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 is the angle between the radius vector and the tangent to a curve of the given family at any point (r, θ), then

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

dr

dr

tan

901

1

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

Hence the DE of the orthogonal trajectories is

(2)

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

Q

P

dr

d

cottan 1

1tantan 1

Q

rP

dr

dr

rP

Q

dr

dr

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

Soln: Differentiating the eq

is the DE of eq (1)

DE of the orthogonal trajectories is

dr

da

cos1

sin1..

cos

sin1

cos

1

raas

radr

d

cos

sin1

dr

dr

sin1

cos

dr

dr

Separating the variables

r

drd

cos

)sin1(

r

drdd tan sec

rc lnlncosln)tanln(sec

r

c

cos

)tan(sec

is the eq of the orthogonal trajectory

r

c

cos

)cossin

cos1

(

r

c

2cos

)sin1(

r

c

2sin1

)sin1(

r

c

)sin1()sin1(

)sin1(

r

c

)sin1(

1

)sin1( cr

Find the eq of orthogonal trajectory of the curve (1)

Soln: Differentiating the eq

)cos2(..

sin

)cos2(

sin)cos2(

)cos2(

sin

)cos2(][

)cos2(

sin1

2

2

2

raas

rrdr

d

adr

d

dr

da

cos2

a

r

is the DE of eq (1)

DE of the orthogonal trajectories is

Separating the variables

r

drd

sin

)cos2(

r

drdd cot csc2

sin

)cos2(

dr

dr

)cos2(

sin

dr

dr

is the eq of the orthogonal trajectory

r

c

2)cot(csc

sin

rc lnlnsinln)cotln(csc2

r

cc

2)

sinos

sin1

(

sin

r

c

2

3

)cos1(

sin

23 )cos1(sin cr

Find the eq of orthogonal trajectory of the curve

2sin2 ar

EQUATION

SOLVABLE FOR P

Equation solvable for Parameter PIf and

then we have and is called

Parameter

Geometrically

cosax

222 ayx

sinay

jaiar ˆsinˆsin)(ˆ

)(ˆ r

y

x

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

will be denoted by p and

will be denoted

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

dx

dy

52

.......

dx

dyand

dx

dy

52 ....... pandp

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:

06 222 yxyppx

0623 222 yxypxyppx

0)2(2)3( yxpyxpxp

0)3)(2( yxpyxp

0)2( yxp 0)3( yxpand

ydx

dyx 2 y

dx

dyx 3

x

dx

y

dy2

x

dx

y

dy3

cxy lnln2ln xcy ln3lnln 2cxy

3x

cy

02 cxy03 cyx

So the Soln is

0))(( 32 cyxcxy

Solve

Soln:

01)(2 pyxxyp

012 ypxpxyp

0)1()1( xpxpyp

0)1)(1( ypxp

1 xp 1... ypand

012 xpypxyp

1dx

dyx 1

dx

dyy

x

dxdy dxydy

)ln(ln cxy xcy

2

2

cxy ln )(22 xcy

So the Soln is 0]2)][ln([ 2 cxycxy

0)ln( cxy 0222 cxy

Solve

Soln:

0)(

)2()( 2222

xyy

pxxyypyx

0)(

)())(( 2222

xyy

pxyxyypyxyx

0)())((

)())(( 2

xyyxyxyp

xypypyxyx

0])()[(

])()[(

yxypxy

yyxpyxp

Solving eq (1)

Put y = vx and

0])(][)([ yxypxyyxp

)2......(0)(....

)1......(0)(

yxypand

xyyxp

)1......(0)( xyyxp

yx

yx

dx

dy

yx

yxp

dx

dvxv

dx

dy

v

v

vxx

vxx

dx

dvxv

1

1

on integration

v

vvvv

v

v

dx

dvx

1

1

1

1 2

v

vv

v

vv

dx

dvx

1

12

1

21 22

x

dx

vv

dvv

12

)1(2

12 lnln)12ln(

2

1cxvv

x

dx

vv

dvv

12

)22(

2

12

12

12 lnln)12ln( cxvv

12

12 )12( cvvx

12

1

2

2

)12( cx

y

x

yx

12

122 )2( cxxyy

21222 )2( ccxxyy

Solving eq (2)

Put y = vx and

)2.....(0)( yxyp

yx

y

dx

dy

yx

yp

dx

dvxv

dx

dy

v

v

vxx

vx

dx

dvxv

1

v

vvvv

v

v

dx

dvx

11

2

on integration

v

vv

v

vv

dx

dvx

1

2

1

2 22

x

dx

vv

dvv

2

)1(2

32 lnln)2ln(

2

1cxvv

x

dx

vv

dvv

2

)22(

2

12

32

12 lnln)2ln( cxvv

32

12 )2( cvvx 3

2

1

2

2

)2( cx

y

x

yx

32

12 )2( cxyy

4322 )2( ccxyy

So the Soln is

cxyyxxyy )2)(2( 222

Solve

Solve

0)(2 xpyxyp

0)( 22 xypyxyp