Polar Coordinates

POLAR COORDINATES

Let f(r, θ, c) = 0 be the equation of family of  curves. The DE of this family can be obtained by elimination of c, which is P dr + Q dθ = 0 (1) Where P and Q are function of r and θ. We know from calculus that if θ  is the angle between the radius vector and the tangent to a curve of the given family at any point (r, θ), then d θ  tan θ  r 

dr 

If θ  is the angle between the radius vector and 9 the tangent to an orthogonal trajectory at (r, θ), then θ 9 θ  9 9 

tan θ 9

cot θ 

tan θ 9 tan θ 

9 For the two curves to be orthogonal. From eq (1), we have

θ  d θ 

dr 

 P 

Q

r 

d θ 

rP 

dr 

Q

Hence the DE of the orthogonal trajectories is

r 

d θ 

Q

dr 

rP 

(2)

Solution of eq (2) is the required family of the orthogonal trajectories of the family f(r, θ, c) = 0

Find the eq eq of orthogonal trajectory of the curve curve r = a (1 + sin θ) (1) Soln: Differentiating the eq 9

d θ 

9

dr 

a cos θ 

r 

d θ 

9

r cos θ 

sin θ 

9

sin θ 

a cos θ 

.as .a

DE of the orthogonal trajectories is

r 

cos θ 

d θ  dr 

9

dr  r  9

is the DE of eq (1)

cos θ 

dr 

d θ 

sin θ 

sin θ 

Separating the variables

(9

sin θ )d θ 

dr 

cos θ 

r 

secθ d θ  ln(sec θ 

tanθ d θ 

dr  r 

tan θ ) ln cos θ  (sec θ 

tan θ )

cos θ 

ln c c r 

ln r 

(

sin θ 

9

cos θ  cos θ  cos θ 

(9

sin θ )

c

9

9

r 

sin

θ 

)

(9

9

c

sin θ )

r 

sin θ ) cos

r 

(9 (9

(9

c

9

θ 

sin θ )

sin θ )(9

sin θ )

c (9

sin θ )

r 

is the eq of the orthogonal trajectory

c r 

c r 

Find the eq eq of orthogonal trajectory of the curve curve

a

r 

(1)

cos θ 

9

Soln: Differentiating the eq 9

a sin θ 

(9

cos θ ) (9

d θ  dr 

.as .a

r ( 9 r ( 9

9

[

d θ  dr 

cos θ )

]

9

cos θ ) sin θ  cos θ )

d θ  dr 

(9

(9

cos θ ) a sin θ 

cos θ ) r sin θ 

9

r 

d θ 

(9

cos θ ) is the DE of eq (1)

sin θ 

dr 

DE of the orthogonal trajectories is

r 

sin θ 

d θ  dr 

(9

cos θ )

Separating the variables

(9

9

cos θ )d θ 

dr 

sin θ 

r 

cscθ d θ 

cotθ d θ 

dr  r 

9

ln(csc θ  cot θ ) sin θ 

sin (9

r sin

9

9

sin θ  cosθ 

c ( sin θ  r 

θ 

θ 

ln r 

r  9

cos θ )

ln c

c

(csc θ  cotθ )

9

ln sin θ 

9

c(9

cos θ )

sin θ 

9

is the eq of the orthogonal trajectory

c

)

9

r 

Find the eq eq of orthogonal trajectory of the curve curve 9

r 

a sin 9 θ 

EQUATION SOLVABLE FOR P

Equation solvable for Parameter P If

 x 

a cos θ 

then we have

and 9

 x 

 y

Parameter  Geometrically

 y

ˆ (θ ) r   y

9

a sin θ  a

9

and θ  is called

a sin θ i ˆ

ˆ a sin θ  j 

ˆ (θ ) r 

 x 

Consider first order DE with degree more than one or higher degree. In this section

dy dx 

will be denoted by p and

dy dx  2

2

...and .... 5

dy dx 

5

will be denoted

by  p ...and .... p where p will be parameter. Here we will solve the first order DE with degree more than one or higher.

with following method: Solve the first order DE by factorizing the right side of the DE and take each factor  seperately and then solve it. After solving each factor, multiply the solution of each factor and place them equal to zero.

Solve Soln:

2

 x   p 2

 x   p

2

6 y

 xyp

2

 xp( xp

2

0 2

0 0

3 xyp

2 xyp

6 y

3 y )

2( xp

2 y )

( xp

2 y )( xp

( xp

2 y )

0

3 y )

0

and ( xp

3 y )

0

 x 

dy dx 

dy  y

ln  y  y  y

2

dy

2 y

 x 

dx 

dy

 x 

 y

3 y

dx 

3

dx   x 

2 ln  x  ln c ln  y

2

cx 

2

cx 

 y

So the Soln is 2

3

cx  )( x   y

3 ln  x 

3

 x  3  x   y c

0 ( y

ln c c

c)

0 0

Solve Soln:

 xyp

2

( x   y ) p 1

0

 xyp

2

 xp  yp 1

0

 xyp

2

 yp  xp 1

0

 yp( xp 1) ( xp 1)

( xp 1)( yp 1)  xp

1

0

0

and ... yp

1

 x 

dy dx 

dy  y  y  y

1

dy

 y

dx 

dx 

 ydy

 x   y

(ln  x  ln c )

ln( cx )

So the Soln is

 y

0 [ y

 y

2

ln( cx )][ y

2

dx 

2

2

ln cx 

1

2

2 x 

c  x 

2(c  x ) 2c

2 x  c ]

0

0

2

Solve Soln:

( x   y )  p

2

( 2 y

 y( y  x )

( x   y )( x   y ) p  y ( y  x )

2

2

 xy  x  ) p

0 2

( y

2

 xy

 y

2

0

( x   y )( x   y ) p

2

 py( y  x )

 p( y  x )( y  x )  y( y  x )  p( x   y )[ p( x   y )  y ]

( y  x )[ p( y  x )  y ]

0

0

2

 x  ) p

[ p( x   y )  y  x ][ p( y  x )  y ]

0

 p( x   y )  y  x  0......(1) and .... p( y  x )  y Solving eq (1)

 p

0......( 2)

 p( x   y )  y  x  0......(1)

 x   y

dy

 x   y

 x   y

dx 

 x   y

Put y = vx and

v  x 

dv

 x  vx 

1 v

dx 

 x  vx 

1 v

dy dx 

v  x 

dv dx 

 x   x 

dv

1 v

dx 

1 v

v

1 2v

dv

1 v

v

v

2v

1 (2 2 v 1 2

ln( v

2

2

1

v

1 v 2

v

1 v dx  (1 v )dv dx  2

v

2

2

2v

1

1 v

 x 

2v )dv

dx 

2v

1

 x 

2v

1)

on integration

ln  x  ln c1

1

ln( v

2

2v

1) 2

ln  x 

ln c1

1

 x (v

2

 y

2

 x (

2

 x 

2v 2

 y  x 

1) 2

c1 1

1) 2

c1

1

( y

2

( y

2

2 xy  x  ) 2 2

2

2 xy  x  )

c1 c

2

1

c2

 p( y  x )  y

Solving eq (2)

 p

 y  x   y

dy

 y

dx 

 x   y dy

Put y = vx and

v  x   x 

dv dx 

dv dx 

vx   x  vx  v

1 v

0.....( 2)

v

v

dx 

v  x 

1 v v

v

1 v

v

2

dv dx 

 x 

dv

2v

dx 

1 v

(1 v )dv v

2

2 1 2

v

2

ln( v

2

v

2

2v

1 v

dx 

2v

1 (2

v

2

 x 

2v )dv

dx 

2v

 x 

2v )

on integration

ln  x  ln c 3

1

ln( v

2

2v ) 2

ln  x  ln c3

1

 x (v

2

2v )

2

c3

 x (

 y

2 2

 x 

2

 y

)2  x 

1

( y

2

( y

2

2 xy ) 2

c3

2 xy )

2

c

3

c4

So the Soln is

( y

2

2

2 xy  x  )( y

2

2 xy )

1

c

c3

Solve

Solve

2

 p  y

2

 p  y

( x   y ) p  x  0

2

( x   y ) p  xy

0