..Polar
coordinates of a point P...From
polar to cartesian coordinates...From
cartesian to polar coordinates...Polar
equation of a curve.
..Direction
of a curve in polar coordinates
..Isogonal
Curves
..Tangent
line parallel to the polar-axis
..Tangent
line orthogonal to the polar-axis
..Investigation of
a curve ; an example
..From a
cartesian equation to a polar equation
..From a
polar equation to a cartesian equation
..More
examples of curves with polar equation..Dot
product in polar coordinates..A line
and its polar equation
..A circle
and its polar equation
..A conic
section and its polar equations..Common
points of two curves
..Rotating
and the polar equation...Asymptotes in
polar coordinates
Polar coordinates of a
point P.
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 the polar axis.
On that x-axis, there is just 1 vector E 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
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.
From polar to cartesian coordinates.
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 coordinates of P.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
From cartesian to polar coordinates.
We start with a cartesian coordinate system. We choose O as pole and the 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 = 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.Starting with that pair, all pairs are
(r, t + 2.k.pi) and (- r, t + (2.k+1).pi )
Polar equation of a curve.
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 explicit form r = f(t).Such equation is a polar equation of a curve.
With each solution (ro,to) of the polar equation, corresponds a point with polar coordinates (ro,to). Generally the equation has an infinity number of such 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 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.
Examples
- 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)
Direction of a curve in polar coordinates
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'
|
Isogonal Curves
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 ekt
These isogonal curves are called the logarithmic spirals or the Bernouilli
spirals.
Tangent line parallel to the polar-axis
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 solve this trigonometric equation.
r
tan(t) = - -----
r'
|
Tangent line orthogonal to the polar-axis
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), 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
|
Investigation of a curve ; an example
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 to is such value and (ro,to) is a solution of r = cos(t/2).
The corresponding point P is P(ro,to).
The pairs (ro,to + 4.k.pi) are also solutions of r = cos(t/2) and with these solutions corresponds the same point P. This means that, if t runs over the interval [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 whole curve.
- Symmetry
If (ro,to) is a solution of r = cos(t/2), then (ro, - to) 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 to the polar axis and we can restrict our investigation to to in [0,2.pi].
- Symmetry again
If (ro,to) is a solution of r = cos(t/2), then2 pi - to to to cos(---------) = cos(pi - --) = - cos ---- = - ro 2 2 2From this, we see that (-ro, 2pi-to) is a solution of r = cos(t/2).
The corresponding points are symmetric with respect to the y-axis through O and orthogonal to x.
C is symmetric with respect to that y-axis and we can restrict our investigation 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 iftan(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 2This 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-axis This can be calculated in the same way as above. It is left as an exercise.
- The curve C
From a cartesian equation to a polar equation
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(xo,yo) is a point of C. Take a pair of polar coordinates (ro,to) of P. Then xo = ro.cos(to) and yo = ro.sin(to).P on C => F(xo,yo) = 0 => F(ro.cos(to), ro.sin(to)) = 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 that F(ro.cos(to), ro.sin(to)) = 0
Now, take xo = ro.cos(to) and yo = ro.sin(to), then (xo,yo) are the cartesian coordinates of P andP 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. |
From a polar equation to a cartesian equation
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 cartesian equation of C.
The transformation from G(r,t) = 0 to F(r.cos(t), r.sin(t)) = 0 is often 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 |
Example 1
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
Example 2
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 we multiply 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 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
More examples of curves with polar equation
There is a site on the net where you can see the graph of all famous 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...
Dot product in polar coordinates
Take a polar coordinate system and two vectors
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 line and its polar equation
A line through the pole.
Since all the points of the line correspond with a constant value of the angle 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
...
A line d not containing the pole.
Draw a line through the pole perpendicular to the given line d.Say N(ro,to) is the intersection point.
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 and its polar equation
A circle with the pole as center.
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 |
An arbitrary circle
Say C(ro,to) 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 - 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
|
A special circle with
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)
|
A conic section and its polar equations
See Polar equation of a not degenerated conic section.Common points of two curves
A problem with common points of two curves
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 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
The cause of this problem
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 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.
A solution of this problem. A special prorety (P).
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 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!
Equations with property (P)
We know that ro r = ----------- (1) cos(t - to)is the equation of a line d.
If P(r1,t1) is a solution of (1), then all other polar coordinates of 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 = -0.927 This gives the points with polar coordinates (5, 1.71) and (5, -0.14)- We know that
r2 - 2 r ro cos(t - to) + ro2 = R2 (2)is the equation of a circle.
If P(r1,t1) is a solution of (2), then all other polar coordinates of 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 not the
property (P) but
r2 = R2is an equation of the same circle and this equation has property (P).Example:
We calculate the intersection points of the curves with equationr = 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. - The equation t = to, of a line containing the pole, has not the
property (P) but
t = to + k.piis an equation of the same line and this equation has property (P).
The pole is a special point
Since the pole has so many polar coordinates, we always have to investigate 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 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. |
An extensive example.
Given:
5
K has equation r = ----------------------------
( 1 - 2 sin(t) + 3 cos(t) )
2
L has equation r = ----------------------
( 2 cos(t) + sin(t) )
Calculate
|
First we investigate the nature of the curves K and L.
- We know from here that (2
sin(t) -3 cos(t)) can be transformed in the form e.cos(t -to).
Then K has equation
5 r = ---------------------- ( 1 - e cos(t-to) )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-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.
Rotating and the polar equation.
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 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. |
Asymptotes in polar coordinates
Visualisation of dr/dt
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.mP = 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- Combining these results, we have
1 l 1 dr d 1 - = - --- = - ----.--- = --- (-) m r2 r2 dt dt r
Asymptotes
Say r = f(t) is a polar equation of a curve c.First we write this equation as 1/r = g(t).
If to is a solution of g(t)=0, then to gives the direction of the asymptote.
From above, we have :
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:
|
Example
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 to = 2pi/3 and t1 = -2pi/3
These values are the directions of the asymptotes.
The corresponding values of m are
mo = - 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.
Information Provided by Johan.Claeys@ping.be