DOUBLE INTEGRAL
BETA AND GAMMA FUNCTION
VOLUME
SURFACE AREA
GRADIENT, DIVERGENCE AND CURL
LINE INTEGRAL
GREENS THEOREM
SURFACE INTEGRAL
GAUSS DIVERGENCE THEOREM
STOKES THEOREM
CONSERVATIVE VECTOR FIELD
LIMITS
CONTINUITY
DIFFERENTIABILITY
APPLICATION OF DERIVATIVES
MEAN VALUE THEOREM
Part B: FUNCTION OF TWO VARIABLES
INTRODUCTION
DIFFERENTIAL EQUATIONS OF FIRST ORDER AND FIRST DEGREE
ORTHOGONAL TRAJECTORY
DIFFERENTIAL EQUATION WITH CONSTANT COEFFICIENTS
CAUCHY EULER EQUATIONS
DIFFERENTIAL EQUATION OF SECOND ORDER
Section 5: LINEAR ALGEBRA
VECTOR SPACE AND LINEAR EQUATIONS
ORTHOGONALITY
EIGENVALUES AND EIGENVECTORS
LINEAR TRANSFORMATION
Section 6: GROUP THEORY
CYCLIC GROUP
PERMUTATION GROUP
ISOMORPHISM OF GROUPS
EXTRNAL DIRECT PRODUCT
HOMOMORPHISM OF GROUP
SET THEORY
REAL NUMBERS
TOPOLOGY ON REAL LINE
REAL SEQUENCES
INFINITE SERIES
POWER SERIES

Problems on Orthogonal Trajectory

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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.