Intuitive definition for curvature

Question

I was reading about curvature of a curve but I didn't understand it.

I'm looking for an easy definition and also I want to know is there any general formula for finding curvature of the all curves ?

Note : I looked at Wolfram Math Wolrd: Curvature but It was really hard to me

Answer 1

In the plane curvature is rate of change of slope angle or rotation of tangent $\phi $ w.r.t. arc $s$, $ \tan \phi = \dfrac{dy}{dx},$ where $ ds^2= dx^2+dy^2$

This is all that is important by way of definition. Physically it is rate at which a curve turns, the rest is just manipulation:

Curvature

$$ = \frac{d\phi}{ds}= = \frac{d\tan^{-1} \frac{dy}{dx}}{ds}=\frac{d\tan^{-1} {y^{\prime}}}{\dfrac{dx\cdot ds}{dx}} =\frac{y''/(1+y'^2)}{\sec\phi } $$

But

$$ \sec \phi = \sqrt{1+y'^2}$$

Plug in, curvature general formula /differential relation in rectangular coordinates becomes

$$\frac{y''}{ (1+y'^2)^{\frac32}} \tag1 $$

It is constant for a circle. $k=1 $ for circle, else for ellipse $ k=b/a $

$$ y = k \sqrt{a^2-x^2} $$

$$ y^{'} = k \frac{-x} {\sqrt{a^2-x^2}} $$

$$ y^{''} = k \frac{-a^2} {({a^2-x^2})^\frac32} $$

Can you now use formula (1)?

Or if you choose $x=0,~$ it simplifies to $ \dfrac{b}{a^2}$.

Answer 2

I was reading about curvature of a curve but I didn't understand it.

Imagine your in a car, travelling a road in the mountains at constant speed. When entering curves, you feel pushed sideways. The more the road is curved, the more you are feeling the effect. This is because, even though you are travelling at constant speed, there is an acceleration when you change directions. This acceleration is directed towards the center of curvature of the curve. For instance, if you are travelling on a circular road, then you are constantly accelerating towards the center of the circle.

The curvature $\kappa$ of a curve at a point is the magnitude of this acceleration when you are travelling at unit speed.

In other words, given a planar curve $\gamma : [0,1] \rightarrow \mathbb{R}^2$ such that $\|\gamma'(t)\| = 1$ (constant unit speed), the curvature of $\gamma$ at $t$ is defined as $$ \kappa(t) = \|\gamma''(t)\|. $$

In general, if you are travelling at a constant positive speed $v = \|\gamma'(t)\| > 0$, then the curve $\alpha(t) = \gamma(t/v)$ has unit speed, so that $$ \kappa(t) = \| \alpha''(t) \| = \frac{\|\gamma''(t/v)\|}{v^2}. $$

is there any general formula for finding curvature of the all curves?

Yes. In general, if $\gamma : [0,1] \rightarrow \mathbb{R}^3$ is a curve that never stops ($\|\gamma'(t)\| > 0$, for all $t$), then the formula of the curvature of $\gamma$ is $$ \kappa(t) = \frac{\|\gamma'(t) \times \gamma''(t)\|}{\|\gamma'(t)\|^3}, $$ where $\times$ denotes the cross product of $\mathbb{R}^3$. The physical interpretation of curvature does not change in this general case.

Answer 3

There are two notions of the curvature of a curve in an $2$-dimensional Euclidean vector space. The first one is due to Euler (1775) who defines the curvature to be the limit of the quotient of the change of the tangent vector compared by the length of the observed interval.

The second one is a cinematic-dynamic measurement of the orientated curvature $\kappa_c$ of curve $c$ given by $$\kappa_c=\frac{\text{orientated normal-component of the acceleration}}{\text{square of $c$’s velocity}}$$

Answer 4

The curvature of a plane curve $C$ at a point $P$ is the reciprocal of the radius of the osculating circle.

The higher the curvature at a point on the curve, the smaller the radius of the osculating circle. A circle has the same nonzero curvature at every point. A line has zero curvature at every point.