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**10.3** Polar Coordinates

### Polar Coordinates

A coordinate system represents a point in the plane by an ordered pair of numbers called coordinates. Usually we use Cartesian coordinates, which are directed distances from two perpendicular axes.

Here we describe a coordinate system introduced by
**Newton, called the polar coordinate system, which is **
more convenient for many purposes.

**We choose a point in the plane that is called the pole **

*(or origin) and is labeled O. Then we draw a ray (half-line) *
**starting at O called the polar axis.**

### Polar Coordinates

This axis is usually drawn horizontally to the right and

*corresponds to the positive x-axis in Cartesian coordinates.*

*If P is any other point in the plane, let r be the distance *
*from O to P and let *θ be the angle (usually measured in
*radians) between the polar axis and the line OP as in *
Figure 1.

**Figure 1**

### Polar Coordinates

*Then the point P is represented by the ordered pair (r, *θ)
*and r, *θ **are called polar coordinates of P.**

We use the convention that an angle is positive if

measured in the counterclockwise direction from the polar axis and negative in the clockwise direction.

*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 by agreeing that, as in Figure 2, *
*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.*

**Figure 2**

### 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).

### Example 1

Plot the points whose polar coordinates are given.

(a) (1, 5π/4) (b) (2, 3π) (c) (2, –2π/3) (d) (–3, 3π/4)

Solution:

The points are plotted in Figure 3.

**Figure 3**

*Example 1 – Solution*

In part (d) the point (–3, 3π/4) is located three units from
the pole in the fourth quadrant because the angle 3π/4 is in
*the second quadrant and r = –3 is negative.*

cont’d

### Polar Coordinates

In the Cartesian coordinate system every point has only

one representation, but in the polar coordinate system each point has many representations. For instance, the point

(1, 5π/4) in Example 1(a) could be written as (1, –3π/4) or (1, 13π/4) or (–1, π/4). (See Figure 4.)

**Figure 4**

### Polar Coordinates

In fact, since a complete counterclockwise rotation is given
by an angle 2π, the point represented by polar coordinates
*(r, *θ) is also represented by

*(r, *θ *+ 2n*π*) and (–r, *θ *+ (2n + 1)*π)
*where n is any integer.*

### Polar Coordinates

The connection between polar and Cartesian coordinates
can be seen from Figure 5, in which the pole corresponds
to the origin and the polar axis coincides with the positive
*x-axis.*

**Figure 5**

### Polar Coordinates

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

and so

Although Equations 1 were deduced from Figure 5, which
*illustrates the case where r* > 0 and 0 < θ < π/2, these

*equations are valid for all values of r and *θ*.*

### Polar Coordinates

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*

which can be deduced from Equations 1 or simply read from Figure 5.

**Figure 5**

### Example 2

Convert the point (2, π/3) from polar to Cartesian coordinates.

Solution:

*Since r = 2 and *θ = π/3, Equations 1 give
*x = r cos *θ

*y = r sin *θ

Therefore the point is (1, ) in Cartesian coordinates.

### Example 3

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

Solution:

*If we choose r to be positive, then Equations 2 give*

Since the point (1, –1) lies in the fourth quadrant, we can choose θ = –π/4 or θ = 7π/4. Thus one possible answer is ( , –π/4); another is ( , 7π/4).

### Polar Coordinates

**Note**

Equations 2 do not uniquely determine θ *when x and y are *
given because, as θ increases through the interval

0 ≤ θ < 2π, each value of tan θ occurs twice.

Therefore, 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.

### Polar Curves

### Polar Curves

* The graph of a polar equation r = f(*θ), or more generally

*F(r,*θ

*) = 0, consists of all points P that have at least one*

*polar representation (r,*θ) whose coordinates satisfy the equation.

### Example 4

*What curve is represented by the polar equation r = 2?*

Solution:

*The curve consists of all points (r, *θ*) with r = 2. Since r*
represents the distance from the point to the pole, the
*curve r = 2 represents the circle with center O*

*and radius 2. In general, the equation r = a represents a *
*circle with center O and radius |a|. (See Figure 6.)*

**Figure 6**

### Example 7

*Sketch the curve r = 1 + sin *θ.

Solution:

*We first sketch the graph of r = 1 + sin *θ *in Cartesian *

coordinates in Figure 10 by shifting the sine curve up one unit.

*r = 1 + sin *θin Cartesian coordinates, 0 ≤ θ ≤ 2π

*Example 7 – Solution*

*This enables us to read at a glance the values of r that *

correspond to increasing values of θ. For instance, we see
that as θ increases from 0 to π*/2, r (the distance from O) *
increases from 1 to 2, so we sketch the corresponding part
of the polar curve in Figure 11(a).

cont’d

**Figure 11(a)**

*Stages in sketching the cardioid r = 1 + sin *θ

*Example 7 – Solution*

As θ increases from π/2 to π*, Figure 10 shows that r*

decreases from 2 to 1, so we sketch the next part of the curve as in Figure 11(b).

cont’d

**Figure 11(b)**

*Stages in sketching the cardioid r = 1 + sin *θ

**Figure 10**

*r = 1 + sin *θin Cartesian coordinates, 0 ≤ θ ≤ 2π

*Example 7 – Solution*

As θ increases from π to 3π*/2, r decreases from 1 to 0 as *
shown in part (c). Finally, as θ increases from 3π/2 to 2π,
*r increases from 0 to 1 as shown in part (d). If we let *θ
increase beyond 2π or decrease beyond 0, we would
simply retrace our path.

cont’d

**Figure 11(c)**

*Stages in sketching the cardioid r = 1 + sin *θ

**Figure 11(d)**

*Example 7 – Solution*

Putting together the parts of the curve from

Figure 11(a)–(d), we sketch the complete curve in part (e).

**It is called a cardioid because it’s shaped like a heart.**

cont’d

**Figure 11(e)**

*Stages in sketching the cardioid r = 1 + sin *θ

### Polar Curves

The remainder of the curve is drawn in a similar fashion,
with the arrows and numbers indicating the order in which
the portions are traced out. The resulting curve has four
**loops and is called a four-leaved rose.**

### Symmetry

### Symmetry

When we sketch polar curves it is sometimes helpful to

take advantage of symmetry. The following three rules are explained by Figure 14.

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

**Figure 14(a)**

### Symmetry

*(b) 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.)

**Figure 14(b)**

### Symmetry

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

**Figure 14(c)**

### Symmetry

The curves in Example 7 is symmetric about θ = π/2 because sin(π – θ) = sin θ and cos 2(π – θ) = cos 2θ.

The four-leaved rose is also symmetric about the pole. These symmetry properties could have been used in sketching the curves.

### Tangents to Polar Curves

### 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 and the Product Rule, we have

### Tangents to Polar Curves

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

Notice that if we are looking for tangent lines at the pole,
*then r = 0 and Equation 3 simplifies to*

### Tangents to Polar Curves

*For instance, we found that r = cos 2*θ = 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 2*θ
at the origin.

### Example 9

*(a) For the cardioid r = 1 + sin*θ, 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.

Solution:

*Using Equation 3 with r = 1 + sin *θ, we have

*Example 9 – Solution*

(a) The slope of the tangent at the point where θ = π/3 is

cont’d

*Example 9 – Solution*

(b) Observe that

cont’d

*Example 9 – Solution*

Therefore there are horizontal tangents at the points and vertical tangents at

When θ = 3π*/2, both dy /d*θ *and dx /d*θ
are 0, so we must be careful. Using l’Hospital’s Rule, we
have

cont’d

*Example 9 – Solution*

By symmetry,

cont’d

*Example 9 – Solution*

Thus there is a vertical tangent line at the pole (see Figure 15).

cont’d

*Tangent lines for r = 1 + sin *θ

### Tangents to Polar Curves

**Note**

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 + sin 2θ
*y = r sin *θ = (1 + sin θ) sin θ = sin θ + sin^{2}θ

Then we have

which is equivalent to our previous expression.

### Graphing Polar Curves with

### Graphing Devices

### Graphing Polar Curves with Graphing Devices

Although it’s useful to be able to sketch simple polar curves by hand, we need to use a graphing calculator or computer when we are faced with a curve as complicated as the

ones shown in Figures 16 and 17.

*r = sin*^{3}(2.5θ) + cos^{3}(2.5θ) *r = 2 + sin*^{3}(2.4θ)

**Figure 16** **Figure 17**

### 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. 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.

### Example 10

*Graph the curve r = sin(8*θ/5).

Solution:

Let’s assume that our graphing device doesn’t have a built-in polar graphing command.

In this case we need to work with the corresponding parametric equations, which are

*x = r cos *θ = sin(8θ/5) cos θ *y = r sin *θ = sin(8θ/5) sin θ

*Example 10 – Solution*

In any case we need to determine the domain for θ. So we
ask ourselves: How many complete rotations are required
*until the curve starts to repeat itself? If the answer is n, then*

*and so we require that 16n*π/5 be an even multiple of π.

*This will first occur when n = 5. Therefore we will graph the *
entire curve if we specify that 0 ≤ θ ≤ 10π.

cont’d

*Example 10 – Solution*

Switching from θ *to t, we have the equations*

*x = sin(8t /5) cos t * *y = sin(8t /5) sin t * 0 ≤ t ≤ 10π

and Figure 18 shows the resulting curve. Notice that this rose has 16 loops.

cont’d

*r = sin(8*θ/ 5)

**Figure 18**