# 8.4 Polar coordinates: graphs  (Page 6/16)

 Page 6 / 16

## Sketching the graph of a rose curve ( n Even)

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

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 $\text{\hspace{0.17em}}\theta =\frac{\pi }{2}\text{\hspace{0.17em}}$ and the pole.

Now we will find the zeros. First make the substitution $\text{\hspace{0.17em}}u=4\theta .$

$\begin{array}{c}\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}}\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}}\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}}0=2\mathrm{cos}\text{\hspace{0.17em}}4\theta \\ \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}}\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}}\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}}0=\mathrm{cos}\text{\hspace{0.17em}}4\theta \\ \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}}\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}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0=\mathrm{cos}\text{\hspace{0.17em}}u\\ {\mathrm{cos}}^{-1}0=u\\ \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}}\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}}u=\frac{\pi }{2}\\ \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}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4\theta =\frac{\pi }{2}\\ \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}}\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}}\theta =\frac{\pi }{8}\end{array}$

The zero is $\text{\hspace{0.17em}}\theta =\frac{\pi }{8}.\text{\hspace{0.17em}}$ The point $\text{\hspace{0.17em}}\left(0,\frac{\pi }{8}\right)\text{\hspace{0.17em}}$ is on the curve.

Next, we find the maximum $\text{\hspace{0.17em}}|r|.\text{\hspace{0.17em}}$ We know that the maximum value of $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}u=1\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}\theta =0.\text{\hspace{0.17em}}$ Thus,

$\begin{array}{l}\\ \begin{array}{l}r=2\mathrm{cos}\left(4\cdot 0\right)\hfill \\ r=2\mathrm{cos}\left(0\right)\hfill \\ r=2\left(1\right)=2\hfill \end{array}\end{array}$

The point $\text{\hspace{0.17em}}\left(2,0\right)\text{\hspace{0.17em}}$ is on the curve.

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

 $\theta$ 0 $\frac{\pi }{8}$ $\frac{\pi }{4}$ $\frac{3\pi }{8}$ $\frac{\pi }{2}$ $\frac{5\pi }{8}$ $\frac{3\pi }{4}$ $r$ 2 0 −2 0 2 0 −2

As $\text{\hspace{0.17em}}r=0\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}\theta =\frac{\pi }{8},\text{\hspace{0.17em}}$ it makes sense to divide values in the table by $\text{\hspace{0.17em}}\frac{\pi }{8}\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}r=0,\text{\hspace{0.17em}}$ each petal extends out a distance of $\text{\hspace{0.17em}}r=2,\text{\hspace{0.17em}}$ and then turns back to zero $\text{\hspace{0.17em}}2n\text{\hspace{0.17em}}$ times for a total of eight petals. See the graph in [link] .

Sketch the graph of $\text{\hspace{0.17em}}r=4\mathrm{sin}\left(2\theta \right).$

The graph is a rose curve, $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ even

## Sketching the graph of a rose curve ( n Odd)

Sketch the graph of $\text{\hspace{0.17em}}r=2\mathrm{sin}\left(5\theta \right).$

The graph of the equation shows symmetry with respect to the line $\text{\hspace{0.17em}}\theta =\frac{\pi }{2}.\text{\hspace{0.17em}}$ Next, find the zeros and maximum. We will want to make the substitution $\text{\hspace{0.17em}}u=5\theta .$

$\begin{array}{c}\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}}\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}}\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}}\text{\hspace{0.17em}}0=2\mathrm{sin}\left(5\theta \right)\\ \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}}\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}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0=\mathrm{sin}\text{\hspace{0.17em}}u\\ {\mathrm{sin}}^{-1}0=0\\ \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}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}u=0\\ \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}}\text{\hspace{0.17em}}5\theta =0\\ \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}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\theta =0\end{array}$

The maximum value is calculated at the angle where $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is a maximum. Therefore,

$\begin{array}{l}\begin{array}{l}\\ r=2\mathrm{sin}\left(5\cdot \frac{\pi }{2}\right)\end{array}\hfill \\ r=2\left(1\right)=2\hfill \end{array}$

Thus, the maximum value of the polar equation is 2. This is the length of each petal. As the curve for $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ odd yields the same number of petals as $\text{\hspace{0.17em}}n,\text{\hspace{0.17em}}$ there will be five petals on the graph. See [link] .

Create a table of values similar to [link] .

 $\theta$ 0 $\frac{\pi }{6}$ $\frac{\pi }{3}$ $\frac{\pi }{2}$ $\frac{2\pi }{3}$ $\frac{5\pi }{6}$ $\pi$ $r$ 0 1 −1.73 2 −1.73 1 0

Sketch the graph of $r=3\mathrm{cos}\left(3\theta \right).$

Rose curve, $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ odd

## 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 $\text{\hspace{0.17em}}r=\theta \text{\hspace{0.17em}}$ for $\text{\hspace{0.17em}}\theta \ge 0.\text{\hspace{0.17em}}$ As $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ increases, $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ increases at a constant rate in an ever-widening, never-ending, spiraling path. See [link] .

Given an Archimedes’ spiral over $\text{\hspace{0.17em}}\left[0,2\pi \right],$ sketch the graph.

1. Make a table of values for $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ over the given domain.
2. Plot the points and sketch the graph.

## Sketching the graph of an archimedes’ spiral

Sketch the graph of $\text{\hspace{0.17em}}r=\theta \text{\hspace{0.17em}}$ over $\text{\hspace{0.17em}}\left[0,2\pi \right].$

As $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ is equal to $\text{\hspace{0.17em}}\theta ,\text{\hspace{0.17em}}$ 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] .

 $\theta$ $\frac{\pi }{4}$ $\frac{\pi }{2}$ $\pi$ $\frac{3\pi }{2}$ $\frac{7\pi }{4}$ $2\pi$ $r$ 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] .

#### Questions & Answers

I've run into this: x = r*cos(angle1 + angle2) Which expands to: x = r(cos(angle1)*cos(angle2) - sin(angle1)*sin(angle2)) The r value confuses me here, because distributing it makes: (r*cos(angle2))(cos(angle1) - (r*sin(angle2))(sin(angle1)) How does this make sense? Why does the r distribute once
How can you tell what type of parent function a graph is ?
generally by how the graph looks and understanding what the base parent functions look like and perform on a graph
William
if you have a graphed line, you can have an idea by how the directions of the line turns, i.e. negative, positive, zero
William
y=x will obviously be a straight line with a zero slope
William
y=x^2 will have a parabolic line opening to positive infinity on both sides of the y axis vice versa with y=-x^2 you'll have both ends of the parabolic line pointing downward heading to negative infinity on both sides of the y axis
William
y=x will be a straight line, but it will have a slope of one. Remember, if y=1 then x=1, so for every unit you rise you move over positively one unit. To get a straight line with a slope of 0, set y=1 or any integer.
Aaron
yes, correction on my end, I meant slope of 1 instead of slope of 0
William
what is f(x)=
I don't understand
Joe
Typically a function 'f' will take 'x' as input, and produce 'y' as output. As 'f(x)=y'. According to Google, "The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain."
Thomas
Sorry, I don't know where the "Â"s came from. They shouldn't be there. Just ignore them. :-)
Thomas
Darius
Thanks.
Thomas
Â
Thomas
It is the Â that should not be there. It doesn't seem to show if encloses in quotation marks. "Â" or 'Â' ... Â
Thomas
Now it shows, go figure?
Thomas
what is this?
i do not understand anything
unknown
lol...it gets better
Darius
I've been struggling so much through all of this. my final is in four weeks 😭
Tiffany
this book is an excellent resource! have you guys ever looked at the online tutoring? there's one that is called "That Tutor Guy" and he goes over a lot of the concepts
Darius
thank you I have heard of him. I should check him out.
Tiffany
is there any question in particular?
Joe
I have always struggled with math. I get lost really easy, if you have any advice for that, it would help tremendously.
Tiffany
Sure, are you in high school or college?
Darius
Hi, apologies for the delayed response. I'm in college.
Tiffany
how to solve polynomial using a calculator
So a horizontal compression by factor of 1/2 is the same as a horizontal stretch by a factor of 2, right?
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Can you please help me. Tomorrow is the deadline of my assignment then I don't know how to solve that
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
divide simplify each answer 3/2÷5/4
divide simplify each answer 25/3÷5/12
Momo
how can are find the domain and range of a relations
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
For Plan A to reach $27/month to surpass Plan B's$26.50 monthly payment, you'll need 3,000 texts which will cost an additional \$10.00. So, for the amount of texts you need to send would need to range between 1-100 texts for the 100th increment, times that by 3 for the additional amount of texts...
Gilbert
...for one text payment for 300 for Plan A. So, that means Plan A; in my opinion is for people with text messaging abilities that their fingers burn the monitor for the cell phone. While Plan B would be for loners that doesn't need their fingers to due the talking; but those texts mean more then...
Gilbert