac{1}{\sqrt{\cosh^2 t + \sinh^2 t}}\langle 0, \cosh t, \sinh t\rangle = 0$$ The unit binormal vector of the curve is \(\mathbf{B}(t)=\langle 1,0,0\rangle\) and its torsion is \(\tau = 0\).

Key concepts

Torsion

Torsion is a measure of how much a curve twists out of the plane of its osculating circle. To imagine this, think of a winding road climbing a hill. The torsion tells us how much the road is twisting as it goes up. In mathematical terms, torsion is defined as \[ \tau = -\frac{\mathbf{B}'(t) \cdot \mathbf{N}(t)}{||\mathbf{r}'(t)||} \]where

• \( \mathbf{B}'(t) \) is the derivative of the unit binormal vector, and
• \( \mathbf{N}(t) \) is the unit normal vector.

In the exercise, the torsion was calculated to be 0, meaning there is no twisting at this point. The road, so to speak, is not twisting away from the plane it's in at this point.

Derivative of the Curve

The derivative of a curve gives us information about the curve's direction. It is essentially the rate of change of the position vector \( \mathbf{r}(t) \). By finding the derivative, \( \mathbf{r}'(t) \), we can determine how the curve is moving in space.
For example, if we have \[ \mathbf{r}(t) = \langle t, \cosh t, -\sinh t \rangle \]the derivative is \[ \mathbf{r}'(t) = \langle 1, \sinh t, -\cosh t \rangle \]indicating that the curve is changing in the direction of this vector. It's like saying if we're making a path with a pencil, the derivative tells us which way the pencil is pointing.

Unit Tangent Vector

The unit tangent vector is a normalized version of the derivative of the curve. It gives us a direction that has a fixed length, usually 1, making it useful for understanding the direction of the curve without worrying about how fast we're going.
It is calculated by dividing the derivative by its magnitude:
\[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||} \]In our exercise \[ \mathbf{r}'(t) = \langle 1, \sinh t, -\cosh t \rangle \]leads to \[ \mathbf{T}(t) = \frac{1}{\sqrt{1 + \sinh^2 t + \cosh^2 t}} \langle 1, \sinh t, -\cosh t \rangle \] This vector maintains the direction of \( \mathbf{r}'(t) \) but scales it to ensure a length of 1.

Unit Normal Vector

The unit normal vector is perpendicular to the unit tangent vector and points towards the curve's concave side. Calculating it involves differentiating the unit tangent vector and normalizing the result. This vector tells us how the curve is turning in space.
Here's how it’s obtained:

• First, differentiate the unit tangent vector \( \mathbf{T}(t) \).
• Find the magnitude of this derivative.
• Normalize by dividing by its magnitude.

Given the unit tangent vector \( \mathbf{T}(t) \), its derivative \( \mathbf{T}'(t) \) can be calculated, and normalizing gives us the unit normal vector
\[ \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{||\mathbf{T}'(t)||} \] This vector helps us understand how sharply the curve is bending.