Polar Coords

8/12/2019 Polar Coords

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Polar coordinates and curves described using polar coordinates

Polar coordinates.

Polar coordinates provide an alternative system of coordinates for describing the location of points in a plane. In this system there is an

origin or pole , O, and a polar axis, which is a half-line or ray extending horizontally to the right of the pole.

The position of a point P in the polar coordinate plane is determined by its polar coordinates ( ),r , where:

ris the "distance" of the point P from the pole (origin), and

is the angle between the ray OP and the polar axis, following the convention that angles measured in the counterclockwise direction are

considered to be positive and angles measured in the clockwise direction are considered to be negative.

Unlike rectangular coordinates, the polar coordinates are not unique. The point ( ),r + 2 k , where kis an integer, is the same as the

point ( ),r . For example,

,2

6 =

,2

13

6

and

,1

5

4 =

,1

3

4.


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It is legitimate to allow negative values of rwhereby the point ( ),r is at a distance of r from the pole on the opposite side of the

pole as ( ),r along the same infinite line through the pole and as ( ),r . Thus

( ),r = ( ),r + .

Because of the familiarity of the standard rectangular cartesian coordinate system, the pole of the polar coordinate system is often identified

with the origin in the standardx-y coordinate plane and then the polar axis can be identified with the section of thex axis associated with

positivex values.

When converting the coordinates of a point from polar form to rectangular form, the following equations may be used.

{=x rcos

=y rsin

When converting the coordinates of a point from rectangular form to polar form, the following equations may be used.

=+x2

y2

r2

=

arctan y

xx 0

If =x 0, then =

2 or =

2 as appropriate.

For example, the point with rectangular coordinates ( ),1 3 has (possible) polar coordinates

,2

6, while the point with rectangular

coordinates ( ),0 2 has (possible) polar coordinates

,2

2 or

,2

2.


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Polar curves.

Typically, an equation relating the variables rand describes a curve in polar coordinates or simply a polar curve.

Example 1:

Sketch the polar curve given by the equation =r 4 cos .

Solution:

We can find the corresponding cartesian equation as follows.

Multiplying both sides of the equation=r 4 cos

by rgives the equation

=r2

4 rcos ,

so that, if a point on the curve has rectangular coordinates ( ),x y we can use the equations

{=+x

2y

2r

2

=x rcos to obtain the cartesian equation:

=+x2

y2

4x.

This equation is equivalent to

= +x2

4x y2

0,

which, in turn, by completing the square for thexterms, is equivalent to

= + +x

2

4x 4 y

2

4,that is,

=+( )x 22

y2

4.

This is the cartesian equation of a circlewith centre at the point ( ),2 0 and radius 2.

Notes:

Alternatively, one can see directly that a point Pwith polar coordinates ( ),r is on the circle with its centre at the point ( ),0 2 and radius

2 exactly when

=r 4 cos

from the fact that the triangle formed from the diameter of the circle that lies along the xaxis, the line segment from the origin to the point

Pand the line segment from P to the point ( ),4 0 always has a right angle at P.


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One can always resort to the construction of a table of values.

|

2

3

4

60

6

4

3

2

2

3

3

4

5

6

=r 4 cos | 0 2 2 2 2 3 4 2 3 2 2 2 0 2 2 2 2 3 4

The pointAhas polar coordinates

,

62 3 and rectangular coordinates ( ),3 3 .

The pointBhas polar coordinates

,


The point Chas polar coordinates

,

32 and rectangular coordinates ( ),1 3 .

The pointsDhas polar coordinates

,

2

32 or

,

32 and rectangular coordinates ( ),1 3 .

The pointsEhas polar coordinates

,

3

42 2 or

,


The points Fhas polar coordinates

,

3

42 2 or

,


A useful aid to to sketching the polar graph of an equation of the form =r ( )f is to first plot of the graph of =r ( )f in cartesian

coordinateswith the axis horizontal and the raxis vertical.

Plotting =r 4 cos in this way indicates that as increases from

2 to +

2, the distance rof the point ( ),r from the origin first

increases from 0 to a maximum value of 4 as increases from

2to 0, and the decreases to 0 as theta increases from 0 to

2.


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One can imagine a ray sweeping around the origin with a point that traces out the polar graph first moving away from the origin and then

back towards the origin.

Example 2:

Sketch the polar curve given by the equation =r cos 2 .

Solution:

The cartesian graph of indicates that r reaches the maximum value of 1 when = 0, and 2 for 0 2 and r has the

minimum value 1 when =

2 and

3

2 for 0 2

Also =r 0 when when =

4,

3

4,

5

4,

3

2 and

7

4 for 0 2 .


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The following pictures indicate how the curve is traced out as increases.

Example 3:

Sketch the polar curve given by the equation =r 1 cos .

Solution:

The following graph and table of values can be used to help sketch the graph.

| 0

4

2

3

4

5

4

3

2

7

42

=r 1 cos | 0 2 22

1 +2 22

2 +2 22

1 2 22

0

2 2

2 ~ 0.29289 and

+2 2

2 ~ 1.7071.


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Note: This curve is called a cardioid.

Example 4:

Sketch the polar curve given by the equation =r 1 sin 2 .

Solution:

The following graph and table of values can be used to help sketch the graph.

| 0

4

2

3

4

5

4

3

2

7

42

=r

1 sin 2

| 1 0 1 2 1 0 1 2 1


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Tangents to polar curves.

Given a polar curve =r ( )f , we can make the substitution =r ( )f in the equations

{=x rcos

=y rsin

to obtain the parametric equations:

{=x ( )f cos

=y ( )f sin for the curve.

Then

d x

d = ( )f ' cos ( )f sin =

d r

d cos rsin

and

d y

d = +( )f ' sin ( )f cos =

d r

d +sin rcos ,

so that

dy

dx =

d y

d

d x

d

=

+

d r

dsin rcos

d r

dcos rsin

------- (i).

If the curve passes though the pole then we can take =r 0 in (i) to see that the slope of the tangent at such a point is given by

=dy

dxtan .

provided that d r

d0.

Example 5:

Find the polar coordinates of the points on the cardioid given by the equation =r 1 cos where the tangent line to the curve is (a)

horizontal (b) vertical.

Solution:

=r 1 cos

implies that

=d r

dsin .

Hence

d x

d =

d r

d cos rsin = sin cos ( )1 cos sin

= sin ( )cos ( )1 cos

= sin ( )2 cos 1 .


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Also

d y

d =

d r

d +sin rcos = +sin

2 ( )1 cos cos

= + 1 cos2

cos cos2

= + 1 cos 2 cos2

= ( )+2 cos 1 ( )cos 1 .

so that

dy

dx =

d y

d

d x

d

= ( )+2 cos 1 ( )cos 1

sin ( )2 cos 1.

Tangent lines to the curve are horizontal where =d y

d0 and

d x

d0, and tangent lines are vertical where =

d x

d0 and

d y

d0.

=d y

d0

=( )+2 cos 1 ( )cos 1 0

=cos 1

2 or =cos 1.

Now =cos 1

2 if and only if =

2

3 or =

4

3 for 0 < 2 and =cos 1 if and only if = 0 for 0 < 2 .

Hence =dy

d0 when = 0,

2

3 or

4

3 for 0 < 2 .

=d x

d0

=sin ( )2 cos 1 0

=sin 0 or =cos 1

2.

Now =sin 0 if and only if = 0 or = for 0 < 2 and =cos 1

2 if and only if =

3 or =

5

3 for 0 < 2 .

Hence =dx

d0 when = 0,

3, , or

5

3 for 0 < 2 .

Thus we see that the tangent line to the curve are horizontalat the points

,3

2

2

3 and

,3

2

4

3 and the tangent line to the curve are

verticalat the points

,

1

2

3, ( ),2 and

,

1

2

5

3.

Notes:

The following picture shows the location of the horizontal and vertical tangent lines.


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The slope of tangent line to the curve tends to 0 as tends to 0, since

lim 0

( )+2 cos 1 ( )cos 1

sin ( )2 cos 1 = lim

0

+2 cos 1

2 cos 1. lim

0

1 cos

sin

= 3 lim 0

+

2

!2

4

!4. . .

+

3

!3. . .

= 0.

This might suggest that there is a tangent line to the curve at the pole with zero slope. However, the picture indicates that this is not

possible and this can be confirmed from the fact that the section of the curve in the first quadrant given by 0

3 is the graph of the

function f given by =( )fx 2 4x 4x

22 1 4x

2 and the section of the curve for

5

3 2 , or equivalently

3 0,

is the graph of =y ( )fx .

The formula for the function f can be confirmed by first obtaining a cartesian equation for the curve.

If r 0, the equation =r 1 cos is equivalent to

=r2

r rcos .

Substituting =rcos x in this equation gives

=r2

r x

which is equivalent to

=+r2

x r.

This equation implies that

=( )+r2

x2

r2,

which then gives the following cartesian equation for the cardioid.

=( )+ +x2

y2

x2

+x2

y2.

This equation is equivalent to

=+ + + + +x4

y4

x2

2x2y

22x

32x y

2+x

2y

2

and to

=+ + + + x4

y4

2x2y

22x

32x y

2y

20 ------- (ii).


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This equation can be arranged as a quadratic equation for y2, that is, in the form

=+ + +y4

( )+ 2x2

2x 1 y2

x4

2x3

0.

The quadratic formula gives

y2 =

1 2x 2x2

+/ ( )+ 2x2

2x 12

4x4

8x2

2 ------- (iii).

Since =( )+ 2x2

2x 12

+ + + 4x4

4x2

1 8x3

4x2

4x = + + 4x4

1 8x3

4x, equation (iii) gives

=y2

1 2x 2x2

+/ 1 4x

2 =

2 4x 4x2

+/2 1 4x

4

which, in turn, gives

y = + 2 4x 4x

2+/2 1 4x

2.

The cardioid can be broken up into four sections given by the four equations:

=y +2 4x 4x

22 1 4x

2, =y

2 4x 4x2

2 1 4x

2,

=y +

2 4x 4x

2

2

1 4x2

, =y

2 4x 4x

2

2

1 4x2

.

The first of these four equations decribes the function f mentioned previously.

The following picture shows the graphs of these four equations in red, green, maroonand bluerespectively.

The function f given by

=( )fx 2 4x 4x2 2 1 4x

2

does not have a derivative at 0, although for x > 0 we have

=( )f ' x

4

1 4x8x 4

4 2 4x 4x2

2 1 4x.

It is possible to check that ( )f ' x tends to 0 as x 0+.


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As a further investigation of the nature of the curve forxclose to 0 note that the binomial expansion

=1 4x 1 2x 2x2

4x3

10x4

28x5

. . .

gives

2 4x 4x2

2 1 4x = + + +8x3

20x4

56x5

. . . ,

so that whenxis close to 0 we have

( )fx ~ 2 x

3

2.

.

The following picture shows the graphs of =y ( )fx and =y ( )fx in redand the graphs of =y 2 x

3

2and =y 2 x

3

2 in blueforx

close to 0.

As a final remark, we can find a formula fordy

dx from equation (iii) by implicit differentiation.

Differentiating both sides of equation (ii) with respect toxgives:

+4x3

4y3 + +dy

dx4x y

24x

2y + + +

dy

dx6x

22y

24x y

dy

dx2y =

dy

dx0.

This implies that

dy

dx ( )+ + 4y

34x

2y 4x y 2y = ( )+ + +4x

34x y

26x

22y

2

so that

dy

dx =

+ + +4x3

4x y2

6x2

2y2

+ + 4y3

4x2y 4x y 2y

= + + +2x

32x y

23x

2y

2

y ( )+ + 2y2

2x2

2x 1.

Note that the derivativedy

dx does not exist at the point ( ),0 0 .

It is possible to use this formula to find the coordinates of the points where =dy

dx0 and where =

dx

dy0, but a lot more work is needed than

when working with polar coordinates.

For example, =dy

dx0 when =+ + +2x

32x y

23x

2y

20, that is, when =y

2

+2x3 3x2

+2x 1. Using this equation to substitute for y

2 in

equation (ii) gives (after some work)

=x

2( )+4x 3

( )+2x 12

0.

Hence =dy

dx0 when =x

3

4 and =y +

3 3

4. (After some more work.)


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Questions:

Q1. Find a cartesian equation for each of the following polar curves and also describe the curve.

(a) =r 3 sin (b) =r +2 sin 2 cos (c) =r csc (d) =r tan sec

Q2. In each of the following cases find a polar equation for the curve with the given cartesian equation.

(a) =y2

x (b) =+x y 4 (c) =+x2

y2

2x (d) =x2

y2

1

Q3. Sketch the curve with the given polar equation.

(a) =r sin (b) =r 1 sin (c) =r sin 2 (d) =r cos 3

(e) =r 1 2 cos (f) =r

sin

2 (g) =r +1 2 cos 2

Q4. Find the slope of the tangent line to the curve =r 2 sin at the point where =

6.

Q5. Find polar coordinates of the points on the curve =r 3 cos where the tangent line is horizontal or vertical.

Answers:

Q1. (a) =+x2

y2

3y =+x2

x

3

2

29

4 ... a circle with centre at

,0

3

2 and radius

3

2.

(b) =+x2

y2

+2y 2x =+( )x 12

( )y 12

2 ... a circle with centre at ( ),1 1 and radius 2.

(c) =y 1 ... a horizontal line through ( ),0 1 .

(d) =y x2 ... a parabola with its axis vertical and vertex at ( ),0 0 .

Q2. (a) =r csc cot (b) =r4

+cos sin (c) =r 2 cos (d) =r

1

cos 2

Q3. (a)

(b)


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(c)

(d)

(e)

(f)


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(g)

Q4. =dy

dxtan 2 so =

dy

dx =

6

tan

3= 3 .

Q5. The tangent line is horizontal at the points

,

3

2

4 and

,

3

2

4, and vertical at the points ( ),3 0 and

,0

2.

Version: 16/4/2008 Peter Stone

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Polar Coords