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Polar coordinates

In the plane we choose a fixed point O, and we call it the pole.
Additionally we choose an axis x through the pole and call it thepolar axis.
On that x-axis, there is just 1 vectorE such that abs(E)=1.
The pole and the polar axis constitute the basis of the polar coordinate system.

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 thatP = 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.

A polar coordinate system is given and point P has polar coordinates (r,t).We choose a y-axis through the pole O and perpendicular to the x-axis.So, we have cartesian axes x and y. Call (x,y) the cartesian coordinatesof P.

According to the previous definition, the cartesian coordinates of U are(cos t, sin t).
SinceP = 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

We start with a cartesian coordinate system. We choose O as pole andthe x-axis as polar-axis.

The cartesian coordinates of a point P are (x,y).
Choose the u-axis such that r > 0. then

P = rU =>P² = r²U²  => 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.
Starting with that pair, all pairs are
(r, t + 2.k.pi) and (- r, t + (2.k+1).pi )

Consider a connection between the polar coordinates of a point and suppose,that connection can be expressed in the form F(r,t)=0 or maybe in the explicitform r = f(t).
Such equation is a polar equation of a curve.
With each solution (r_o,t_o) of the polar equation, corresponds a point withpolar coordinates (r_o,t_o). Generally the equation has an infinity number ofsuch solutions and so, we have an infinity number of points.The set of all these points is the curve of the equation.

Each point P of that curve has at least one pair of polar coordinates whosatisfy the equation. Note that, in general, not all pairs of polarcoordinates of P are solutions of the equation.

Note that one curve can have different polar equations.

r= t is a spiral line (red)
t= pi/4 is a line through the pole and slope 1
r = 2 is a circle with radius = 2 (green)

Suppose a curve c has polar equation r = f(t).
At a variable point P of the curve, we draw a tangent line andon 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                  xThus,                   dy   y                   -- - -                   dx   x       x dy - y dx        tan(n) = ---------- = --------------                     dy   y     x dx + y dy                 1 + -- . -                     dx   xFrom 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) . dtFrom this we calculate        x dy - y dx = r² dtFrom    x²  + y²  = r²  we find        2 x dx + 2 y dy = 2 r dr     or         x dx +  y dy =  r drNow, 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 ) ateach point.

              rtan(n) = --------             r'

We look for the curves such that the direction n (see figure above )of that curve is a constant k in each point.

     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 e^kt

These isogonal curves are called the logarithmic spirals or theBernouilli spirals.

In previous figure we see that

         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 solvethis trigonometric equation.

              rtan(t) = - -----             r'

In previous figure we see that

         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), wherethe 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 solvethis trigonometric equation.

            r'tan(t) = -----           r

We take the curve with polar equation r = cos(t/2).

Restriction of the domain.
The period of cos(t/2) is 4.pi .
Therefore, we take a closer look at the interval [-2.pi, 2.pi] for t.
With each t-value in that interval, corresponds one point of the curve C.
Say t_o is such value and (r_o,t_o) is a solution of r = cos(t/2).
The corresponding point P is P(r_o,t_o).
The pairs (r_o,t_o + 4.k.pi) are also solutions of r = cos(t/2)and with thesesolutions corresponds the same point P. This means that, if t runs over theinterval [2 pi, 6 pi], we don't find any new points of the curve C.
This conclusion also holds for each next interval of length 4.pi.
Thus, it is sufficient that t runs through [-2.pi, 2.pi] to know the wholecurve.
Symmetry
If (r_o,t_o) is a solution of r = cos(t/2),then (r_o, - t_o) is a solution of r = cos(t/2).
The corresponding points are symmetric with respect to the polar axis.
Hence, the curve C is symmetric with respect t_o the polaraxis and we can restrict our investigation to t_o in [0,2.pi].
Symmetry again
If (r_o,t_o) is a solution of r = cos(t/2), then
```
                 2 pi - t_o             t_o           t_o            cos(---------) = cos(pi - --) = - cos ---- = - r_o                    2                 2             2
```
From this, we see that (-r_o, 2pi-t_o) is a solution of r = cos(t/2).
The corresponding points are symmetric with respect to the y-axis throughO and orthogonal to x.
C is symmetric with respect to that y-axis and we can restrict ourinvestigation to [0,pi].
If t increases from 0 to pi, r decreases from 1 to 0.
All the rest of C follows from the symmetry.

Tangent line parallel to the polar-axis

         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
In point P(0.816;1.23) the tangent line is parallel to the polar-axis.

Tangent line orthogonal to the polar-axisThis can be calculated in the same way as above. It is left as an exercise.
The curve C

Suppose a curve C has a cartesian equation F(x,y) = 0.
If we replace, without thinking, 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 a curve C'.
We'll show that C = C'.

First we'll show that each point of C is a point of C'.
Say P(x_o,y_o) is a point of C. Take a pair of polar coordinates(r_o,t_o) of P.Then x_o = r_o.cos(t_o) and y_o = r_o.sin(t_o).
```
 P on C => F(x_o,y_o) = 0 => F(r_o.cos(t_o), r_o.sin(t_o)) = 0 => P on C'
```
Now we'll show that each point of C' is a point of C.
If P is on C', then P has a pair of polar coordinates (ro,to) such thatF(r_o.cos(t_o), r_o.sin(t_o)) = 0
Now, take x_o = r_o.cos(t_o) and y_o = r_o.sin(t_o),then (x_o,y_o) are the cartesiancoordinates of P and
```
 P on C' => F(r_o.cos(t_o), r_o.sin(t_o)) = 0 => F(x_o,y_o) = 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.
Sometimes that equation can be transformed to an equation
F(r.cos(t), r.sin(t)) = 0.
In the same way as above you can show that F(x,y) = 0 is the cartesianequation of C.
The transformation from G(r,t) = 0 to F(r.cos(t), r.sin(t)) = 0 isoften very difficult or impossible!

Suppose a curve C has a polar equation G(r,t) = 0.

If that equation can be transformed to an equation
F(r.cos(t), r.sin(t))=0 then F(x,y)=0 is the cartesian equation of C.

Suppose a curve C has a polar equation r.r = 4 . We can transform this in

         r²  cos² (t)  + r²  sin² (t)  - 4 = 0  <=>  x²  + y²  = 4

Suppose a curve C has a polar equation r = 1 + 2.cos(t).
If we choose t such that cos(t) = 0.5, then r = 0.Thus, the pole is part of C and we don't add a point to C if wemultiply both sides of r = 1 + 2.cos(t) by r.

 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 writeC 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

There is a site on the net where you can see the graph of allfamous curves with the polar equation and the cartesian equivalent (if any).

         Cardioid        Cissoid of Diocles        Cochleoid        Conchoid        Trifolium        Folium        Folium of Descartes        Hyperbolic Spiral        Lemniscate of Bernoulli        Right Strophoid        etc...

Go and look at

Famous Curves Index

There you will find the names of the curve, its history and some of its associated curves.

Take a polar coordinate system and two vectors

u(r₁ , t₁ ) andv(r₂ , t₂ )In the corresponding  cartesian coordinate system the two  vectorshave cartesian coordinatesu(x₁ , y₁ ) andv(x₂ , y₂ )with        x₁  = r₁  cos(t₁ )   x₂  = r₂  cos(t₂ )        y₁  = r₁  sin(t₁ )   y₂  = r₂  sin(t₂ )The dot productu.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 vectorsu (r₁ , t₁ ) andv (r₂ , t₂ ) isu .v = r₁ r₂  cos(t₁ - t₂ )

Since all the points of the line correspond with a constant value of theangle t, the equation of such a line is t = constant.
Example :
t = 0 is the line of the polar axis
t = pi/2 is the y-axis
t = 1
...

Draw a line through the pole perpendicular to the given line d.
Say N(r_o,t_o) is the intersection point.

         A pointP(r,t) is on line d<=>PN is orthogonal toON<=>PN .ON = 0<=>     (N -P).N = 0<=>N.N -P.N = 0<=>     r_o.r_o.cos(0) - r.r_o.cos(t - t_o) = 0Since r_o and cos(t - t_o) are not 0                r_o<=>     r = -----------            cos(t - t_o)

The equation of a line l through the pole perpendicular to a line d,and with intersection point N(r_o,t_o) is

             r_o      r = -----------          cos(t - t_o)

In that case the polar equation is obvious.
Examples :
r = 2
r = 5
r = -5 (Same circle as r = 5)
r = 0 The pole
...

A circle with the pole as center and radius R has a polar equation r = R

Say C(r_o,t_o) is the center and R is the radius.

P(r,t) is on the circle<=>     ||CP ||  = R<=>     ||CP ||²  = R²<=>     (P -C)² = R²<=>P²  - 2PC +C²  = R²<=>     r²  - 2 r r_o cos(t - t_o) + r_o²  = R²

A circle, with C(r_o,t_o) as center and R as radius, has has a polar equation

         r²  - 2 r r_o cos(t - t_o) + r_o²  = R²

Take the circle with center C(R,0) and R is the radius.
For the equation we take the previous formula with ro = R and to = 0.
The equation is

         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)

See Polar equation of a not degenerated conic section.

Say curve l is a line with a polar equation t = 1.
Say curve c is a circle with a polar equation r = 2.

In cartesian coordinates we find the coordinates of the commonpoints by solving the system of the two equations.
But if we solve the system of the polar equations we find only onepoint with polar coordinates (2,1).

Now we'll show

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.

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 acommon point of c and c', BUT it is possible that V contains asolution (r_o,t_o) and that V' contains a different solution (r₁,t₁)and that both solutions correspond with a common pointof c and c'.Then (r_o,t_o) 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 acommon point of the line and the circle.

There are polar equations with the following special property!

 We say that an equation F(r,t)=0 of a curve c has property (P)                if and only ifALL 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), andcurve c' has a polar equation F'(r,t)=0.Now the set of solutions of the system of the two polar equationscontains the coordinates of ALL the common points of the two curvesand the problem has disappeared!

 We know that                r_o        r = -----------         (1)            cos(t - t_o)

is the equation of a line d.
If P(r1,t1) is a solution of (1), then all other polar coordinatesof P are solutions of (1). So, the equation (1) has property (P).

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 = -0.927This gives the points with polar coordinates        (5, 1.71)  and  (5, -0.14)

We know that
```
         r²  - 2 r r_o cos(t - to) + r_o²  = R²       (2)
```
is the equation of a circle.
If P(r1,t1) is a solution of (2), then all other polar coordinatesof P are solutions of (2). So, the equation (2) has property (P).
This is also true for the circle with the equation r = 2R cos(t)

The equation r = R, of the circle with the pole as center, has notthe property (P) but

         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 = 1/2We 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.

The equation t = t_o, of a line containing the pole, has notthe property (P) but
```
         t = t_o + k.pi
```
is an equation of the same line and this equation has property (P).

Since the pole has so many polar coordinates, we always have toinvestigate this case separately.

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 pointsare the solutions of the system

         r = 4 cos(t)        r = 4 sin(t)

The coordinates of the pole are not a solution of that systembut the pole IS an intersection point of the curves.

To calculate the common points of curve c with equation F(r,t)=0 andc' with equation F'(r,t)=0, it is sufficient to solve the system ofthe equations F(r,t)=0 and F'(r,t)=0 if at least one of the equationshas the property (P). But even then you have to investigate separatelyif 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

The intersection points of K and L in polar coordinates.
The equation of K and L in cartesian coordinates.
The intersection points of K and L in cartesian coordinates.

First we investigate the nature of the curves K and L.

We know from herethat (2 sin(t) -3 cos(t)) can be transformed in the form e.cos(t -t_o).
Then K has equation
```
              5    r = ----------------------         ( 1 - e cos(t-t_o) )
```
And this is the equation of a conic section.
Similarly, the equation of L can be transformed in the form
```
               2    r = -------------        A.cos(t-t_o)
```
This is the equation of a line

Since the equation of L has the property (P), the intersection pointsof K and L are the solutions of the system

            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 herewe solve this equation.

 <=>  ....<=>  t₁ = 2.518876  and t₂ = -0.2137328The corresponding values of r are :     r₁ = -1.92058   and  r₂ = 1.14785

To convert the equation of K to a cartesian equation, we appeal onthe 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 = 2

To 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 = 2

With 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.

Say alpha is a constant value. Take the curves

         c with equation F(r,t)=0                (1)and        c' with equation F(r, t - alpha)=0      (2)Now we haveP(r_o,t_o) is a solution of (1)  <=> P(r_o,t_o + alpha) is a solution of (2)

This means that we obtain the curve c' by rotating the curve cclockwise 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 andon 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)

 SayN = lS  andT = mS  (this defines the numbers l and m)NP.PT = 0        (P -N).(T -P) = 0P.T -P.P -N.T +N.P = 0         0  -P.P -N.T +  0  = 0P.P = -N.T        r.r = -l.m

P =N +NPP.B =N.B +NP.B        rU.B = lS.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

Combining these results, we have

         1       l        1    dr    d   1        -  = - --- =  - ----.--- = --- (-)        m      r²        r²   dt   dt   r

Say r = f(t) is a polar equation of a curve c.
First we write this equation as 1/r = g(t).
If t_o is a solution of g(t)=0, then t_o gives the directionof the asymptote.
From above, we have :

         1      d  1        -  =  -- (-) = g'(t)  <=> m = 1/ g'(t)        m     dt  rSo,        m_o = 1/ g'(t_o)

From m_o we denote point T and then we draw the asymptote.

The equation of the asymptote is

                 m_o        r = ---------------            cos(t-(t_o+pi/2))                 m_o      <=> r = ----------              sin(t -t_o)

To calculate an asymptotes of the curve c with polar equation r = f(t),we take four steps:

Write g(t) = 1/f(t).
Calculate t_o = a solution of g(t)=0.
t_o gives the directionof the asymptote.
Calculate m_o = 1/ g'(t_o).

The asymptote is the line through point (m_o,t_o+pi/2) and with direction t_o. Its equation is

                  m_o          r = -----------              sin(t - t_o)

Take the curve c with equation r = 3/(1+2cos(t)).
Then g(t) = (1+2cos(t))/3 and g'(t) = -2sin(t)/3
g(t) = 0 for t_o = 2pi/3 and t₁ = -2pi/3
These values are the directions of the asymptotes.
The corresponding values of m are
m_o = - sqrt(3) and m₁ = sqrt(3).
The asymptotes are

                 - 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.

Topics and Problems

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