# 10.4 Polar coordinates: graphs  (Page 5/16)

 Page 5 / 16

Sketch the graph of $\text{\hspace{0.17em}}r=3-2\mathrm{cos}\text{\hspace{0.17em}}\theta .$

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 $\text{\hspace{0.17em}}r=a±b\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}r=a±b\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}a>0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}b>0,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\text{\hspace{0.17em}}a 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.

## Sketching the graph of an inner-loop limaçon

Sketch the graph of $\text{\hspace{0.17em}}r=2+5\text{cos}\text{\hspace{0.17em}}\theta .$

Testing for symmetry, we find that the graph of the equation is symmetric about the polar axis. Next, finding the zeros reveals that when $\text{\hspace{0.17em}}r=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\theta =1.98.\text{\hspace{0.17em}}$ The maximum $\text{\hspace{0.17em}}|r|\text{\hspace{0.17em}}$ is found when $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta =1\text{\hspace{0.17em}}$ or when $\text{\hspace{0.17em}}\theta =0.\text{\hspace{0.17em}}$ 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.

 $\theta$ $0$ $\frac{\pi }{6}$ $\frac{\pi }{3}$ $\frac{\pi }{2}$ $\frac{2\pi }{3}$ $\frac{5\pi }{6}$ $\pi$ $\frac{7\pi }{6}$ $\frac{4\pi }{3}$ $\frac{3\pi }{2}$ $\frac{5\pi }{3}$ $\frac{11\pi }{6}$ $2\pi$ $r$ 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 $\text{\hspace{0.17em}}\theta =\pi .\text{\hspace{0.17em}}$ The graph is shown in [link] .

## Investigating lemniscates

The lemniscate is a polar curve resembling the infinity symbol $\text{\hspace{0.17em}}\infty \text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}{r}^{2}={a}^{2}\mathrm{cos}\text{\hspace{0.17em}}2\theta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}{r}^{2}={a}^{2}\mathrm{sin}\text{\hspace{0.17em}}2\theta \text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}a\ne 0.\text{\hspace{0.17em}}$ The formula $\text{\hspace{0.17em}}{r}^{2}={a}^{2}\mathrm{sin}\text{\hspace{0.17em}}2\theta \text{\hspace{0.17em}}$ is symmetric with respect to the pole. The formula $\text{\hspace{0.17em}}{r}^{2}={a}^{2}\mathrm{cos}\text{\hspace{0.17em}}2\theta \text{\hspace{0.17em}}$ is symmetric with respect to the pole, the line $\text{\hspace{0.17em}}\theta =\frac{\pi }{2},\text{\hspace{0.17em}}$ and the polar axis. See [link] for the graphs.

## Sketching the graph of a lemniscate

Sketch the graph of $\text{\hspace{0.17em}}{r}^{2}=4\mathrm{cos}\text{\hspace{0.17em}}2\theta .$

The equation exhibits symmetry with respect to the line $\text{\hspace{0.17em}}\theta =\frac{\pi }{2},\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}u=2\theta .$

So, the point $\left(0,\frac{\pi }{4}\right)$ is a zero of the equation.

Now let’s find the maximum value. Since the maximum of $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}u=1\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}u=0,\text{\hspace{0.17em}}$ the maximum $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}2\theta =1\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}2\theta =0.\text{\hspace{0.17em}}$ Thus,

$\begin{array}{c}\text{\hspace{0.17em}}{r}^{2}=4\mathrm{cos}\left(0\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{r}^{2}=4\left(1\right)=4\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}r=±\sqrt{4}\text{\hspace{0.17em}}=2\end{array}$

We have a maximum at (2, 0). Since this graph is symmetric with respect to the pole, the line $\text{\hspace{0.17em}}\theta =\frac{\pi }{2},$ and the polar axis, we only need to plot points in the first quadrant.

Make a table similar to [link] .

 $\theta$ 0 $\frac{\pi }{6}$ $\frac{\pi }{4}$ $\frac{\pi }{3}$ $\frac{\pi }{2}$ $r$ 2 $\sqrt{2}$ 0 $\sqrt{2}$ 0

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

## 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 $\text{\hspace{0.17em}}r=a\mathrm{cos}\text{\hspace{0.17em}}n\theta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}r=a\mathrm{sin}\text{\hspace{0.17em}}n\theta \text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}a\ne 0.\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ is even, the curve has $\text{\hspace{0.17em}}2n\text{\hspace{0.17em}}$ petals. If $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ is odd, the curve has $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ petals. See [link] .

How look for the general solution of a trig function
stock therom F=(x2+y2) i-2xy J jaha x=a y=o y=b
root under 3-root under 2 by 5 y square
The sum of the first n terms of a certain series is 2^n-1, Show that , this series is Geometric and Find the formula of the n^th
cosA\1+sinA=secA-tanA
why two x + seven is equal to nineteen.
The numbers cannot be combined with the x
Othman
2x + 7 =19
humberto
2x +7=19. 2x=19 - 7 2x=12 x=6
Yvonne
because x is 6
SAIDI
what is the best practice that will address the issue on this topic? anyone who can help me. i'm working on my action research.
simplify each radical by removing as many factors as possible (a) √75
how is infinity bidder from undefined?
what is the value of x in 4x-2+3
give the complete question
Shanky
4x=3-2 4x=1 x=1+4 x=5 5x
Olaiya
hi can you give another equation I'd like to solve it
Daniel
what is the value of x in 4x-2+3
Olaiya
if 4x-2+3 = 0 then 4x = 2-3 4x = -1 x = -(1÷4) is the answer.
Jacob
4x-2+3 4x=-3+2 4×=-1 4×/4=-1/4
LUTHO
then x=-1/4
LUTHO
4x-2+3 4x=-3+2 4x=-1 4x÷4=-1÷4 x=-1÷4
LUTHO
A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was  1350  bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after  3  hours?
v=lbh calculate the volume if i.l=5cm, b=2cm ,h=3cm
Need help with math
Peya
can you help me on this topic of Geometry if l help you
litshani
( cosec Q _ cot Q ) whole spuare = 1_cosQ / 1+cosQ
A guy wire for a suspension bridge runs from the ground diagonally to the top of the closest pylon to make a triangle. We can use the Pythagorean Theorem to find the length of guy wire needed. The square of the distance between the wire on the ground and the pylon on the ground is 90,000 feet. The square of the height of the pylon is 160,000 feet. So, the length of the guy wire can be found by evaluating √(90000+160000). What is the length of the guy wire?
the indicated sum of a sequence is known as