Curve tracing

POLAR COORDINATESPOLAR COORDINATES
and CURVE TRACINGand CURVE TRACING
10

A coordinate system represents
a point in the plane by an ordered
pair of numbers called coordinates.
POLAR COORDINATES

Usually, we use Cartesian coordinates,
which are directed distances from two
perpendicular axes.
POLAR COORDINATES

Here, we describe a coordinate system
introduced by Newton, called the polar
coordinate system.
 It is more convenient for many purposes.
POLAR COORDINATES

Polar Coordinates
In this section, we will learn:
How to represent points in polar coordinates.
POLAR COORDINATES

POLE
We choose a point in the plane
that is called the pole (or origin)
and is labeled O.

POLAR AXIS
Then, we draw a ray (half-line) starting
at O called the polar axis.
 This axis is usually drawn horizontally to the right
corresponding to the positive x-axis in Cartesian
coordinates.

ANOTHER POINT
If P is any other point in the plane, let:
 r be the distance from O to P.
 θ be the angle (usually measured in radians)
between the polar axis and the line OP.

POLAR COORDINATES
P is represented by the ordered pair (r, θ).
r, θ are called polar coordinates of P.

POLAR COORDINATES
We use the convention that
an angle is:
 Positive—if measured in the counterclockwise
direction from the polar axis.
 Negative—if measured in the clockwise
direction from the polar axis.

If P = O, then r = 0, and we agree that
(0, θ) represents the pole for any value
of θ.
POLAR COORDINATES

We extend the meaning of polar
coordinates (r, θ) to the case in which
r is negative—as follows.
POLAR COORDINATES

POLAR COORDINATES
We agree that, as shown, the points (–r, θ)
and (r, θ) lie on the same line through O
and at the same distance | r | from O, but
on opposite sides of O.

POLAR COORDINATES
If r > 0, the point (r, θ) lies in the same
quadrant as θ.
If r < 0, it lies in the quadrant on the opposite
side of the pole.
 Notice that (–r, θ)
represents
the same point
as (r, θ + π).

POLAR COORDINATES
Plot the points whose polar coordinates
are given.
a. (1, 5π/4)
b. (2, 3π)
c. (2, –2π/3)
d. (–3, 3π/4)
Example 1

POLAR COORDINATES
The point (1, 5π/4) is plotted here.
Example 1 a

The point (2, 3π) is plotted.
Example 1 bPOLAR COORDINATES

POLAR COORDINATES
The point (2, –2π/3) is plotted.
Example 1 c

POLAR COORDINATES
The point (–3, 3π/4) is plotted.
 It is is located three
units from the pole
in the fourth quadrant.
 This is because
the angle 3π/4 is in
the second quadrant
and r = -3 is negative.
Example 1 d

CARTESIAN VS. POLAR COORDINATES
In the Cartesian coordinate system, every
point has only one representation.
However, in the polar coordinate system,
each point has many representations.

CARTESIAN VS. POLAR COORDINATES
For instance, the point (1, 5π/4) in
Example 1 a could be written as:
 (1, –3π/4), (1, 13π/4), or (–1, π/4).

CARTESIAN & POLAR COORDINATES
In fact, as a complete counterclockwise
rotation is given by an angle 2π, the point
represented by polar coordinates (r, θ) is
also represented by
(r, θ + 2nπ) and (-r, θ + (2n + 1)π)
where n is any integer.

The connection between polar and Cartesian
coordinates can be seen here.
 The pole corresponds to the origin.
 The polar axis coincides with the positive x-axis.

If the point P has Cartesian coordinates (x, y)
and polar coordinates (r, θ), then, from
the figure, we have: cos sin
x y
r r
θ θ= =

cos
sin
x r
y r
θ
θ
=
=
Equations 1
Therefore,
CARTESIAN & POLAR COORDS.

Although Equations 1 were deduced from
the figure (which illustrates the case where
r > 0 and 0 < θ < π/2), these equations are
valid for all values of r and θ.
 See the general
definition of sin θ
and cos θ
in Appendix D.

Equations 1 allow us to find
the Cartesian coordinates of a point
when the polar coordinates are known.

To find r and θ when x and y are known,
we use the equations
 These can be
deduced from
Equations 1 or
simply read from
the figure.
2 2 2
tan
y
r x y
x
θ= + =
Equations 2

Convert the point (2, π/3) from polar to
Cartesian coordinates.
 Since r = 2 and θ = π/3,
Equations 1 give:
 Thus, the point is (1, ) in Cartesian coordinates.
Example 2
1
cos 2cos 2 1
3 2
3
sin 2sin 2. 3
3 2
x r
y r
π
θ
π
θ
= = = × =
= = = =
3

Represent the point with Cartesian
coordinates (1, –1) in terms of polar
coordinates.
Example 3

If we choose r to be positive, then
Equations 2 give:
 As the point (1, –1) lies in the fourth quadrant,
we can choose θ = –π/4 or θ = 7π/4.
Example 3
2 2 2 2
1 ( 1) 2
tan 1
r x y
y
x
θ
= + = + − =
= = −

Thus, one possible answer is:
( , –π/4)
Another possible answer is:
( , 7π/4)
Example 3
2
2

Equations 2 do not uniquely determine θ
when x and y are given.
 This is because, as θ increases through the interval
0 ≤ θ ≤ 2π, each value of tan θ occurs twice.
Note

So, in converting from Cartesian to polar
coordinates, it’s not good enough just to find r
and θ that satisfy Equations 2.
 As in Example 3, we must choose θ so that
the point (r, θ) lies in the correct quadrant.
Note

POLAR CURVES
The graph of a polar equation r = f(θ)
[or, more generally, F(r, θ) = 0] consists
of all points that have at least one polar
representation (r, θ), whose coordinates
satisfy the equation.

POLAR CURVES
What curve is represented by
the polar equation r = 2 ?
 The curve consists of all points (r, θ)
with r = 2.
 r represents the distance from the point
to the pole.
Example 4

POLAR CURVES
 Thus, the curve r = 2 represents
the circle with center O and radius 2.
Example 4

POLAR CURVES
In general, the equation r = a represents
a circle O with center and radius |a|.
Example 4

POLAR CURVES
Sketch the polar curve θ = 1.
 This curve consists of all points (r, θ)
such that the polar angle θ is 1 radian.
Example 5

POLAR CURVES
It is the straight line that passes through O
and makes an angle of 1 radian with the polar
axis.
Example 5

POLAR CURVES
Notice that:
 The points (r, 1) on
the line with r > 0 are
in the first quadrant.
 The points (r, 1) on
the line with r < 0 are
in the third quadrant.
Example 5

POLAR CURVES
a. Sketch the curve with polar equation
r = 2 cos θ.
b. Find a Cartesian equation for this curve.
Example 6

POLAR CURVES
First, we find the values of r for
some convenient values of θ.
Example 6 a

POLAR CURVES
We plot the corresponding points (r, θ).
Then, we join these points
to sketch the curve—as
follows.
Example 6 a

The curve appears to be a circle.
Example 6 aPOLAR CURVES

POLAR CURVES
We have used only values of θ between 0
and π—since, if we let θ increase beyond π,
we obtain the same points again.
Example 6 a

POLAR CURVES
To convert the given equation to a Cartesian
equation, we use Equations 1 and 2.
 From x = r cos θ, we have cos θ = x/r.
 So, the equation r = 2 cos θ becomes r = 2x/r.
 This gives:
2x = r2
= x2
+ y2
or x2
+ y2
– 2x = 0
Example 6 b

POLAR CURVES
Completing the square,
we obtain:
(x – 1)2
+ y2
= 1
 The equation is of a circle with center (1, 0)
and radius 1.
Example 6 b

POLAR CURVES
The figure shows a geometrical illustration
that the circle in Example 6 has the equation
r = 2 cos θ.
 The angle OPQ is
a right angle,
and so r/2 = cos θ.
 Why is OPQ
a right angle?

POLAR CURVES
Sketch the curve r = 1 + sin θ.
 Here, we do not plot points as in Example 6.
 Rather, we first sketch the graph of r = 1 + sin θ
in Cartesian coordinates by shifting the sine curve
up one unit—as follows.
Example 7

POLAR CURVES
This enables us to read at a glance the
values of r that correspond to increasing
values of θ.
Example 7

POLAR CURVES
For instance, we see that, as θ increases
from 0 to π/2, r (the distance from O)
increases from 1 to 2.
Example 7

POLAR CURVES
So, we sketch the corresponding part
of the polar curve.
Example 7

POLAR CURVES
As θ increases from π/2 to π,
the figure shows that r decreases
from 2 to 1.
Example 7

POLAR CURVES
So, we sketch the next part of
the curve.
Example 7

POLAR CURVES
As θ increases from to π to 3π/2,
r decreases from 1 to 0, as shown.
Example 7

POLAR CURVES
Finally, as θ increases from 3π/2 to 2π,
r increases from 0 to 1, as shown.
Example 7

POLAR CURVES
If we let θ increase beyond 2π or
decrease beyond 0, we would simply
retrace our path.
Example 7

POLAR CURVES
Putting together the various parts of the curve,
we sketch the complete curve—as shown
next.
Example 7

CARDIOID
It is called a cardioid—because it’s
shaped like a heart.
Example 7

POLAR CURVES
Sketch the curve r = cos 2θ.
 As in Example 7, we first sketch r = cos 2θ,
0 ≤ θ ≤2π, in Cartesian coordinates.
Example 8

POLAR CURVES
As θ increases from 0 to π/4,
the figure shows that r decreases
from 1 to 0.
Example 8

POLAR CURVES
So, we draw the
corresponding portion
of the polar curve
(indicated by ).1
Example 8

POLAR CURVES
As θ increases from π/4 to π/2, r goes
from 0 to –1.
 This means that the distance from O increases
from 0 to 1.
Example 8

POLAR CURVES
 However, instead of being
in the first quadrant,
this portion of the polar curve
(indicated by ) lies on
the opposite side of the pole
in the third quadrant.
2
Example 8

POLAR CURVES
The rest of the curve is
drawn in a similar fashion.
 The arrows and numbers
indicate the order in which
the portions are traced out.
Example 8

POLAR CURVES
The resulting curve
has four loops
and is called a
four-leaved rose.
Example 8

SYMMETRY
When we sketch polar curves, it is
sometimes helpful to take advantage
of symmetry.

RULES
The following three rules
are explained by figures.

RULE 1
If a polar equation is unchanged when θ
is replaced by –θ, the curve is symmetric
about the polar axis.

RULE 2
If the equation is unchanged when r is
replaced by –r, or when θ is replaced by
θ + π, the curve is symmetric about the pole.
 This means that
the curve remains
unchanged if we rotate
it through 180° about
the origin.

RULE 3
If the equation is unchanged when θ is
replaced by π – θ, the curve is symmetric
about the vertical line θ = π/2.

SYMMETRY
The curves sketched in Examples 6 and 8
are symmetric about the polar axis, since
cos(–θ) = cos θ.

SYMMETRY
The curves in Examples 7 and 8 are
symmetric about θ = π/2, because
sin(π – θ) = sin θ and cos 2(π – θ) = cos 2θ.

SYMMETRY
The four-leaved rose is also symmetric
about the pole.

SYMMETRY
These symmetry properties
could have been used in sketching
the curves.

SYMMETRY
For instance, in Example 6, we need only
have plotted points for 0 ≤ θ ≤ π/2 and then
reflected about the polar axis to obtain
the complete circle.

TANGENTS TO POLAR CURVES
To find a tangent line to a polar curve r = f(θ),
we regard θ as a parameter and write its
parametric equations as:
x = r cos θ = f (θ) cos θ
y = r sin θ = f (θ) sin θ

Then, using the method for finding slopes of
parametric curves (Equation 2 in Section 10.2)
and the Product Rule, we have:
Equation 3
sin cos
cos sin
dy dr
r
dy d d
dx drdx r
d d
θ θ
θ θ
θ θ
θ θ
+
= =
−

We locate horizontal tangents by finding
the points where dy/dθ = 0 (provided that
dx/dθ ≠ 0).
Likewise, we locate vertical tangents at
the points where dx/dθ = 0 (provided that
dy/dθ ≠ 0).

Notice that, if we are looking for tangent
lines at the pole, then r = 0 and Equation 3
simplifies to:
tan if 0
dy dr
dx d
θ
θ
= ≠

For instance, in Example 8, we found that
r = cos 2θ = 0 when θ = π/4 or 3π/4.
 This means that the lines θ = π/4 and θ = 3π/4
(or y = x and y = –x) are tangent lines to r = cos 2θ
at the origin.

a. For the cardioid r = 1 + sin θ of Example 7,
find the slope of the tangent line when
θ = π/3.
b. Find the points on the cardioid where
the tangent line is horizontal or vertical.
Example 9

Using Equation 3 with r = 1 + sin θ, we have:
2
sin cos
cos sin
cos sin (1 sin )cos
cos cos (1 sin )sin
cos (1 2sin ) cos (1 2sin )
1 2sin sin (1 sin )(1 2sin )
dr
r
dy d
drdx r
d
θ θ
θ
θ θ
θ
θ θ θ θ
θ θ θ θ
θ θ θ θ
θ θ θ θ
+
=
−
+ +
=
− +
+ +
= =
− − + −
Example 9

The slope of the tangent at the point where
θ = π/3 is:
Example 9 a
3
1
2
cos( /3)(1 2sin( /3))
(1 sin( /3))(1 2sin( /3)
(1 3)
(1 3 / 2)(1 3)
1 3 1 3
1
(2 3)(1 3) 1 3
dy
dx θ π
π π
π π=
+
=
+ −
+
=
+ −
+ +
= = = −
+ − − −

Observe that:
Example 9 b
dy
dθ
= cosθ(1+ 2sinθ) = 0 whenθ =
π
2
,
3π
2
,
7π
6
,
11π
6
dx
dθ
= (1+ sinθ)(1− 2sinθ) = 0 whenθ =
3π
2
,
π
6
,
5π
6

Hence, there are horizontal tangents at
the points
(2, π/2), (½, 7π/6), (½, 11π/6)
and vertical tangents at
(3/2, π/6), (3/2, 5π/6)
 When θ = 3π/2, both dy/dθ and dx/dθ are 0.
 So, we must be careful.
Example 9 b

Using l’Hospital’s Rule, we have:
(3 / 2)
(3 / 2) (3 / 2)
(3 /2)
(3 /2)
lim
1 2sin cos
lim lim
1 2sin 1 sin
1 cos
lim
3 1 sin
1 sin
lim
3 cos
dy
dxθ π
θ π θ π
θ π
θ π
θ θ
θ θ
θ
θ
θ
θ
−
− −
−
−
→
→ →
→
→
+  
=  ÷ ÷
− +  
= −
+
−
= − = ∞
Example 9 b

By symmetry,
(3 / 2)
lim
dy
dxθ π +
→
= −∞
Example 9 b

Thus, there is a vertical tangent line
at the pole.
Example 9 b

Instead of having to remember Equation 3,
we could employ the method used to
derive it.
 For instance, in Example 9, we could have written:
x = r cos θ = (1 + sin θ) cos θ = cos θ + ½ sin 2θ
y = r sin θ = (1 + sin θ) sin θ = sin θ + sin2
θ
Note

 Then, we would have
which is equivalent to our previous
expression.
Note
/ cos 2sin cos
/ sin cos2
cos sin 2
sin cos2
dy dy d
dx dx d
θ θ θ θ
θ θ θ
θ θ
θ θ
+
= =
− +
+
=
− +

GRAPHING POLAR CURVES
It’s useful to be able to
sketch simple polar curves
by hand.

GRAPHING POLAR CURVES
However, we need to use a graphing
calculator or computer when faced with curves
as complicated as shown.

GRAPHING POLAR CURVES WITH GRAPHING DEVICES
Some graphing devices have commands that
enable us to graph polar curves directly.
With other machines, we need to convert to
parametric equations first.

GRAPHING WITH DEVICES
In this case, we take the polar equation r = f(θ)
and write its parametric equations as:
x = r cos θ = f(θ) cos θ
y = r sin θ = f(θ) sin θ
 Some machines require that the parameter
be called t rather than θ.

Graph the curve r = sin(8θ / 5).
 Let’s assume that our graphing device doesn’t
have a built-in polar graphing command.
Example 10

In this case, we need to work with
the corresponding parametric equations,
which are:
 In any case, we need to determine
the domain for θ.
Example 10
cos sin(8 /5)cos
sin sin(8 /5)sin
x r
y r
θ θ θ
θ θ θ
= =
= =

So, we ask ourselves:
 How many complete rotations are required
until the curve starts to repeat itself ?
Example 10

If the answer is n, then
 So, we require that 16nπ/5
be an even multiple of π.
8( 2 ) 8 16
sin sin
5 5 5
8
sin
5
n nθ π θ π
θ
+  
= + ÷
 
=
Example 10

This will first occur when n = 5.
 Hence, we will graph the entire curve
if we specify that 0 ≤ θ ≤ 10π.
Example 10

Switching from θ to t,
we have the equations
sin(8 /5)cos
sin(8 /5)sin
0 10
x t t
y t t
t π
=
=
≤ ≤
Example 10

This is the resulting curve.
 Notice that
this rose has
16 loops.
Example 10

Investigate the family of polar curves given
by r = 1 + c sin θ.
How does the shape change as c changes?
 These curves are called limaçons—after a French word
for snail, because of the shape of the curves for certain
values of c.
Example 11

The figures show computer-drawn
graphs for various values of c.
Example 11

For c > 1, there is a loop that decreases
in size as decreases.
Example 11

When c = 1, the loop disappears and
the curve becomes the cardioid that we
sketched in Example 7.
Example 11

For c between 1 and ½, the cardioid’s cusp
is smoothed out and becomes a “dimple.”
Example 11

When c decreases from ½ to 0,
the limaçon is shaped like an oval.
Example 11

This oval becomes more circular as c → 0.
When c = 0, the curve is just the circle r = 1.
Example 11

The remaining parts show that, as c
becomes negative, the shapes change
in reverse order.
Example 11

In fact, these curves are reflections about
the horizontal axis of the corresponding
curves with positive c.
Example 11

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