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Sketch the graph of r = 3 2 cos θ .

Graph of the limaçon r=3-2cos(theta). Extending to the left.
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Another type of limaçon, the inner-loop limaçon , is named for the loop formed inside the general limaçon shape. It was discovered by the German artist Albrecht Dürer (1471-1528), who revealed a method for drawing the inner-loop limaçon in his 1525 book Underweysung der Messing . A century later, the father of mathematician Blaise Pascal , Étienne Pascal(1588-1651), rediscovered it.

Formulas for inner-loop limaçons

The formulas that generate the inner-loop limaçons are given by r = a ± b cos θ and r = a ± b sin θ where a > 0 , b > 0 , and a < b . The graph of the inner-loop limaçon passes through the pole twice: once for the outer loop, and once for the inner loop. See [link] for the graphs.

Graph of four inner loop limaçons side by side. (A) is r=a+bcos(theta),a<b. Extended to the right. (B) is a-bcos(theta), a<b. Extends to the left. (C) is r=a+bsin(theta), a<b. Extends up. (D) is r=a-bsin(theta), a<b. Extends down.

Sketching the graph of an inner-loop limaçon

Sketch the graph of r = 2 + 5 cos θ .

Testing for symmetry, we find that the graph of the equation is symmetric about the polar axis. Next, finding the zeros reveals that when r = 0 , θ = 1.98. The maximum | r | is found when cos θ = 1 or when θ = 0. Thus, the maximum is found at the point (7, 0).

Even though we have found symmetry, the zero, and the maximum, plotting more points will help to define the shape, and then a pattern will emerge.

See [link] .

θ 0 π 6 π 3 π 2 2 π 3 5 π 6 π 7 π 6 4 π 3 3 π 2 5 π 3 11 π 6 2 π
r 7 6.3 4.5 2 −0.5 −2.3 −3 −2.3 −0.5 2 4.5 6.3 7

As expected, the values begin to repeat after θ = π . The graph is shown in [link] .

Graph of inner loop limaçon r=2+5cos(theta). Extends to the right. Points on edge plotted are (7,0), (4.5, pi/3), (2, pi/2), and (-3, pi).
Inner-loop limaçon
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Investigating lemniscates

The lemniscate is a polar curve resembling the infinity symbol or a figure 8. Centered at the pole, a lemniscate is symmetrical by definition.

Formulas for lemniscates

The formulas that generate the graph of a lemniscate    are given by r 2 = a 2 cos 2 θ and r 2 = a 2 sin 2 θ where a 0. The formula r 2 = a 2 sin 2 θ is symmetric with respect to the pole. The formula r 2 = a 2 cos 2 θ is symmetric with respect to the pole, the line θ = π 2 , and the polar axis. See [link] for the graphs.

Four graphs of lemniscates side by side. (A) is r^2 = a^2 * cos(2theta). Horizonatal figure eight, on x-axis. (B) is r^2 = - a^2 * cos(2theta). Vertical figure eight, on y axis. (C) is r^2 = a^2 * sin(2theta). Diagonal figure eight on line y=x. (D) is r^2 = -a^2 *sin(2theta). Diagonal figure eight on line y=-x.

Sketching the graph of a lemniscate

Sketch the graph of r 2 = 4 cos 2 θ .

The equation exhibits symmetry with respect to the line θ = π 2 , the polar axis, and the pole.

Let’s find the zeros. It should be routine by now, but we will approach this equation a little differently by making the substitution u = 2 θ .

0 = 4 cos 2 θ 0 = 4 cos u 0 = cos u cos 1 0 = π 2 u = π 2 Substitute  2 θ  back in for  u . 2 θ = π 2 θ = π 4

So, the point ( 0 , π 4 ) is a zero of the equation.

Now let’s find the maximum value. Since the maximum of cos u = 1 when u = 0 , the maximum cos 2 θ = 1 when 2 θ = 0. Thus,

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

We have a maximum at (2, 0). Since this graph is symmetric with respect to the pole, the line θ = π 2 , and the polar axis, we only need to plot points in the first quadrant.

Make a table similar to [link] .

θ 0 π 6 π 4 π 3 π 2
r 2 2 0 2 0

Plot the points on the graph, such as the one shown in [link] .

Graph of r^2 = 4cos(2theta). Horizontal lemniscate, along x-axis. Points on edge plotted are (2,0), (rad2, pi/6), (rad2 7pi/6).
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Investigating rose curves

The next type of polar equation produces a petal-like shape called a rose curve. Although the graphs look complex, a simple polar equation generates the pattern.

Rose curves

The formulas that generate the graph of a rose curve    are given by r = a cos n θ and r = a sin n θ where a 0. If n is even, the curve has 2 n petals. If n is odd, the curve has n petals. See [link] .

Graph of two rose curves side by side. (A) is r=acos(ntheta), where n is even. Eight petals extending from origin, equally spaced. (B) is r=asin(ntheta) where n is odd. Three petals extending from the origin, equally spaced.

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 ?
Anwar Reply
how do you get the (1.4427)^t in the carp problem?
Gabrielle Reply
A hedge is contrusted to be in the shape of hyperbola near a fountain at the center of yard.the hedge will follow the asymptotes y=x and y=-x and closest distance near the distance to the centre fountain at 5 yards find the eqution of the hyperbola
ayesha Reply
A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour. To the nearest hour, what is the half-life of the drug?
Sandra Reply
Find the domain of the function in interval or inequality notation f(x)=4-9x+3x^2
prince Reply
Jessica Reply
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?
Jean Reply
rotation by 80 of (x^2/9)-(y^2/16)=1
Garrett Reply
thanks the domain is good but a i would like to get some other examples of how to find the range of a function
bashiir Reply
what is the standard form if the focus is at (0,2) ?
Lorejean Reply
Roy Reply
Practice Key Terms 9

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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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