# How to find Orthogonal Trajectories of Family of Curves

Differential Equation / Friday, November 1st, 2019

# Orthogonal Trajectories and Family of Curves

Before learning about orthogonal trajectories we have to first understand the meaning of family of curves.

## Family of curves

The equation of a curve generally involves, besides the current co-ordinates x and y, one or more parameters, upon which the size, shape and position of a particular curve in relation to the co-ordinate axes depend.

Let us consider the equation x2 + y2 = a2, which represents a circle with center at the origin and radius ‘a’. If now the parameter ‘a’ changes its value, the size of the circle changes also. So, x2 + y2 = a2 represents a system or a family of circles, all circles of the system or family have their centers at the origin, but their radii differ.

Again let us consider the equation (x – a)2 + y2 = r2 , where ‘r’ is constant and ‘a’ is variable parameter. The equation represents a system of circles of fixed radius ‘r’, but their centers move on the x-axis as ‘a’ change its value.

A system of curves formed in this way is called a family of curves, and the quantity ‘a’ which remains fixed for one particular member of the family, but changes in passing from one member to another member of the family, is called a parameter.

## Orthogonal Trajectories

If two families of curves be such that each member of one family cuts each member of the other family orthogonally (i.e., at right angles), then each family is said to be a set of orthogonal trajectories of the other.

## Orthogonal Trajectories in Rectangular Cartesian Co-ordinates

Let, the equation of the family of curves be

$f\left( x,y,a \right)=0……….(i)$

‘a’ being the variable parameter.

Differentiating (i) with respect to x and eliminating ‘a’, we have the differential equation of the family of curves as

$\phi \left( x,y,\frac{dy}{dx} \right)=0……….(ii)$

If two curves cut at right angles and if ‘m’ be the slope of one curve at the point of intersection then

$m=\tan \psi =\frac{dy}{dx}$

Thus the slope of the other curve be m’, where

$mm’=-1~~i.e.,m’=\frac{1}{m}=-\frac{dx}{dy}$

Thus the differential equation of the system of orthogonal trajectories of (i) will be given by

$\phi \left( x,y,-\frac{dy}{dx} \right)=0……….(iii)$

Integration of (iii) will lead to equation of the family of orthogonal trajectories in Cartesian form.

## Orthogonal Trajectories in Polar Co-ordinates

Let the equation of the family of curves in polar co-ordinates be

$f\left( r,\theta ,a \right)=0………..(i)$

where ‘a’ is the variable parameter.

Differentiating (i) with respect to θ and eliminating ‘a’, we get the differential equation of the family of curves,

$\phi \left( r,\theta ,\frac{dr}{d\theta } \right)=0………..(ii)$

If φ and φ’ denote the angles which the tangent to the given curve and the trajectory at the point of intersection (r, θ), make with the radius vector to the common point, φ ~ φ’ = (π/2) and so tanφ = – cotφ’.

Now since tanφ = r(dθ/dr), obviously the differential equation to the system of orthogonal trajectories is obtained by substituting

$-\frac{1}{r}\frac{dr}{d\theta }~~for~~r\frac{d\theta }{dr}~~i.e.,~~-{{r}^{2}}\frac{d\theta }{dr}~~for~~\frac{dr}{d\theta }~~in~~(ii).$

Hence the differential equation to the required family of orthogonal trajectories is

$\phi \left( r,\theta ,-{{r}^{2}}\frac{d\theta }{dr} \right)=0………..(iii)$

The equation of the family of orthogonal trajectories will be obtained by integrating (iii).

## Working rule for the determination of Orthogonal Trajectories

### Cartesian Co-ordinates

Given

$f\left( x,y,a \right)=0……….(i)$

Where ‘a’ is a parameter, as the equation of the family of curves.

Step 1: Differentiating (i) with respect to x and eliminate the parameter ‘a’ to get the differential equation of the family of curves:

$\phi \left( x,y,\frac{dy}{dx} \right)=0……….(ii)$

Step 2: Replace (dy/dx) by (- dx/dy) in (ii) to get the differential equation of the family of the orthogonal trajectories:

$\phi \left( x,y,-\frac{dy}{dx} \right)=0……….(iii)$

Step 3: Integrate (iii) to get the Cartesian equation of the family of orthogonal trajectories. Obviously, the final equation will involve an arbitrary constant.

### Polar Co-ordinates

Given

$f\left( r,\theta ,a \right)=0………..(i)$

as the family of curves; here ‘a’ is parameter.

Step 1: Form the differential equation

$\phi \left( r,\theta ,\frac{dr}{d\theta } \right)=0………..(ii)$

by eliminating ‘a’ from the given equation of the curve.

Step 2: Replace

$-\frac{1}{r}\frac{dr}{d\theta }~~for~~r\frac{d\theta }{dr}~~i.e.,~~-{{r}^{2}}\frac{d\theta }{dr}~~for~~\frac{dr}{d\theta }~~in~~(ii).$

To get the differential equation of the family of orthogonal trajectories:

$\phi \left( r,\theta ,-{{r}^{2}}\frac{d\theta }{dr} \right)=0………..(iii)$

Step 3: Integrate (iii) to get the polar equation of the family of orthogonal trajectories. This equation will obviously involve an arbitrary constant.

 Example 01

Find the equation of the orthogonal trajectories of the family of curves

${{x}^{\frac{2}{3}}}+{{y}^{\frac{2}{3}}}={{a}^{\frac{2}{3}}},~~where~~a~~is~~parameter$

Solution:

Equation of the given family of curve is

${{x}^{\frac{2}{3}}}+{{y}^{\frac{2}{3}}}={{a}^{\frac{2}{3}}}……….(i)$

Differentiating with respect to x, we have

$\frac{2}{3}{{x}^{-\frac{1}{3}}}+\frac{2}{3}{{y}^{-\frac{1}{3}}}\frac{dy}{dx}=0$

$\Rightarrow {{y}^{\frac{1}{3}}}+{{x}^{\frac{1}{3}}}\frac{dy}{dx}=0……….(ii)$

is the differential equation of the given family of curves. The differential equation of the orthogonal trajectories is

${{y}^{\frac{1}{3}}}+{{x}^{\frac{1}{3}}}\left( -\frac{dx}{dy} \right)=0$

$\Rightarrow {{y}^{\frac{1}{3}}}dy-{{x}^{\frac{1}{3}}}dx=0……….(iii)$

Integrating (iii) we get the equation of the required orthogonal trajectories as

$\frac{3}{4}{{y}^{\frac{4}{3}}}-\frac{3}{4}{{x}^{\frac{4}{3}}}=\frac{3}{4}c$

$\Rightarrow {{y}^{\frac{4}{3}}}-{{x}^{\frac{4}{3}}}=c$

Where c is arbitrary constant.

 Example 02

Find the orthogonal trajectories of the family of cardioids

$r=a\left( 1-\cos \theta \right)$

Solution:

Here

$r=a\left( 1-\cos \theta \right)$

Differentiating with respect to θ, we have

$\frac{dr}{d\theta }=a\sin \theta$

Eliminating ‘a’, we have

$\frac{dr}{d\theta }=\frac{r\sin \theta }{\left( 1-\cos \theta \right)}=\frac{r.2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{{\sin }^{2}}\frac{\theta }{2}}$

$\Rightarrow \frac{dr}{d\theta }=r\cot \frac{\theta }{2}$

Which is the differential equation of the given family of curves.

Replacing

$\frac{dr}{d\theta }~~by~~-{{r}^{2}}\frac{d\theta }{dr}$

We get the differential equation of orthogonal trajectories as

$-{{r}^{2}}\frac{d\theta }{dr}=r\cot \frac{\theta }{2}$

$\Rightarrow \frac{dr}{r}=-\tan \frac{\theta }{2}d\theta$

Integrating, we get

$\log r=2\log \cos \frac{\theta }{2}+\log 2c$

$\Rightarrow r=2c{{\cos }^{2}}\frac{\theta }{2}$

$\Rightarrow r=c\left( 1+\cos \theta \right)$

Where c is arbitrary, is the equation of the required family of orthogonal trajectories.

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