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Sketching the graph of a rose curve ( n Even)

Sketch the graph of r = 2 cos 4 θ .

Testing for symmetry, we find again that the symmetry tests do not tell the whole story. The graph is not only symmetric with respect to the polar axis, but also with respect to the line θ = π 2 and the pole.

Now we will find the zeros. First make the substitution u = 4 θ .

0 = 2 cos 4 θ 0 = cos 4 θ 0 = cos u cos 1 0 = u u = π 2 4 θ = π 2 θ = π 8

The zero is θ = π 8 . The point ( 0 , π 8 ) is on the curve.

Next, we find the maximum | r | . We know that the maximum value of cos u = 1 when θ = 0. Thus,

r = 2 cos ( 4 0 ) r = 2 cos ( 0 ) r = 2 ( 1 ) = 2

The point ( 2 , 0 ) is on the curve.

The graph of the rose curve has unique properties, which are revealed in [link] .

θ 0 π 8 π 4 3 π 8 π 2 5 π 8 3 π 4
r 2 0 −2 0 2 0 −2

As r = 0 when θ = π 8 , it makes sense to divide values in the table by π 8 units. A definite pattern emerges. Look at the range of r -values: 2, 0, −2, 0, 2, 0, −2, and so on. This represents the development of the curve one petal at a time. Starting at r = 0 , each petal extends out a distance of r = 2 , and then turns back to zero 2 n times for a total of eight petals. See the graph in [link] .

Sketch of rose curve r=2*cos(4 theta). Goes out distance of 2 for each petal 2n times (here 2*4=8 times).
Rose curve, n even
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Sketch the graph of r = 4 sin ( 2 θ ) .

The graph is a rose curve, n even
Graph of rose curve r=4 sin(2 theta). Even - four petals equally spaced, each of length 4.

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Sketching the graph of a rose curve ( n Odd)

Sketch the graph of r = 2 sin ( 5 θ ) .

The graph of the equation shows symmetry with respect to the line θ = π 2 . Next, find the zeros and maximum. We will want to make the substitution u = 5 θ .

0 = 2 sin ( 5 θ ) 0 = sin u sin 1 0 = 0 u = 0 5 θ = 0 θ = 0

The maximum value is calculated at the angle where sin θ is a maximum. Therefore,

r = 2 sin ( 5 π 2 ) r = 2 ( 1 ) = 2

Thus, the maximum value of the polar equation is 2. This is the length of each petal. As the curve for n odd yields the same number of petals as n , there will be five petals on the graph. See [link] .

Create a table of values similar to [link] .

θ 0 π 6 π 3 π 2 2 π 3 5 π 6 π
r 0 1 −1.73 2 −1.73 1 0
Graph of rose curve r=2sin(5theta). Five petals equally spaced around origin. Point (2, pi/2) on edge is marked.
Rose curve, n odd
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Sketch the graph of r = 3 cos ( 3 θ ).

Graph of rose curve r=3cos(3theta). Three petals equally spaced from origin.

Rose curve, n odd

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Investigating the archimedes’ spiral

The final polar equation we will discuss is the Archimedes’ spiral, named for its discoverer, the Greek mathematician Archimedes (c. 287 BCE - c. 212 BCE), who is credited with numerous discoveries in the fields of geometry and mechanics.

Archimedes’ spiral

The formula that generates the graph of the Archimedes’ spiral    is given by r = θ for θ 0. As θ increases, r increases at a constant rate in an ever-widening, never-ending, spiraling path. See [link] .

Two graphs side by side of Archimedes' spiral. (A) is r= theta, [0, 2pi]. (B) is r=theta, [0, 4pi]. Both start at origin and spiral out counterclockwise. The second has two spirals out while the first has one.

Given an Archimedes’ spiral over [ 0 , 2 π ] , sketch the graph.

  1. Make a table of values for r and θ over the given domain.
  2. Plot the points and sketch the graph.

Sketching the graph of an archimedes’ spiral

Sketch the graph of r = θ over [ 0 , 2 π ] .

As r is equal to θ , the plot of the Archimedes’ spiral begins at the pole at the point (0, 0). While the graph hints of symmetry, there is no formal symmetry with regard to passing the symmetry tests. Further, there is no maximum value, unless the domain is restricted.

Create a table such as [link] .

θ π 4 π 2 π 3 π 2 7 π 4 2 π
r 0.785 1.57 3.14 4.71 5.50 6.28

Notice that the r -values are just the decimal form of the angle measured in radians. We can see them on a graph in [link] .

Graph of Archimedes' spiral r=theta over [0,2pi]. Starts at origin and spirals out in one loop counterclockwise. Points (pi/4, pi/4), (pi/2,pi/2), (pi,pi), (5pi/4, 5pi/4), (7pi/4, pi/4), and (2pi, 2pi) are marked.
Archimedes’ spiral
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Questions & Answers

For each year t, the population of a forest of trees is represented by the function A(t) = 117(1.029)t. In a neighboring forest, the population of the same type of tree is represented by the function B(t) = 86(1.025)t.
Shakeena Reply
by how many trees did forest "A" have a greater number?
how solve standard form of polar
Rhudy Reply
what is a complex number used for?
Drew Reply
It's just like any other number. The important thing to know is that they exist and can be used in computations like any number.
I would like to add that they are used in AC signal analysis for one thing
Good call Scott. Also radar signals I believe.
Is there any rule we can use to get the nth term ?
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how do you get the (1.4427)^t in the carp problem?
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Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the high temperature of ?105°F??105°F? occurs at 5PM and the average temperature for the day is ?85°F.??85°F.? Find the temperature, to the nearest degree, at 9AM.
Karlee Reply
if you have the amplitude and the period and the phase shift ho would you know where to start and where to end?
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rotation by 80 of (x^2/9)-(y^2/16)=1
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thanks the domain is good but a i would like to get some other examples of how to find the range of a function
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Practice Key Terms 9

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