MULTIPLE INTEGRAL
VECTOR CALCULUS
DIFFERENTIAL EQUATIONS
MATRICES
DIFFERENTIAL CALCULUS

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.