Radius of curvature: Definition, Formula, Derivation

The curvature is the concept in geometry that indicates the change in direction of the curve at a certain point. While the radius of curvature gives the radius of the approximate circle that matches the curve at a particular point. In this article, we are discussing the radius of curvature along with its derivations and some of the practice numerical.

What is Radius of curvature?

The radius of curvature gives the extent of bend in the curve at a certain point which is equal to the reciprocal of the curvature (`\kappa`).

∴ `\rho = \frac{1}{\kappa}`

Where,
`\rho` = Radius of curvature
`\kappa` = Curvature

Thus we can say that the curve with higher curvature has a lower radius of curvature and the curve with lower curvature has a higher radius of curvature.

Let us explain first what actually radius of curvature indicates.

radius of curvature graph

The above figure shows the curve of the arc length of ‘S’. The tangent drawn to the curve at point ‘P’ makes an angle `\psi` with the horizontal axis.

Now consider a point ‘Q’ on the curve which is very close to point ‘P’. The arc ‘PQ’ indicates the smallest portion of the curve at ‘P’ of length dS.

The tangent drawn to the curve at point Q makes an angle of (`\psi + d\psi`) with the horizontal axis.

The normal drawn to the tangent NP and tangent QT meets at point C. Here point ‘C’ indicates the center of curvature and length ‘CP’ or ‘CQ’ indicates the radius of curvature at point ‘P’.

From the above figure, the curvature of the curve at point P is given by,

`\kappa = \frac{d\psi}{dS}`

While the radius of curvature at point ‘P’ is given by,

`\rho = \frac{1}{\kappa} = \frac{dS}{d\psi}`

Radius of curvature equations:

Following are the formulae used to calculate the radius of curvature in different forms:-

1] Cartesian form:-

If the curve is expressed in a cartesian form, then the radius of curvature is given by,

`\rho = \frac{(1+y_{1}^{2})^{\frac{3}{2}}}{y_{2}}`

Where,
`y_{1} = \frac{dy}{dx}`
`y_{2} = \frac{d^{2}y}{dx^{2}}`


2] Parametric form:-

When the curve is expressed by the parametric equations x = f(t), y = g(t), Then in such case, the radius of curvature is given by,

`\rho = \frac{[x’^{2}+y’^{2}]^{\frac{3}{2}}}{y”x’-x”y’}`

Where,
x’ = `\frac{dx}{dt}`, x” = `\frac{d^{2}x}{dt^{2}}`
y’ = `\frac{dy}{dt}`, y” = `\frac{d^{2}y}{dt^{2}}`


3] Polar form:-

If the curve is expressed in the polar form [r = f(θ)], then the radius of curvature is given by,

`\rho = \frac{(r^{2}+r_{1}^{2})^{\frac{3}{2}}}{(r^{2}+2r_{1}^{2}-rr_{2})}`

Where,
`r_{1} = \frac{dr}{d\theta}`
`r_{2} = \frac{d^{2}r}{d\theta^{2}}`

Derivation:

Following are the derivations for the radius of curvature in different forms:-

1] Cartesian form:-

Radius of curvature cartesian form derivation

The above figure indicates the curve ‘S’ in a cartesian form [y = f(x)]. The tangent drawn to the curve at point ‘P’ makes an angle of `\psi` with the horizontal axis.

Now the slope of this tangent is given by,

Slope = `tan\psi = \frac{dy}{dx}`

∴ `tan\psi = \frac{dy}{dx} \cdots \text{[Equation-1]}`

Differentiate the equation with respect to S.

`\frac{d}{dS}(tan\psi) = \frac{d}{dS}(\frac{dy}{dx})`

`sec^{2}\psi.\frac{d\psi}{dS} = \frac{d}{dx}(\frac{dy}{dx}).\frac{dx}{dS}`

`(1+tan^{2}\psi)\frac{d\psi}{dS} = \frac{d^{2}y}{dx^{2}}.\frac{dx}{dS}`

We know that `\frac{d\psi}{dS} = \frac{1}{\rho}`

∴ `(1+tan^{2}\psi)\frac{1}{\rho} = \frac{d^{2}y}{dx^{2}}.\frac{dx}{dS}`

By putting the value of `tan\psi` in the above equation we get,

`(1+{\frac{dy}{dx}}^{2})\frac{1}{\rho} = \frac{d^{2}y}{dx^{2}}.\frac{1}{\frac{dS}{dx}}`

Now assume, `\frac{dy}{dx} = y_{1} and \frac{d^{2}y}{dx^{2}} = y_{2}`

The equation becomes,

`(1+y_{1}^{2})\frac{1}{rho} = y_{2}.\frac{1}{\frac{dS}{dx}} \cdots \text{[Equation-2]}`

Now let’s find the value of `\frac{dS}{dx}`, to get the equation for the radius of curvature.

Value of `\mathbf{\frac{dS}{dx}}`:-

Value of dS/dx (Cartesian form derivation)

The above figure indicates a small portion of curve ‘dS’. By applying Pythagoras theorem to the above triangle, we get

`dS^{2}=dx^{2}+dy^{2}`

`dS^{2}=dx^{2}(1+(\frac{dy}{dx})^{2})`

`\frac{dS^{2}}{dx^{2}}=1+(\frac{dy}{dx})^{2}`

`\frac{dS}{dx}=\sqrt{1+(\frac{dy}{dx})^{2}}`

Now by putting `\frac{dy}{dx} = y_{1}`, the equation becomes,

`\frac{dS}{dx}=\sqrt{1+y_{1}^{2}}\cdots \text{[Equation-3]}`

Put this value of `\frac{dS}{dx}` in above equation 2.

`(1+y_{1}^{2})\frac{1}{\rho}=y_{2}.\frac{1}{\sqrt{1+y_{1}^{2}}}`

`\rho = \frac{(1+y_{1}^{2})\sqrt{1+y_{1}^{2}}}{y_{2}}`

`\rho = \frac{(1+y_{1}^{2})^{\frac{3}{2}}}{y_{2}}`

This is the equation to find the radius of curvature in cartesian form.


2] Parametric form:-

The parametric equation consists of variables that never depend on each other but depend on another same independent variable (t). The curve in parametric form is expressed as,

x = f(t), y = g(t)

By differentiation with respect to ‘t’, we get,

`x’ = \frac{dx}{dt}, y’ = \frac{dy}{dt}`

The radius of curvature in cartesian form is given by,

`\rho = \frac{(1+y_{1}^{2})^{\frac{3}{2}}}{y_{2}}`

Where, `y_{1} = \frac{dy}{dx}`

∴ `y_{1} = \frac{dy}{dt}.\frac{dt}{dx}`

`y_{1}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}`

`\mathbf{y_{1}=\frac{y’}{x’}}`

and `y_{2}=\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}y_{1}`

`y_{2}=(\frac{d}{dt}.y_{1}).\frac{dt}{dx}`

`y_{2}={\frac{d}{dt}.(\frac{y’}{x’})}\frac{1}{\frac{dx}{dt}}`

`y_{2}={\frac{y”x’-x”y’}{x’^{2}}}\frac{1}{x’}`

`\mathbf{y_{2}={\frac{y”x’-x”y’}{x’^{3}}}}`

Put these values of `y_{1}` and `y_{2}` in the equation of `\rho`

`\rho=\frac{(1+(\frac{y’}{x’})^{2})^{3/2}}{\frac{y”x’-x”y’}{x’^{3}}}`

`\rho=\frac{x’^{3}(\frac{x’^{2}+y’^{2}}{x’^{2}})^{\frac{3}{2}}}{y”x’-x”y’}`

`\rho=\frac{(x’^{2}+y’^{2})^\frac{3}{2}}{y”x’-x”y’}`

This is the equation to find the radius of curvature in parametric form.


3] Polar form:-

Polar form derivation

The curves shown in the above figure are expressed in a polar form as `r = f(\theta)`. The line ‘OX’ represents the initial line and ‘OP’ is the radius vector that touches the curve at point ‘P’ and makes an angle of `\theta` with the initial line.

The tangent ‘PT’ drawn to the curve makes an angle of `\psi` with the initial line and it makes an angle `\phi` with the radius vector.

The angle `\psi` in the above figure can be written as,

`\psi=\theta+\phi`

By differentiating it with respect to S, we get,

`\frac{d}{dS}(\psi) = \frac{d}{ds}(\theta + \phi)`

`\frac{d\psi}{dS}=\frac{d\theta}{dS}+\frac{d\phi}{dS}`

`\frac{d\psi}{dS}=\frac{d\theta}{dS}+\frac{d\phi}{d\theta}.\frac{d\theta}{dS}`

`\frac{d\psi}{dS}=\frac{d\theta}{dS}(1+\frac{d\phi}{d\theta})`

`\frac{1}{\rho}=\frac{1}{\frac{dS}{d\theta}}(1+\frac{d\phi}{d\theta}) \cdots \text{[Equation-4]}`

Now let’s find the `\frac{dS}{d\theta}` and `\frac{d\phi}{d\theta}`, to get the equation for the radius of curvature.

1] Value for `\mathbf{\frac{dS}{d\theta}}`:-

radius of curvature in Polar form

The above figure indicates the smaller portion of the curve `dS` with the coordinates as follows,

`P=(r,\theta)`, `Q=(r+dr, \theta+d\theta)`

From the above figure,
OP = r
OQ = r + dr
`\anglePOQ = d\theta`
PQ = dS

The line PN indicates the perpendicular drawn to the line OQ from ‘P’.

From the above figure,

`PN=POsind\theta=rsind\theta`

and `NQ=OQ-ON = (r+dr)-rcosd\theta`

`NQ=r(1-cosd\theta)+dr`

`NQ=r(2sin^{2}\frac{d\theta}{2})+dr \cdots [\because 1-cos\theta=2sin^{2}\frac{\theta}{2}]`

Now for `\trianglePQN`

`PQ^{2}=PN^{2}+NQ^{2}`
`dS^{2}=(rsind\theta)^{2}+(2rsin^{2}\frac{d\theta}{2}+dr)^{2}`

Divide both sides with `d\theta^{2}`

`(\frac{dS}{d\theta})^{2}=(\frac{rsind\theta}{d\theta})^{2}+(\frac{2rsin^{2}\frac{d\theta}{2}}{d\theta}+\frac{dr}{d\theta})^{2}`

`(\frac{dS}{d\theta})^{2} = r^{2}(\frac{sind\theta}{d\theta})^{2}+(2r(\frac{d\theta}{4})(\frac{sin\frac{d\theta}{2}}{\frac{d\theta}{2}})^{2}+\frac{dr}{d\theta})^{2}`

`(\frac{dS}{d\theta})^{2} = r^{2}(\frac{sind\theta}{d\theta})^{2}+(\frac{rd\theta}{2}(\frac{sin\frac{d\theta}{2}}{\frac{d\theta}{2}})^{2}+\frac{dr}{d\theta})^{2}`

Taking limits on both sides,

`lim_{d\theta \to 0}(\frac{dS}{d\theta})^{2} = lim_{d\theta \to 0} r^{2}(\frac{sind\theta}{d\theta})^{2}+(lim_{d\theta \to 0}\frac{rd\theta}{2}(\frac{sin\frac{d\theta}{2}}{\frac{d\theta}{2}})^{2}+lim_{d\theta \to 0}\frac{dr}{d\theta})^{2}`

`(\frac{dS}{d\theta})^{2}=r^{2} lim_{d\theta \to 0} (\frac{Sind\theta}{d\theta})^{2}+{0+\frac{dr}{d\theta}}^{2}`

`(\frac{dS}{d\theta})^{2}=r^{2}+(\frac{dr}{d\theta})^{2} \cdots [\because lim_{d\theta \to 0} \frac{sind\theta}{d\theta} = 1]`

`\frac{dS}{d\theta}=\sqrt{r^{2}+(\frac{dr}{d\theta})^{2}}`

By putting `\frac{dr}{d\theta} = r_{1}` in above equation, we get,

`\mathbf{\frac{dS}{d\theta}=\sqrt{r^{2}+r_{1}^{2}}}`

2] Value of `\mathbf{\frac{d\phi}{d\theta}}`:-

radius of curvature in the Polar form

In cartesian form, the slope of the tangent is given by,

`\text{Slope}=\frac{dy}{dx}=tan\psi`

`\frac{dy}{dx}=tan(\theta+\phi)\cdots(\because psi=\theta+\phi)`

`\frac{dy}{dx}=\frac{tan\phi+tan\theta}{1-tan\phitan\theta} \cdots \text{[Equation-5]}`

In cartesian form, the x and y coordinates of point P shown in the above figure can be given by,

`x=rcos\theta, y=rsin\theta`

By differentiating it with respect to `\theta`, we get,

`\frac{dx}{d\theta}=\frac{d}{d\theta}[rcos\theta]`

`\frac{dx}{d\theta}=-r sin\theta+r_{1}cos\theta`

`\text{and} \frac{dy}{d\theta}=\frac{d}{d\theta}[rsin\theta]`

∴`\frac{dy}{d\theta}=rcos\theta+r_{1}sin\theta`

The term `\frac{dy}{dx}` can be also written as,

`\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}`

By putting above values of `\frac{dx}{d\theta}` and `\frac{dy}{d\theta}`, we get,

`\frac{dy}{dx}=\frac{rcos\theta+r_{1}sin\theta}{-rsin\theta+r_{1}cos\theta}`

`\frac{dy}{dx}=\frac{rcos\theta+r_{1}sin\theta}{-rsin\theta+r_{1}cos\theta} \times \frac{\frac{1}{r_{1}}cos\theta}{\frac{1}{r_{1}}cos\theta}`

`\frac{dy}{dx}= \frac{\frac{r}{r_{1}}+tan\theta}{-\frac{r}{r_{1}}tan\theta+1}`

`\frac{dy}{dx}= \frac{(\frac{r}{r_{1}})+tan\theta}{1-(\frac{r}{r_{1}})tan\theta}\cdots \text{[Equation-6]}`

Now compare the right-hand sides of equations 5 and 6,

`\frac{tan\phi+tan\theta}{1-tan\phi tan\theta}=\frac{\frac{r}{r_{1}}+tan\theta}{1-(\frac{r}{r_{1}})tan\theta}`

From the above comparison, we can conclude that,

`tan\phi = \frac{r}{r_{1}}`

`\therefore \phi=tan^{-1}(\frac{r}{r_{1}})`

By differentiating with respect to `\theta`, we get,

`\frac{d\phi}{d\theta}=\frac{d}{d\theta}(tan^{-1}(\frac{r}{r_{1}}))`

`=\frac{1}{1+(\frac{r}{r_{1}})^{2}}.\frac{d}{d\theta}(\frac{r}{r_{1}})`

`=\frac{1}{1+(\frac{r}{r_{1}})^{2}}. \frac{r_{1}r_{1}-rr_{2}}{r_{1}^{2}}`

`=\frac{r_{1}^{2}-rr_{2}}{r_{1}^{2}(1+(\frac{r}{r_{1}})^{2})}`

`\mathbf{\frac{d\phi}{d\theta}=\frac{r_{1}^{2}-rr_{2}}{r_{1}^{2}+r^{2}}}`

Now put these values of `\frac{d\phi}{d\theta}` and `\frac{dS}{d\theta}` in equation 4.

`\frac{1}{\rho}=\frac{1}{\frac{dS}{d\theta}}[1+\frac{d\phi}{d\theta}]`

`\frac{1}{\rho}=\frac{1}{\sqrt{r^{2}+r_{1}^{2}}} [1+\frac{r_{1}^{2}-rr_{2}}{r_{1}^{2}+r^{2}}]`

`\frac{1}{\rho}=\frac{1}{\sqrt{r^{2}+r_{1}^{2}}} [\frac{r_{1}^{2}+r^{2}+r_{1}^{2}-rr_{2}}{r_{1}^{2}+r^{2}}]`

`\frac{1}{\rho}=\frac{1}{\sqrt{r_{1}^{2}+r^{2}}} [\frac{r^{2}+2r_{1}^{2}-rr_{2}}{r_{1}^{2}+r^{2}}]`

`\frac{1}{\rho}=\frac{r^{2}+2r_{1}^{2}-rr_{2}}{(r_{1}^{2}+r^{2})^{\frac{3}{2}}}`

`\therefore \mathbf{\rho=\frac{(r^{2}+r_{1}^{2})^\frac{3}{2}}{r^{2}+2r_{1}^{2}-rr_{2}}}`

This is the equation to find the radius of curvature in polar form.

Radius of curvature solved examples:

1] The curve is represented as, y = a³x⁴. If a = 2, Find the radius of curvature at (2.5).

Solution:-

The above curve is given in the cartesian form.

The radius of curvature is given by,

`\rho=\frac{(1+y_{1}^{2})^{\frac{3}{2}}}{y_{2}}`

Where,

`y_{1}=\frac{dy}{dx}=\frac{d}{dx}(a^{3}x^{4})=4a^{3}x^{3}`

`y_{2}=\frac{d^{2}y}{dx^{2}}=\frac{dy_{1}}{dx}=\frac{d}{dx}(4a^{3}x^{3})=12a^{3}x^{2}`

Now put `y_{1}` and `y_{2}` in above equation of `\rho`.

`\rho = \frac{(1+(4a^{3}x^{3})^{2})^{\frac{3}{2}}}{12a^{3}x^{2}}`

`\rho=\frac{(1+16a^{6}x^{6})^\frac{3}{2}}{12a^{3}x^{2}}`

Now the radius of curvature for a = 2 and x = 2.5 is given by,

`\rho=\frac{(1+16(2)^{6}(2.5)^{6})^{\frac{3}{2}}}{12(2)^{3}(2.5)^{2}}`

`\rho=208334`


2] Find the radius of curvature at any point for the below curve,

`x=t^{2}+5, y= 2t+2`

Solution:-

The curve is represented in a parametric form.

The radius of curvature in the parametric form is given by,

`\rho=\frac{x’^{2}+y’^{2}}{y”x’-x”y’}`

Where

`x’=\frac{dx}{dt}=2t, x”=\frac{dx’}{dt}=2`

`y’=\frac{dy}{dt}=2, y”=\frac{dy’}{dt}=0`

Put these values in the formula of the radius of curvature.

`\rho=\frac{x’^{2}+y’^{2}}{y”x’-x”y’}`

`\rho=\frac{4t^{2}+4}{-4}`

`\rho=-(t^{2}+1)`


3] For the below curve, find the radius of curvature,

`\frac{a}{2r}=cos^{2}\theta`

Solution:-

The curve is described in a polar form.

`\frac{a}{2r}=cos^{2}\theta`

`\thereforer=\frac{a}{2cos^{2}\theta}`

Step 1] Find `r_{1}` and `r_{2}`

`\therefore r_{1}=\frac{dr}{d\theta}=\frac{d}{d\theta}[\frac{a}{2cos^{2}\theta}]`

`r_{1}=\frac{d}{d\theta}[\frac{a}{2}sec^{2}\theta]`

`r_{1}=\frac{a}{2}\frac{d}{d\theta}sec^{2}\theta`

`r_{1}=\frac{a}{2}(2sec^{2}\thetatan\theta)`

`\mathbf{r_{1}=asec^{2}\thetatan\theta}`

`r_{2}=\frac{d^{2}r}{d\theta^{2}}=\frac{dr_{1}}{d\theta}=\frac{d}{d\theta}[asec^{2}\theta.tan\theta]`

`r_{2}=a[sec^{2}\theta \frac{d}{d\theta} tan\theta +tan\theta\frac{d}{d\theta}sec^{2}\theta]`

`r_{2} = a[sec^{2}\theta sec^{2}\theta+tan\theta (2sec^{2}\theta tan\theta)]`

`r_{2} = a[sec^{4}\theta+2sec^{2}\theta tan^{2}\theta]`

`r_{2} = a sec^{2}\theta [sec^{2}\theta+2tan^{2}\theta]`

Step 2] Find the radius of curvature

The radius of curvature in polar form is given by,

`\rho=\frac{(r^{2}+r_{1}^{2})^{\frac{3}{2}}}{r^{2}+2r_{1}^{2}-rr_{2}}`

`\rho=\frac{{(\frac{a}{2cos^{2}\theta})^{2}+(asec^{2}\theta tan\theta)^{2}}^{\frac{3}{2}}}{(\frac{a}{2cos^{2}\theta})^{2}+2(asec^{2}\theta tan\theta)^{2}-[\frac{a}{2cos^{2}\theta}(asec^{2}\theta [sec^{2}\theta+2tan^{2}\theta])]}`

`\rho=\frac{a^{3}[\frac{sec^{4}\theta}{4}+sec^{4}\theta tan^{2}\theta]^{\frac{3}{2}}}{a^{2}[\frac{sec^{4}\theta}{4}+2sec^{4}\theta tan^{2}\theta-\frac{sec^{6}\theta}{2}-sec^{4}\theta tan^{2}\theta]}`

`\rho=\frac{a sec^{6}\theta[\frac{1}{4}+tan^{2}\theta]^{\frac{3}{2}}}{[\frac{sec^{4}\theta}{4}+sec^{4}\theta tan^{2}\theta-\frac{sec^{6}\theta}{2}]}`

`\rho=\frac{a sec^{6}\theta[\frac{1}{4}+tan^{2}\theta]^{\frac{3}{2}}}{sec^{4}\theta[\frac{1}{4}+tan^{2}\theta-\frac{sec^{2}\theta}{2}]}`

`\rho=\frac{a sec^{2}\theta[\frac{1}{4}+tan^{2}\theta]^{\frac{3}{2}}}{[\frac{1}{4}+tan^{2}\theta-\frac{sec^{2}\theta}{2}]}`

FAQs:

  1. Are the center of curvature and radius of curvature similar?

    No, both are different things.

  2. What is the relation between the curvature and the radius of curvature?

    The radius of curvature is the reciprocal of the curvature of the curve.

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