PARAMETRIC EQUATIONS AND POLAR COORDINATES

RoboCup is an international competition of robot soccer teams. These are not the remote-controlled attack robots you may have seen on television. Instead, the robots are engineered and programmed to automatically respond to the positions of the ball, goal and other players.

As programmers of such robots, we would be frustrated by the limitations of the x-y system (called rectangular coordinates) we have used in the first six sections of this chapter. First of all, the path followed by a robot is unlikely to pass the vertical line test for a function.

Even if it did, the path of the robot is irrelevant unless we know when the robot was at each point. Further, what a robot needs to know about a competitor are its distance and direction, not its x- and y-coordinates.

In this section, we introduce two alternatives to functions in rectangular coordinates.

Parametric equations will enable us to track the position of an object as a function of time.

Polar coordinates enable us to keep track of points in terms of distance and direction. These alternative descriptions give us a great deal of needed flexibility for attacking a variety of problems. Many complicated looking graphs have a very simple description in parametric equations or polar coordinates.

Parametric Equations

To describe the position of an object (such as a robot) in terms of time, we need functions g(t) and h(t) that give the object’s x- and y-coordinates, respectively, at any given time t.

So, from this description, we not only know the path the object follows, but we also know when it passes through each point. This is the motivation for the following general definition of parametric equations.

Given any pair of functions g(t) and h(t) defined on the same domain D, the equations

x= g(t), y = h(t)

are called parametric equations. Notice that for each choice of t, the parametric equations specify a point (x, y) = (g(t), h(t)) in the xy-plane. The collection of all such points is called the graph of the parametric equations. The graph is a curve in the x y-plane, referred to as a parametric curve.

The choice of the letter t to denote the independent variable (called the parameter) should make you think of time. In many situations, this is exactly what the parameter represents. In some applications, the parameter has an interpretation other than time; in others, it has no physical meaning at all. In general, the parameter can be any quantity that is convenient for describing the relationship between x and y. In our first example, we can simplify our discussion by eliminating the parameter.

EXAMPLE 7.1 Graphing a Parametric Curve

Sketch the curve defined by the parametric equations x= 6 − t², y= t/2, for −2 ≤ t ≤ 4.

Solution In the table that follows, we list a number of values of the parameter t and the corresponding values of x and y.

t −2 −1 0 1 2 3 4

x 2 5 6 5 2 −3 −10

y −1 −¹₂ 0 ¹₂ 1 ³₂ 2

y

4

2 2

4

6

8

10

1 1 (10, 2) 2

(

3, w

)

(

^5, q

)

(2, 1)

(2, 1)

(

^{5, q}

)

x

FIGURE 0.78 x= 6 − t², y = t

2, −2 ≤ t ≤ 4.

We have plotted these points and connected them with a smooth curve in Figure 0.78.

You might also notice that we can easily eliminate the parameter here, by solving for t in terms of y. We have t= 2y, so that x = 6 − 4y². The graph of this last equation is a parabola opening to the left. The plane curve we’re looking for is the portion of this parabola corresponding to−2 ≤ t ≤ 4. From the table, notice that this corresponds to

−1 ≤ y ≤ 2, so that the plane curve is the portion of the parabola indicated in Figure 0.78, where we have also indicated a number of points on the curve. 

You probably noticed the small arrows drawn on top of the plane curve in Figure 0.78.

These indicate the orientation of the curve (i.e., the direction of increasing t). In the case where t represents time and the curve represents the path of an object, notice that the orientation indicates the direction followed by the object as it traverses the path, as in the following example.

EXAMPLE 7.2 The Path of a Projectile

The path of a projectile thrown horizontally with initial speed of 20 ft/s from a height of 64 feet, has parametric equations x= 20t, y = 64 − 16t², for 0≤ t ≤ 2. Sketch the path.

y

x

20 40

20 40 60

FIGURE 0.79 Path of projectile.

Solution As in example 7.1, we can solve for t, to get t = x/20, so that y = 64 − x² 25. With 0≤ t ≤ 2, we get 0 ≤ x ≤ 40. The portion of the parabola y = 64 −x²

25 with 0≤ x ≤ 40 is shown in Figure 0.79. Note that in this case, the orientation indicated in the graph gives the direction of motion. Although we used the x-y equation for graphing, observe that the parametric equations also tell us when the object is located at a given point and indicates the direction of motion. 

Graphing calculators and computer algebra systems sketch a parametric graph by plot-ting points corresponding to a large number of values of the parameter t and then connecplot-ting the plotted points with a smooth curve. The appearance of the resulting graph depends greatly on the graphing window used and also on the particular choice of t-values. This can be seen in the following example.

EXAMPLE 7.3 Parametric Equations Involving Sines and Cosines Sketch the curve defined by the parametric equations

x= 2 cos t, y = 2 sin t, for (a) 0 ≤ t ≤ 2π and (b) 0 ≤ t ≤ π (7.1)

Solution The default graph produced by most graphing calculators looks something like the curve shown in Figure 0.80a (where we have added arrows indicating the orientation). With some thought, we can improve this sketch. First, notice that since x= 2 cos t, x ranges between −2 and 2. Similarly, y ranges between −2 and 2. Changing the graphing window to−2.1 ≤ x ≤ 2.1 and −2.1 ≤ y ≤ 2.1 produces the curve shown in Figure 0.80b. The curve still looks like an ellipse, but with some more thought we can identify it as a circle. Rather than eliminate the parameter by solving for t in terms of either x or y, instead notice from (7.1) that

x²+ y²= 4 cos²t+ 4 sin²t = 4(cos²t+ sin²t)= 4.

So, the plane curve lies on the circle of radius 2 centered at the origin. In fact, it’s the whole circle, as we can see by recognizing what the parameter represents in this case.

Recall from the definition of sine and cosine that if (x, y) is a point on the unit circle and θ is the angle from the positive x-axis to the line segment joining (x, y) and the origin, then we define cosθ = x and sin θ = y. Since we have x = 2 cos t and y = 2 sin t, the parameter t corresponds to the angleθ. Further, the curve is the entire circle of radius 2, traced out as the angle t ranges from 0 to 2π. A “square” graphing window is one with the same scale on the x- and y-axes (not necessarily the same x and y ranges, but the same scales!). Some programs generate such a window when you specify an aspect ratio of 1. Such a square window gives us the circle seen in Figure 0.80c. Finally, what would change if the domain were limited to 0≤ t ≤ π? Since we’ve identified t as the angle as measured from the positive x-axis, it should be clear that you will now get the top half of the circle of radius 2, as shown in Figure 0.80d.

y

x

4 4

4 4

FIGURE 0.80a x= 2 cos t, y = 2 sin t.

y

x

2 2

2 2

FIGURE 0.80b x= 2 cos t, y = 2 sin t.

y

x

2 2

2

2 y

x

2 2

2

FIGURE 0.80c A circle.

FIGURE 0.80d Top semicircle.



REMARK 7.1

To sketch a parametric graph on a CAS, you may need to write the equations in vector format.

For instance, in the case of example 7.3, instead of entering x= 2 cos t and y = 2 sin t, you would enter the ordered pair of functions (2 cos t, 2 sin t).

Simple modifications to the parametric equations in example 7.3 will produce a variety of circles and ellipses. We explore this in example 7.4 and the exercises.

EXAMPLE 7.4 More Circles and Ellipses Defined by Parametric Equations

Identify the plane curves (a) x = 2 cos t, y = 3 sin t, (b) x = 2 + 4 cos t, y = 3 + 4 sin t and (c) x = 3 cos 2t, y = 3 sin 2t, all for 0 ≤ t ≤ 2π.

y

x

2 2

3 3

FIGURE 0.81a

x= 2 cos t, y = 3 sin t.

Solution A computer-generated sketch of (a) is shown in Figure 0.81a. It’s difficult to determine from the sketch whether the curve is an ellipse or simply a distorted graph of a circle. You should rule out a circle, since the parametric equations produce x-values be-tween−2 and 2 and y-values between −3 and 3. To verify that this is an ellipse, notice that

x² 4 + y²

9 =4 cos²t

4 +9 sin²t

9 = cos²t+ sin²t = 1.

A computer-generated sketch of (b) is shown in Figure 0.81b. You should verify that this is the circle (x− 2)²+ (y − 3)²= 16. Finally, a computer sketch of (c) is shown in Figure 0.81c. You should verify that this is the circle x²+ y²= 9. So, what is the role of the 2 in the argument of cosine and sine? If you sketched this on a calculator, you may have noticed that the circle was completed long before the calculator finished graphing.

Because of the 2, a complete circle corresponds to 0≤ 2t ≤ 2π or 0 ≤ t ≤ π. With the domain 0≤ t ≤ 2π, the circle is traced out twice. You might say that the factor of 2 in the argument doubles the speed with which the curve is traced.

y

x

2 6

2 2 4

y

x

3 3

3 3

FIGURE 0.81b

x= 2 + 4 cos t, y = 3 + 4 sin t. FIGURE 0.81c

x= 3 cos 2t, y = 3 sin 2t.



In example 7.5, we see how to find parametric equations for a line segment.

REMARK 7.2

Look carefully at the plane curves in examples 7.3 and 7.4 until you can identify the roles of each of the constants in the equations x= a+ b cos ct, y(t) = d + e sin ct.

These interpretations are important in applications. In particular, (a, d) is the center of an ellipse with axes of length 2b and 2e.

EXAMPLE 7.5 Parametric Equations for a Line Segment

Find parametric equations for the line segment joining the points (1, 2) and (4, 7).

Solution For a line segment, notice that the parametric equations can be linear functions.

That is,

x= a + bt, y = c + dt,

for some constants a, b, c and d. (Eliminate the parameter t to see why this generates a line.) The simplest way to choose these constants is to have t= 0 correspond to the starting point (1, 2). Note that if t= 0, the equations reduce to x = a and y = c. To start our segment at x= 1 and y = 2, we set a = 1 and c = 2. Now note that with t = 1, the equations are x= a + b and y = c + d. To produce the endpoint (4, 7), we must have a+ b = 4 and c + d = 7. With a = 1 and c = 2, solve to get b = 3 and d = 5. We now have that

x= 1 + 3t, y = 2 + 5t, for 0 ≤ t ≤ 1 is a pair of parametric equations describing the line segment. 

In general, for parametric equations of the form x= a + bt, y = c + dt, notice that you can always choose a and c to be the x- and y-coordinates, respectively, of the starting point (since x= a, y = b corresponds to t = 0). Then b is the difference in x-coordinates (endpoint minus starting point) and d is the difference in y-coordinates. With these choices, the line segment is always sketched out for 0≤ t ≤ 1.

REMARK 7.3

Note that in example 7.5, as for all parametric equations, there are an infinite number of choices of parameters that produce a given curve. For instance, you can easily verify that the parametric equations

x= −2 + 3t, y = −3 + 5t, for 1 ≤ t ≤ 2 and

x = t, y = 1+ 5t

3 , for 1 ≤ t ≤ 4

also produce the line segment of example 7.5. We say that each of these pairs of parametric equations is a different parameterization of the curve.

As we illustrate with example 7.6, every equation of the form y = f (x) can be simply written in parametric equations.

EXAMPLE 7.6 Parametric Equations from an x-yEquation

Find parametric equations for the portion of the parabola y= x²from (−1, 1) to (3, 9).

Solution Any equation of the form y= f (x) can be converted to parametric form by simply letting x equal t. Here, this gives us y= x²= t². Then

x= t, y = t², for − 1 ≤ t ≤ 3,

is a parametric representation of the curve. (Of course, you can use the letter x as the parameter instead of the letter t if you prefer.) 

We can describe many plane curves parametrically that are unlike anything you’ve seen so far in your studies. Many of these are difficult to draw by hand, but can be easily plotted with a graphing calculator or CAS.

y

x

2 4

2

4

4

2 2 4

FIGURE 0.82a x= t²− 2, y = t³− t.

y

x

4 8

4

8

6 4 2

FIGURE 0.82b

x= t³− t, y = t⁴− 5t²+ 4.

EXAMPLE 7.7 Some Unusual Plane Curves

Sketch the plane curves (a) x = t²− 2, y = t³− t and (b) x = t³− t, y = t⁴− 5t²+ 4.

Solution A sketch of (a) is shown in Figure 0.82a. This is clearly not the graph of any function (consider the vertical line test to see why), so converting to an x-y equation would not be particularly helpful. You should, however, examine the parametric equations to see if important portions of the graph have been left out (e.g., is there supposed to be anything to the left of x= −2?). In this case, x = t²− 2 ≥ −2 for all t and y = t³− t has no maximum or minimum (think about why). It seems that most of the graph is indeed shown in Figure 0.82a. A computer sketch of (b) is shown in Figure 0.82b. Again, this is not the graph of a function. To get an idea of the scope of the graph, note that x= t³− t has no maximum or minimum. To find the minimum of y = t⁴− 5t²+ 4, note that

t⁴− 5t²+ 4 = (t²− 5/2)²−9 4. You should conclude that y≥ −⁹₄, as indicated in Figure 0.82b. 

In section 0.3, we found the relationship between graphs of one-to-one functions and their inverses. Recall that you can reflect the graph of y= f (x) across the line y = x to get the graph of y= f⁻¹(x). (See Figure 0.83a.) Using parametric equations, this can be easily implemented on a graphing calculator.

y

x y  f(x)

y  x

y  f^1(x)

FIGURE 0.83a

y= f (x) and y = f⁻¹(x).

2

1 1

2

20 20

10 10 y

x

FIGURE 0.83b y= x⁵+ 8x³+ x + 1.

EXAMPLE 7.8 Drawing the Graph of an Unknown Inverse Function Draw a graph of the inverse function of f (x)= x⁵+ 8x³+ x + 1.

Solution Given the graph of f (x) (see Figure 0.83b), we could certainly draw by hand part of the graph of the inverse. However, recall that if (x, y) is a point on the graph of f (x) [i.e., y= f (x)], then (y, x) is a point on the graph of f⁻¹(x) [i.e., x= f⁻¹(y)].

The simplest choice of parametric equations for y= f (x) is x = t

y= f (t).

Swapping x and y now gives parametric equations for f⁻¹(x), namely, x= f (t)

y= t.

In this case, we want x= t⁵+ 8t³+ t + 1 and y = t. The graph of f⁻¹(x) along with the graphs of f (x) and y= x are shown in Figures 0.83c and 0.83d.

4

2 2

4

4 4

2 2 y

x 4 2 2 4

4 4

2 2 y

x

FIGURE 0.83c y= f⁻¹(x).

FIGURE 0.83d

y= f (x) and y = f⁻¹(x).



You should now have some idea of the flexibility of parametric equations. This flexibility alone increases our ability to solve problems and makes the study of parametric equations worthwhile.

Polar Coordinates

The search for a more convenient description of a graph leads us to some important ideas.

Recall that in our discussion of robot soccer, we observed that in the heat of the action, information about the distance and direction to the goal is more important than the x- and y-coordinates of the goal. This kind of description is the idea behind polar coordinates.

You are familiar with the rectangular coordinates (x, y) of a point, where we identify a point by its horizontal displacement x and vertical displacement y from the origin.

An alternative description of a point in the x y-plane consists of specifying the distance r from the point to the origin and the angleθ (in radians) measured from the positive x-axis counterclockwise to the ray connecting the point and the origin (see Figure 0.84). We describe the point by the ordered pair (r, θ) and refer to r and θ as polar coordinates for the point.

y

x (r, u )

r

u

FIGURE 0.84 Polar coordinates.

EXAMPLE 7.9 Converting from Polar to Rectangular Coordinates Plot the points with the indicated polar coordinates and determine the corresponding rectangular coordinates (x, y), for: (a) (2, 0), (b) (3,^π₂), (c) (−3,^π₂) and (d) (2, π).

Solution (a) Notice that the angleθ = 0 locates the point on the positive x-axis. At a distance of r = 2 units from the origin, this corresponds to the point (2, 0) in rectangular coordinates (see Figure 0.85a).

(b) The angleθ = ^π₂ locates points on the positive y-axis. At a distance of r= 3 units from the origin, this corresponds to the point (0, 3) in rectangular coordinates (see Figure 0.85b).

(c) The angle is the same as in (b), but a negative value of r indicates that the point is located 3 units in the opposite direction, at the point (0,−3) in rectangular coordinates (see Figure 0.85b).

(d) The angle θ = π corresponds to the negative x-axis. The distance of r = 2 units from the origin gives us the point (−2, 0) in rectangular coordinates (see Figure 0.85c).

y

(2, 0) x 2

y

x 3

3

(

^3, q

)

(

3, q

)

q

2 y

(2, p) p x

FIGURE 0.85a

The point (2, 0) in polar coordinates.

FIGURE 0.85b The points

 3,π

2

 and



−3,π 2



in polar coordinates.

FIGURE 0.85c

The point (2,π) in polar coordinates.



EXAMPLE 7.10 Converting from Rectangular to Polar Coordinates Find all polar coordinate representations of the rectangular point (1, 1).

Solution From Figure 0.86a (on the following page), notice that the point lies on the line y= x, which makes an angle of ^π₄ with the positive x-axis. From the distance formula, we get that r =√

1²+ 1²=√

2. This says that we can write the point as√ 2,^π₄

in polar coordinates. Referring to Figure 0.86b, notice that we can specify the same point by using a negative value of r, r = −√

2 with the angle ⁵₄^π. (Think about this some.) Notice further, that the angle ⁹₄^π =^π₄ + 2π corresponds to the same ray shown in Figure 0.86a

(see Figure 0.86c). In fact, all of the polar points√

2,^π₄ + 2nπ and

−√

2,^5π₄ + 2nπ for any integer n correspond to the same point in the x y-plane.

y

x 1

1

d

2

y

x 1

1

h

2

y

x 1

1

,  d  2p

2

FIGURE 0.86a

Polar coordinates for the point (1, 1).

FIGURE 0.86b

An alternative polar representation of (1, 1).

FIGURE 0.86c

Another polar representation of the point (1, 1).



REMARK 7.4

Given a distance r and angleθ, there is exactly one point in the xy-plane with the polar coordinates (r, θ). However, as we saw in example 7.10, for a given point (x, y) in the plane, there are an infinite number of possible polar coordinate representations.

In particular, you can use both positive and negative values of r . Also, for a given angleθ, the angles θ ± 2π, θ ± 4π and so on, all correspond to the same ray and can also be used. For convenience, we use the notationθ + 2nπ (for any integer n) to represent all of these possible angles.

y

x (r, u )

r

x  r cos u

y  r sin u

u

FIGURE 0.87

Converting from polar to rectangular coordinates.

Referring to Figure 0.87, notice that it is a simple matter to find the rectangular co-ordinates (x, y) of a point specified in polar coordinates as (r, θ). From the usual definitions for sinθ and cos θ, we get

x= r cos θ and y = r sin θ. (7.2)

As we’ve already observed, a given point (x, y) in the plane will have infinitely many polar coordinate representations. From equations (7.2), notice that

x²+ y²= r²cos²θ + r²sin²θ = r²(cos²θ + sin²θ) = r² and for x= 0,

y

x = r sinθ

r cosθ = sinθ

cosθ = tan θ.

That is, every polar coordinate representation (r, θ) of the point (x, y), where x = 0 must satisfy

r² = x²+ y² and tanθ = y

x. (7.3)

Notice that since there’s more than one choice of r andθ, we cannot actually solve equations (7.3) to produce formulas for r andθ. In particular, while you might be tempted to write θ = tan⁻¹_y

x

, this is not the only possible choice. Remember that for (r, θ) to be a polar representation of the point (x, y), θ can be any angle for which tan θ = _x^y, while tan⁻¹_y

x

 gives you only one angleθ in the interval

−^π₂,^π₂

. Finding polar coordinates for a given point is typically a process involving some graphing and some thought.

EXAMPLE 7.11 Converting from Rectangular to Polar Coordinates Find all polar coordinate representations for the rectangular points (a) (2, 3) and (b) (−3, 1).

Solution (a) With x = 2 and y = 3, we have from (7.3) that r²= x²+ y²= 2²+ 3²= 13, so that r = ±√

13. Also,

tanθ = y x = 3

2.

One solution of this (the most obvious solution) is θ = tan⁻¹₃

2

≈ 0.98 radians. To determine which choice of r corresponds to this angle, note that (2, 3) is located in the first quadrant (see Figure 0.88a). Since 0.98 radians also puts you in the first quadrant, this angle corresponds to the positive value of r , so that √

13, tan⁻¹₃

2

 is one po-lar representation of the point. The negative choice of r corresponds to an angle one half-circle (i.e.,π radians), away (see Figure 0.88b), so that another representation is −√

13, tan⁻¹₃

2

+ π

. Every other polar representation is found by adding multiples of 2π to the two angles used above. That is, every polar representation of the point (2, 3) must have the form√

13, tan⁻¹₃

2

+ 2nπ or

−√

13, tan⁻¹₃

2

+ 2nπ

, for some integer choice of n.

y

x

13

u  tan^1

(

^w

)

3

2

(2, 3)

y

x u  tan^1

(

w

)

u  tan^1

(

^w

)

^{ p}

(2, 3)

FIGURE 0.88a The point (2, 3).

FIGURE 0.88b Negative value of r .

(b) For the point (−3, 1), we have x = −3 and y = 1. From (7.3), we have r²= x²+ y²= (−3)²+ 1²= 10,

so that r= ±√

10. Further,

tanθ = y x = 1

−3, so that the most obvious choice for the polar angle isθ = tan⁻¹

−¹₃

≈ −0.32, which lies in the fourth quadrant. Since the point (−3, 1) is in the second quadrant, this choice of the angle corresponds to the negative value of r (see Figure 0.89). The positive value of r then corresponds to the angle θ = tan⁻¹

−¹₃

+ π. Observe that all po-lar coordinate representations must then be of the form

−√

10, tan⁻¹

−¹₃

+ 2nπ or√

10, tan⁻¹

−¹₃

+ π + 2nπ

, for some integer choice of n. 

y

x

u  tan^1

(

a

)

(3, 1) u  tan^1

(

^a

)

 p

FIGURE 0.89 The point (−3, 1).

Observe that the conversion from polar coordinates to rectangular coordinates is com-pletely straightforward, as in example 7.12.

EXAMPLE 7.12 Converting from Polar to Rectangular Coordinates Find the rectangular coordinates for the polar points (a)

3,^π₆

and (b) (−2, 3).

Solution For (a), we have from (7.2) that

x= r cos θ = 3 cosπ 6 =3√

3 2 and

y= r sin θ = 3 sinπ 6 = 3

2. The rectangular point is then

3√ 3 2 ,³₂

. For (b), we have x= r cos θ = −2 cos 3 ≈ 1.98 and

y= r sin θ = −2 sin 3 ≈ −0.28.

The rectangular point is (−2 cos 3, −2 sin 3), which is located at approximately (1.98, −0.28). 

y

2 2

2 2

x r  2

FIGURE 0.90a The circle r = 2.

The graph of a polar equation r= f (θ) is the set of all points (x, y) for which x= r cos θ, y = r sin θ and r = f (θ). In other words, the graph of a polar equation is a graph in the x y-plane of all those points whose polar coordinates satisfy the given equa-tion. We begin by sketching two very simple (and familiar) graphs. The key to draw-ing the graph of a polar equation is to always keep in mind what the polar coordinates represent.

REMARK 7.5

Notice that for any point (x, y) specified in rectangular coordinates (x= 0), we can always write the point in polar coordinates using either of the polar angles tan⁻¹_y

x

or tan⁻¹y

x

+ π. You can determine which angle corresponds to r =

x²+ y²and which corresponds to

r= −

x²+ y²by looking at the quadrant in which the point lies.

EXAMPLE 7.13 Some Simple Graphs in Polar Coordinates Sketch the graphs of (a) r= 2 and (b) θ = π/3.

Solution For (a), notice that 2= r =

x²+ y², and so, we want all points whose distance from the origin is 2 (with any polar angleθ). Of course, this is the definition of a circle of radius 2 with center at the origin (see Figure 0.90a). For (b), notice that θ = π/3 specifies all points with a polar angle of π/3 from the positive x-axis (at any

distance r from the origin). Including negative values for r , this defines a line with slope tanπ/3 =√

3 (see Figure 0.90b). 

REMARK 7.6

Calculators sketch polar graphs of r = f (θ) by computing the value of f (θ) for numerous values ofθ at regularly spaced values of θ and then plotting the resulting points (x, y). You should be aware that the appearance of a calculator plot depends on the x-y graphing window specified and also on the range of displayed values ofθ. When drawing polar graphs, you should identify any values ofθ corresponding to r = 0 or to where r reaches a maximum or minimum. In addition, you should identify the range of values ofθ that produces one copy of the polar curve, when this is appropriate.

y

x u

FIGURE 0.90b The lineθ = ^π₃.

It turns out that many familiar curves have simple polar equations.

y

x

1 1

q p w 2p

FIGURE 0.91

y= sin x plotted in rectangular coordinates.

EXAMPLE 7.14 A Surprisingly Simple Polar Graph Sketch the graph of the polar equation r = sin θ.

Solution For reference, we first sketch a graph of the sine function in rectangular coordinates on the interval [0, 2π] (see Figure 0.91). Notice that on the interval 0 ≤ θ ≤

π

2, sin θ increases from 0 to its maximum value of 1. Then, on the interval ^π₂ ≤ θ ≤ π, sin θ decreases from 1 to 0. When plotting the polar graph, keep in mind that r = 0 corresponds to the origin. Next, on the intervalπ ≤ θ ≤^3π₂ , sin θ decreases from 0 to its minimum value of−1. Since the values of r are negative, remember that this means that the points plotted are in the opposite quadrant (i.e., the first quadrant). Notice that this traces out the same curve in the first quadrant as we’ve already drawn for 0≤ θ ≤ ^π₂. Likewise, takingθ in the interval ^3π₂ ≤ θ ≤ 2π retraces the portion of the curve in the second quadrant. Since sinθ is periodic of period 2π, taking further values of θ simply retraces portions of the curve that we have already traced. A sketch of the polar graph is shown in Figure 0.92. We now verify that this curve is actually a circle. Notice that if we multiply the equation r = sin θ through by r, we get

r²= r sin θ.

You should immediately recognize from (7.2) and (7.3) that y= r sin θ and r²= x²+ y². This gives us the rectangular equation

x²+ y²= y or

0= x²+ y²− y.

Completing the square, we get

0= x²+



y²− y + 1 4



−1 4 and, adding ¹₄ to both sides,

1 2

2

= x²+

 y−1

2

2

.

This is the rectangular equation for the circle of radius ¹₂ centered at the point 0,¹₂

, which is what we see in Figure 0.92. 

y

x

1 1

1

FIGURE 0.92 The circle r = sin θ.

The graphs of many polar equations are not the graphs of any functions of the form y= f (x), as in example 7.15.

y

x

20 20

20 20

FIGURE 0.93

The spiral r = θ, θ ≥ 0.

EXAMPLE 7.15 An Archimedian Spiral

Sketch the graph of the polar equation r = θ, for θ ≥ 0.

Solution Notice that here, asθ increases, so too does r. That is, as the polar angle increases, the distance from the origin also increases accordingly. This produces the spiral (an example of an Archimedian spiral) seen in Figure 0.93. 

The graph in example 7.16 is in the general class known as lima¸cons. This class of graphs is defined by r= a ± b sin θ or r = a ± b cos θ for positive constants a and b.

Additional graphs of lima¸cons are shown in Appendix A.

y

x 1

2 3 4 5

q p w 2p

FIGURE 0.94

y= 3 + 2 cos x in rectangular coordinates.

EXAMPLE 7.16 A Limac¸ on

Sketch the graph of the polar equation r = 3 + 2 cos θ.

Solution We begin by sketching the graph of y = 3 + 2 cos x in rectangular coordinates on the interval [0, 2π] to use as a reference (see Figure 0.94). Notice that in this case, we have r= 3 + 2 cos θ > 0 for all values of θ. Further, the maximum value of r is 5 (corresponding to when cosθ = 1 at θ = 0, 2π, etc.) and the minimum value of r is 1 (corresponding to when cosθ = −1 at θ = π, 3π, etc.). In this case, the graph is traced out with 0≤ θ ≤ 2π. We summarize the intervals of increase and decrease for r in the following table.

Interval cosθ r 3  2 cos θ

0,^π₂

Decreases from 1 to 0 Decreases from 5 to 3

_π

2, π

Decreases from 0 to−1 Decreases from 3 to 1

π,³₂^π

Increases from−1 to 0 Increases from 1 to 3

_3π

2 , 2π

Increases from 0 to 1 Increases from 3 to 5

In Figures 0.95a–0.95d, we show how the sketch progresses through each interval indi-cated in the table, with the completed figure (called a limac¸on) shown in Figure 0.95d.

1 2 3 4 5

3

2

1 1 2 3 y

1 x 1 1 2 3 4 5

3

2

1 1 2 3 y

x

FIGURE 0.95a 0≤ θ ≤ ^π₂.

FIGURE 0.95b 0≤ θ ≤ π.

在文檔中 The Real Number System (頁 67-79)