### IIT JAM PHYSICS

NEWTON LAWS OF MOTION
Tension
FRICTION
Friction
VELOCITY AND ACCELERATION
CENTRAL FORCES
Gravitation
UNIFORMLY ROTATING FRAME- CENTRIFUGAL AND CORIOLIS FORCES
CONSERVATION LAWS
Collision
CENTRE OF MASS AND VRIABLE MASS SYSTEMS
RIGID BODY DYNAMICS
FLUID DYNAMICS
COULOMB LAW AND ELECTRIC FIELD
GAUSS LAW OF ELECTROSTATICS AND APPLICATIONS
Capacitance
POLARIZATION OF DIELECTRICS
WORK AND ENERGY IN ELECTROSTATICS
Capacitors
BOUNDARY VALUE PROBLEMS
CURRENT ELECTRICITY
MAGNETOSTATICS
Ampere Law
FARADAY LAW OF ELECTROMAGNETIC INDUCTION
MAGNETIC MATERIALS
DC CIRCUITS
RC Circuit
LR circuit
LC Circuit
AC CIRCUITS
AC Circuit
MAXWELL EQUATIONS and poynting vector
ELECTROMAGNETIC WAVES
REFLECTION AND REFRACTION OF EM WAVES AT THE INTERFACE OF TWO DIELECTRICS
Section 3: MATHEMATICAL PHYSICS
MULTIPLE INTEGRAL
Gradient
VECTOR CALCULUS
DIFFERENTIAL EQUATIONS
MATRICES
Determinant
DIFFERENTIAL CALCULUS
Jacobian
FOURIER SERIES
PARTICLE NATURE OF WAVE
WAVE NATURE OF PARTICLE
H ATOM
POSTULATES OF QUANTUM MECHANICS
SCHRONDINGER WAVE EQUATION
NUCLEAR PHYSICS
SPECIAL THEORY OF RELATIVITY
SIMPLE HARMONIC OSCILLATION
DAMPED AND FORCED OSCILLATION
WAVES
Waves
GEOMETRICAL OPTICS
Thin Lens
INTERFERENCE
Thin Films
DIFFRACTION
Single Slit
Double Slit
POLARIZATION OF LIGHT
THERMAL EXPANSION
CALORIMETRY
Calorimetry
TRANSMISSION OF HEAT
1 of 2

# Problems on Orthogonal Trajectory

Example 1 :

Find the orthogonal trajectories of family of circles $x^2 + y^2 + 2gx + c = 0,$, where g is the parameter.

Solution :

The given family of curves

$x^2 + y^2 + 2gx + c = 0$ …(1)

Differentiate both sides w.r.t. x,

$2x + 2y \dfrac{{dy}}{{dx}} + 2g = 0 \Rightarrow g = -\left( {x + y \dfrac{{dy}}{{dx}}} \right)$ …(2)

Put the value of g in (1)

${x^2} + {y^2} + 2x\left[ {- x - y \dfrac{{dy}}{{dx}}} \right] + c = 0$

${y^2} - {x^2} + 2xy \dfrac{{dy}}{{dx}} + c = 0$ …(3)

which is the differential equation of given family.

Now replace $\dfrac{{dy}}{{dx}}$ by $-\dfrac{{dx}}{{dy}}$ in (3)

So ${y^2} - {x^2} + 2xy \dfrac{{dx}}{{dy}} + c = 0$

$2xy \dfrac{{dx}}{{dy}} - {x^2} = -c - {y^2}$

$2x \dfrac{{dx}}{{dy}} - \dfrac{1}{y}{x^2} = - \dfrac{c}{y} - y$

put ${x^2} = v \Rightarrow 2x \dfrac{{dx}}{{dy}} = \dfrac{{dv}}{{dy}}$

put in (4) $\dfrac{{dv}}{{dy}} - \dfrac{1}{y}v = - \dfrac{c}{y} - y$ …(5)

which is linear differential equation

Now ${\rm{IF}} = {e^{-\int {\dfrac{1}{y}dy} }} = {e^{ - \log y}} = \dfrac{1}{y}$

Multiply both sides (5) with IF, it becomes.

$v \cdot \dfrac{1}{y} = \int {\left( {- \dfrac{c}{{{y^2}}} - 1} \right)} dy$

$\dfrac{v}{y} = \dfrac{c}{y}- y + d$

$\dfrac{{{x^2}}}{y} = \dfrac{{c - {y^2} + dy}}{y}$

$x^2 + y^2 - dy - c = 0$ where d is the parameter.

Example 2 :

Show that one parameter family of curves $y^2 = 4c(c + x)$ are self orthogonal.

Solution :

We are given $y^2 = 4c(c + x)$ …(1)

Differentiate $2y{y_1} = 4c(1) \Rightarrow c = \dfrac{{y{y_1}}}{2}$ …(2)

Put value of c in (1)

${y^2} = 2y{y_1} \left( {\dfrac{{y{y_1}}}{2} + x} \right) \Rightarrow {y^2} = y{y_1}(y{y_1} + 2x)$

(3) gives the differential equation of given family

Now replace y’ by $\dfrac{-1}{y'}$ in (3), we get the differential equation of orthogonal trajectory.

${y^2} = y\left( {-\dfrac{1}{{{y_1}}}} \right) \cdot \left[ {y\left( {-\dfrac{1}{{{y_1}}}} \right) + 2x} \right] \Rightarrow {y^2}y_1^2 = {y^2} - 2xy{y_1}$

${y^2} = {y^2}y_1^2 + 2xy{y_1}$

$y^2 = y y_1 [yy_1 + 2x]$

Which is same as in equation (3)
So, differential equation of given curve and differential equation of its orthogonal trajectaries are same. So, the family of curves is self orthogonal.

Example 3 :

Find the orthogonal trajectories of the family of curves $\dfrac{{{x^2}}}{{{a^2} + \lambda }} + \dfrac{{{y^2}}}{{{b^2} + \lambda }} = 1$ , where $\lambda$ is a parameter.

Solution :

We are given $\dfrac{{{x^2}}}{{{a^2} + \lambda }} + \dfrac{{{y^2}}}{{{b^2} + \lambda }} = 1$

Differentiate both sides w.r.t x

$\dfrac{{2x}}{{{a^2} + \lambda }} + \dfrac{{2y}}{{{b^2} + \lambda }} \cdot \dfrac{{dy}}{{dx}} = 0 \Rightarrow \dfrac{x}{{{a^2} + \lambda }} + \dfrac{y}{{{b^2} + \lambda }} \cdot \dfrac{{dy}}{{dx}} = 0$ …(1)

On solving $x({b^2} + \lambda ) + y({a^2} + \lambda ) \dfrac{{dy}}{{dx}} = 0$

$\lambda \left[ {x + y \dfrac{{dy}}{{dx}}} \right] = -\left[ {{b^2}x + {a^2}y \dfrac{{dy}}{{dx}}} \right]$

$\lambda = - \dfrac{{\left[ {{b^2}x + {a^2}y \dfrac{{dy}}{{dx}}} \right]}}{{x + y \dfrac{{dy}}{{dx}}}}$

$a^2 + \lambda = a^2 - \dfrac{b^2 x + a^2 y \ \ dy/dx}{x + y \ dy/dx} = \dfrac{(a^2 - b^2)x}{x + y \ dy/dx}$

$b^2 + \lambda = b^6 - \left [ \dfrac{b^2 x + a^2 y \dfrac{dy}{dx}}{x + y \dfrac{dy}{dx}} \right ] = \dfrac{(-a^2 + b^2)y (dy/dx)}{x + y \ dy/dx}$

Now on substituting values of $a^2 + \lambda$ and $a^2 + \lambda$ in (i)

$x^2 \cdot \dfrac{x + y dy/dx}{(a^2 - b^2)x} + y^2 \cdot \dfrac{x + y dy/dx}{(-a^2 + b^2)y \dfrac{dy}{dx}} = 0$

$\left [x + y \dfrac{dy}{dx} \right ] \left [ \dfrac{x}{a^2 - b^2} - \dfrac{y}{(a^2 - y^2) \dfrac{dy}{dx}} \right ] = 1$

$\left [x + y \dfrac{dy}{dx} \right ] \left [x - y \dfrac{dx}{dy} \right ] = (a^2 - b^2),$ which is the differential equation is given family. …(2)

Now replace $\dfrac{dy}{dx}$ with $-\dfrac{dx}{dy}$ to obtain differential equation of orthogonal trajectory

$\left [x - y \dfrac{dx}{dy} \right ] \left [x + y \dfrac{dy}{dx} \right ] = (a^2 - b^2 )$

Differential equation (2) and (3) are same, which gives the differential equation of family. It’s orthogonal trajectories are same. So the family of curves are self orthogonal.

Example 4 :

Show that the families of curves given by the equation $r = a(1 - \cos \theta)$ and $r = b(1 + \cos \theta)$ intersect orthogonally.

Solution :

Here we have to show that the family of orthogonal trajectory of the family of curves $r = a(1 - \cos \theta)$

$r = b(1 + \cos \theta )$

We have $r = a(1 - \cos \theta )$ …(1)

On taking logrithm both sides

$\log r = \log a + \log(1 - \cos \theta)$

Differentiate both sides w.r.t $\theta$

$\dfrac{1}{r} \cdot \dfrac{dr}{d \theta} = \dfrac{\sin \theta}{1 - \cos \theta}$ …(2)

which is free from parameter. So, it is the differential equation of given family.

Now replace $\dfrac{dr}{d \theta}$ with $- r^2 \dfrac{d \theta}{d r}$ in (2)

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

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

On integrating $\log r = 2 \log \cos \dfrac{\theta}{2} + \log c$

$\log r = \log \left( {c \ {{\cos }^2} \dfrac{\theta }{2}} \right)$

$r = c\left( {\dfrac{{1 + \cos \theta }}{2}} \right)$

$r = b(1 + \cos \theta )$

which is the required orthogonal trajectory.