Now, we take a point P.
On the line OP we choose an axis u.
The number t is a value of the angle from the x-axis to the u-axis.
The number r is such that P = r.U
The numbers r and t define unambiguous the point P.
We say that (r,t) is a pair of polar coordinates of P.
One point P has many pairs of polar coordinates.
If (r,t) is a pair of polar coordinates,
(r, t + 2.k.pi) is also a pair of polar coordinates and additionally
(- r, t + (2.k+1).pi ) is a pair of polar coordinates too.
Of course, k is an integer.
The polar coordinates of the pole O are by definition (0,t) with t perfectly arbitrary.
According to the previous definition, the cartesian coordinates of U are
(cos t, sin t).
Since P = r.U, the cartesian coordinates of p are (r.cos t, r.sin t).
The transformation formulas are x = r.cos t, y = r.sin t
The cartesian coordinates of a point P are (x,y).
Choose the u-axis such that r > 0. then
P = r U => P2 = r2 U2 => x2 + y2 = r2
r = sqrt(x2 + y2)
Now, choose a t-value such that x = r.cos t and y = r.sin t
In that way we have a pair of polar coordinates (r,t) of P.Each point P of that curve has at least one pair of polar coordinates who satisfy the equation. Note that, in general, not all pairs of polar coordinates of P are solutions of the equation.
Note that one curve can have different polar equations.
Suppose a curve c has polar equation r = f(t).
At a variable point P of the curve, we draw a tangent line and
on that line we denote the axis b in the direction of increasing t-values.
This defines the angle n and the angle a.
The tan(n) defines the direction of the curve c in point P. This tan(n)
is somewhat similar to the notion of slope in cartesian coordinates.
We'll calculate tan(t).
t + n = a
=> n = a - t
tan(a) - tan(t)
=> tan(n) = -----------------
1 + tan(a).tan(t)
Say the variable point P has cartesian coordinates (x,y).
Then we know that
dy y
tan(a) = --- and tan(t) = ---
dx x
Thus,
dy y
-- - -
dx x x dy - y dx
tan(n) = ---------- = --------------
dy y x dx + y dy
1 + -- . -
dx x
From x = r cos(t) and y = r sin(t)
we have
dx = dr.cos(t) - r.sin(t) . dt
dy = dr.sin(t) + r.cos(t) . dt
From this we calculate
x dy - y dx = r2 dt
From x2 + y2 = r2 we find
2 x dx + 2 y dy = 2 r dr or
x dx + y dy = r dr
Now, we can simplify tan(n)
r2 dt r
tan(n) = ------- = ------- =
r dr dr/dt
r
tan(n) = -----
r'
The following formula gives the direction of the curve c at any point P.
Here r = f(t) is the equation of the curve and r' stands for (dr/dt) = f'(t).
From r and r' you can calculate the direction n (see figure above ) at
each point.
r
tan(n) = --------
r'
|
cot(n) = r'/r = (ln(r))'
Now
(ln(r))' = k for all t.
First take k = 0. Then ln(r) = constant and r is constant.
The curves are the circles with midpoint in the origin.
Now, take k not 0. Then
(ln(r))' = k
dln(r)
<=> ------- = k
dt
<=> ln(r) = k t + constant
we denote the constant as ln(m)
<=> ln(r) = k t + ln(m)
<=> r = m ekt
These isogonal curves are called the logarithmic spirals or the
Bernouilli spirals.
the tangent line is parallel to the polar axis
<=> t + n = k.pi
<=> tan(t) = - tan(n)
r
<=> tan(t) = - ---
r'
|
The following formula gives the t-values of all points of a curve r=f(t),
where the tangent line is parallel to the polar-axis. The value r' stands for (dr/dt) = f'(t). To calculate the t-values, you have to solve this trigonometric equation.
r
tan(t) = - -----
r'
|
the tangent line is orthogonal to the polar axis
<=> t + n = pi/2 + k.pi
<=> tan(t) = cot(n)
r'
<=> tan(t) = ---
r
|
The following formula gives the t-values of all points of a curve r=f(t), where
the tangent line is orthogonal to the polar-axis. The value r' stands for (dr/dt) = f'(t). To calculate the t-values, you have to solve this trigonometric equation.
r'
tan(t) = -----
r
|
2 pi - to to to
cos(---------) = cos(pi - --) = - cos ---- = - ro
2 2 2
From this, we see that (-ro, 2pi-to) is a solution of r = cos(t/2).
r' = - (1/2) sin(t/2)
The tangent line is parallel to the polar-axis if and only if
tan(t) = - r/r'
<=> tan(t) = 2 cot(t/2)
To solve this equation we let u = tan(t/2) , then
2 u 2
<=> --------- = -------
1 - u2 u
<=> ...
___ ___
V 2 V 2
<=> tan(t/2) = ---- or tan(t/2) = - ----
2 2
This gives the value t = 1.23 in [0,pi]. Then r = 0.816
P on C => F(xo,yo) = 0 => F(ro.cos(to), ro.sin(to)) = 0 => P on C'
P on C' => F(ro.cos(to), ro.sin(to)) = 0 => F(xo,yo) = 0 => P on K'
|
Suppose a curve C has a cartesian equation F(x,y) = 0.
If we replace x by r.cos(t) an y by r.sin(t),
then we have a polar equation
F(r.cos(t),r.sin(t))=0 of the curve C. |
|
Suppose a curve C has a polar equation G(r,t) = 0.
If that equation can be transformed to an equation |
r2 cos2 (t) + r2 sin2 (t) - 4 = 0 <=> x2 + y2 = 4
C has a polar equation r2 = r + 2.r.cos(t) <=> r = r2 - 2.r.cos(t) Now take the curve C' : r = -( r2 - 2.r.cos(t)) . It is easy to prove that each point of C is on C' and so is the reverse. Hence, we can write C has a polar equation r = (+1 or -1).( r2 - 2.r.cos(t)) <=> C has a polar equation r2 = (r2 - 2 r cos(t))2 <=> C has a cartesian equation x2 + y2 = ( x2 + y2 -2 x )2
Cardioid
Cissoid of Diocles
Cochleoid
Conchoid
Trifolium
Folium
Folium of Descartes
Hyperbolic Spiral
Lemniscate of Bernoulli
Right Strophoid
etc...
Go and look at
u(r1 , t1 ) and v(r2 , t2 )
In the corresponding cartesian coordinate system the two vectors
have cartesian coordinates
u(x1 , y1 ) and v(x2 , y2 )
with
x1 = r1 cos(t1 ) x2 = r2 cos(t2 )
y1 = r1 sin(t1 ) y2 = r2 sin(t2 )
The dot product u.v
= x1 x2 + y1 y2
= r1 cos(t1 ) . r2 cos(t2 ) + r1 sin(t1 ).r2 sin(t2 )
= r1 r2 (cos(t1 )cos(t2 ) + sin(t1)sin(t2 ))
= r1 r2 cos(t1 - t2 )
The dot product of two vectors u (r1 , t1 ) and v (r2 , t2 ) is u . v = r1 r2 cos(t1 - t2 ) |
A point P(r,t) is on line d
<=> PN is orthogonal to ON
<=> PN . ON = 0
<=> (N - P). N = 0
<=> N.N - P.N = 0
<=> ro.ro.cos(0) - r.ro.cos(t - to) = 0
Since ro and cos(t - to) are not 0
ro
<=> r = -----------
cos(t - to)
The equation of a line l through the pole perpendicular to a line d,
and with intersection point N(ro,to) is
ro
r = -----------
cos(t - to)
|
| A circle with the pole as center and radius R has a polar equation r = R |
P(r,t) is on the circle
<=> || CP || = R
<=> || CP ||2 = R2
<=> (P - C)2 = R2
<=> P2 - 2 P C + C2 = R2
<=> r2 - 2 r ro cos(t - to) + ro2 = R2
A circle, with C(ro,to) as center and R as radius, has has a polar equation
r2 - 2 r ro cos(t - to) + ro2 = R2
|
r2 - 2 r R cos(t) = 0
<=> r = 0 or r = 2R cos(t)
<=> r = 2R cos(t)
(since this curve already contains the pole for t = pi/2)
A circle, with C(R,0) as center and R as radius, has has a polar equation
r = 2R cos(t)
|
In cartesian coordinates we find the coordinates of the common
points by solving the system of the two equations.
But if we solve the system of the polar equations we find only one
point with polar coordinates (2,1).
Now we'll show
Similarly,
Say curve c' has a polar equation F'(r,t)=0.
Call V' the set of all solutions of the equation F'(r,t)=0.
Then c' is the set of all points corresponding with the set V'.
With each element of V', there corresponds one and only one point of c'.
With an element of the intersection of V and V' corresponds a common point of c and c', BUT it is possible that V contains a solution (ro,to) and that V' contains a different solution (r1,t1) and that both solutions correspond with a common point of c and c'. Then (ro,to) and (r1,t1) are different polar coordinates of that same point.
In the example above we have
(-2,1) is a solution of t = 1 and this gives a point of the line l.
(2,1+pi) is a solution of r = 2 and this gives a point of the circle c.
Although these solutions are different, the corresponding point is a
common point of the line and the circle.
We say that an equation F(r,t)=0 of a curve c has property (P)
if and only if
ALL polar coordinates of EACH point of c are solutions of F(r,t)=0
Corollary:
Say curve c has a polar equation F(r,t)=0 with the property (P), and curve c' has a polar equation F'(r,t)=0. Now the set of solutions of the system of the two polar equations contains the coordinates of ALL the common points of the two curves and the problem has disappeared!
We know that
ro
r = ----------- (1)
cos(t - to)
is the equation of a line d.
Example:
A line d contains point D(3, pi/4) and stands perpendicular to line OD.
A circle c has center O and radius = 5.
The intersection points are the solutions of the system
/ r = 5
| 3
| r = -------------
\ cos(t - pi/4)
We have cos(t - pi/4) = 3/5
cos(t - pi/4) = cos(0.927)
t - pi/4 = 0.927 or t - pi/4 = -0.927
This gives the points with polar coordinates
(5, 1.71) and (5, -0.14)
r2 - 2 r ro cos(t - to) + ro2 = R2 (2)
is the equation of a circle.This is also true for the circle with the equation r = 2R cos(t)
r2 = R2
is an equation of the same circle and this equation has property (P).
Example:
We calculate the intersection points of the curves with equation
r = cos(t/2) and r = 1/2
We have to solve the system
/
| r2 = 1/4
|
| r = cos(t/2)
\
<=>
/
| r = 1/2 or r = -1/2
|
| r = cos(t/2)
\
<=>
/ /
| r = 1/2 | r = - 1/2
| or |
| cos(t/2)=1/2 | cos(t/2)= - 1/2
\ \
<=> .....
We find four solutions:
(1/2, 2 pi / 3) or (- 1/2, 4 pi / 3) or
(1/2, - 2 pi / 3) or (- 1/2, - 4 pi / 3)
The circle and the curve have four common points.
t = to + k.pi
is an equation of the same line and this equation has property (P).
Example: We calculate the intersection points of the curves with equation
r = 4 cos(t) and r = 4 sin(t)Since the first equation has property (P), the intersection points are the solutions of the system
r = 4 cos(t)
r = 4 sin(t)
The coordinates of the pole are not a solution of that system
but the pole IS an intersection point of the curves.
| To calculate the common points of curve c with equation F(r,t)=0 and c' with equation F'(r,t)=0, it is sufficient to solve the system of the equations F(r,t)=0 and F'(r,t)=0 if at least one of the equations has the property (P). But even then you have to investigate separately if the pole is a common point. |
Given:
5
K has equation r = ----------------------------
( 1 - 2 sin(t) + 3 cos(t) )
2
L has equation r = ----------------------
( 2 cos(t) + sin(t) )
Calculate
|
Then K has equation
5
r = ----------------------
( 1 - e cos(t-to) )
And this is the equation of a conic section.
2
r = -------------
A.cos(t-to)
This is the equation of a line
5
r = -----------------------------
( 1 - 2 sin(t) + 3 cos(t) )
2
r = ----------------------
( 2 cos(t) + sin(t) )
From these equations we have
5 2
--------------------------- = ----------------------
( 1 - 2 sin(t) + 3 cos(t) ) ( 2 cos(t) + sin(t) )
<=> ....
<=> 4 cos(t) + 9 sin(t) = 2
and with the method from
here
we solve this equation.
<=> ....
<=> t1 = 2.518876 and t2 = -0.2137328
The corresponding values of r are :
r1 = -1.92058 and r2 = 1.14785
To convert the equation of K to a cartesian equation, we appeal on the properties of Polar equation of a conic section.
The conic section K has polar equations
5 -5
r = -------------------------- and r = ----------------------------
( 1 - 2 sin(t) + 3 cos(t) ) ( 1 + 2 sin(t) - 3 cos(t) )
<=>
r - 2r sin(t) + 3 r cos(t) = 5 and r + 2r sin(t) - 3 r cos(t) = -5
<=>
r = 5 + 2 r sin(t) - 3 r cos(t) and r = -( 5 + 2 r sin(t) - 3 r cos(t))
<=>
r2 = ( 5 + 2 r sin(t) - 3 r cos(t))2
<=>
x2 + y2 = (5 + 2 y - 3 x)2
<=>
8x2 - 12 xy + 3 y2 - 30 x + 20 y + 25 = 0
For the line L, the transformation is easy.
L has equation 2 r cos(t) + r sin(t) = 2 <=> 2 x + y = 2To calculate the intersection points in cartesian coordinates, we solve the system
8x2 - 12 xy + 3 y2 - 30 x + 20 y + 25 = 0 2 x + y = 2With simple algebra we find
( 1.1217 ; -0.24347) and (1.560 ; -1.120 )
It is easy to verify that these points are the same points as above.
c with equation F(r,t)=0 (1)
and
c' with equation F(r, t - alpha)=0 (2)
Now we have
P(ro,to) is a solution of (1) <=> P(ro,to + alpha) is a solution of (2)
This means that we obtain the curve c' by rotating the curve c
clockwise by an angle alpha.
Example:
If we rotate the circle r = 4cos(t) by an angle of pi/2 radians, the new circle has equation
r = 4 cos(t - pi/2)
<=> r = 4 sin(t)
| We rotate the curve c with equation F(r,t)=0 clockwise by an angle alpha. The new curve has equation F(r, t - alpha)=0. |
Suppose a curve c has polar equation r = f(t).
At a variable point P of the curve, we draw a tangent line and
on that line we denote the axis b in the direction of increasing t-values.
This defines the angle n and the angle a.
Now rotate axis u clockwise by pi/2 radians. This gives axis s.
The intersection point of b and s is T.
Denote N on s such that PN is perpendicular to TP. (see figure)
B is the unit vector with abscis 1 with respect to the axis b.
S is the unit vector with abscis 1 with respect to the axis s.
Now we have (vectors in bold)
Say N = l S and T = m S (this defines the numbers l and m)
NP.PT = 0
(P - N).(T - P) = 0
P.T - P.P - N.T + N.P = 0
0 - P.P - N.T + 0 = 0
P.P = - N.T
r.r = -l.m
P = N + NP
P.B = N.B + NP.B
r U.B = l S.B + 0
r cos(n) = l sin(n)
l = r cot(n)
1 dr dr dr
l = r -.--- = --- (l is the visualisation of ---- )
r dt dt dt
1 l 1 dr d 1
- = - --- = - ----.--- = --- (-)
m r2 r2 dt dt r
1 d 1
- = -- (-) = g'(t) <=> m = 1/ g'(t)
m dt r
So,
mo = 1/ g'(to)
From mo we denote point T and then we draw the asymptote.
The equation of the asymptote is
mo
r = ---------------
cos(t-(to+pi/2))
mo
<=> r = ----------
sin(t -to)
To calculate an asymptotes of the curve c with polar equation r = f(t),
we take four steps:
|
- sqrt(3) sqrt(3)
r = --------------- and r = ---------------
sin(t - 2pi/3) sin(t + 2pi/3)
In a graph this gives a hyperbola and his two asymptotes.