Vector Curvature

Curvature is a very intuitive concept.

If you have a straight line, does it have curvature? No, of course not.

Does a circle have a curvature? Yeah, it has some curvature.

It is our job to quantify how “much” curvature is in a given problem.

Recalling Vectors

Lets have a look at a simple circle.

Does a smaller circle have more curvature than a bigger one?

Yes, it has more curvature, but why?

Python Visual:

import micropip
await micropip.install("matplotlib")
await micropip.install("numpy")
 
import matplotlib.pyplot as plt
import numpy as np
 
# Create a figure and axis
fig, ax = plt.subplots()
 
# Define the center and radius of the larger circle
center_large = (0, 0)
radius_large = 1.5
 
# Define the center and radius of the smaller circle
center_small = (1, 0)  # Positioned to intersect the larger circle
radius_small = 0.5
 
# Draw the larger circle
large_circle = plt.Circle(center_large, radius_large, fill=False, color='blue', linewidth=2)
ax.add_patch(large_circle)
 
# Draw the smaller circle
small_circle = plt.Circle(center_small, radius_small, fill=False, color='red', linewidth=2)
ax.add_patch(small_circle)
 
# Label the circles
ax.text(center_large[0] - 0.8, center_large[1] + 1.7, 'some curvature', color='blue', fontsize=12)
ax.text(center_small[0] + 0.2, center_small[1] - 0.7, 'less curvature', color='red', fontsize=12)
 
# Set the limits of the plot
ax.set_xlim(-2, 3)
ax.set_ylim(-2, 2)
 
# Set aspect of the plot to be equal
ax.set_aspect('equal')
 
# Remove axes
ax.axis('off')
 
# Display the plot
plt.show()
 

So we know the smaller the circle, the more curvature it has.

Definition: Curvature is a measure of how much the direction vector changes

How Does the Derivative of a direction/tangent Vector Change?

Essentially, this is talking about the speed/velocity of the curve. But the thing is, the velocity at which we traverse the curve has no effect on the actual curvature itself.

Lets examine an equation.

$$curvature=K=\frac{dT}{ds}$$

Finding the Curvature

To find the curvature, parametrize the unit tangent vector, which is

$T=\frac{r\prime (t)}{|r\prime (t)|}$

Where we differentiate with respect to arc length

This yields something like

$r\prime (s)$

Which always equals 1, meaning r'(s) is the unit tangent vector

Curvature of a Circle

Find the curvature of a circle with radius R

Q: how do you parameterize a circle?

Answer: use the following formula

$\vec{r}=⟨R\cos \theta ,R\sin \theta ⟩$

which becomes

$\vec{r}(s)=⟨R\cos \frac{s}{R},R\sin \frac{s}{R}⟩$

and thus

${\vec{r}}^{\prime }=⟨-\sin \frac{s}{R},\cos \frac{s}{R}⟩$

and then second derivative

${\vec{r}}^{\prime \prime }=⟨-\frac{1}{R}\cos \frac{s}{R},-\frac{1}{R}\sin \frac{s}{R}⟩$

and then finding the magnitude

$\Vert {\vec{r}}^{\prime \prime }\Vert =\sqrt{{\left(-\frac{1}{R}\cos \frac{s}{R}\right)}^2+{\left(-\frac{1}{R}\sin \frac{s}{R}\right)}^2}$

which yields

$=\frac{1}{R}$