prove that the curvature of a curve is :

Question

κ = ΙdΦ/dsΙ and Φ is the angle between Τ and i

Answer 1

This proof is shorter than people expect, and the whole thing turns on one observation: T is a unit vector, so in the plane it has only one degree of freedom left, and that degree of freedom is exactly the angle Φ.

Start from the definition of curvature, which is the rate at which the unit tangent turns with respect to arc length:

κ = |dT/ds|

Now, since Φ is the angle between T and i, and since T has length 1, we can write T in component form just by reading it off the unit circle:

T = cosΦ i + sinΦ j

Differentiate both sides with respect to s. The only thing on the right that depends on s is Φ, so the chain rule gives us a dΦ/ds on each term:

dT/ds = (−sinΦ i + cosΦ j)(dΦ/ds)

Here is the step that finishes it. Take the magnitude of both sides. The scalar dΦ/ds comes out front as its absolute value, and the vector left behind has length

|−sinΦ i + cosΦ j| = √(sin²Φ + cos²Φ) = 1

by the Pythagorean identity. So we are multiplying |dΦ/ds| by 1, which leaves

κ = |dT/ds| = |dΦ/ds|

and that is the result.

It is worth pausing on what this is really saying, because the formula is more intuitive than it looks. Curvature is supposed to measure how sharply the curve is bending. Φ is literally the direction you are facing as you travel along the curve, and s is the distance you have travelled. So dΦ/ds is how many radians you turn per unit of distance, which is exactly what bending means. The derivative of the unit tangent and the derivative of the heading angle are the same thing, just written two different ways.

A quick check that makes it concrete. On a circle of radius r, the arc length and the central angle are related by s = rΦ, so dΦ/ds = 1/r, which gives κ = 1/r. That matches what we already know, that a small circle bends hard and a large circle bends gently, and that a straight line, where Φ never changes, has κ = 0.