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Polar CoordinatesGood Luck.... Regards WASEEM AKHTER
Citation preview
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
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