# 8.3 Polar coordinates  (Page 2/8)

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$\begin{array}{l}\begin{array}{l}\\ \mathrm{cos}\text{\hspace{0.17em}}\theta =\frac{x}{r}\to x=r\mathrm{cos}\text{\hspace{0.17em}}\theta \end{array}\hfill \\ \mathrm{sin}\text{\hspace{0.17em}}\theta =\frac{y}{r}\to y=r\mathrm{sin}\text{\hspace{0.17em}}\theta \hfill \end{array}$

Dropping a perpendicular from the point in the plane to the x- axis forms a right triangle, as illustrated in [link] . An easy way to remember the equations above is to think of $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ as the adjacent side over the hypotenuse and $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ as the opposite side over the hypotenuse.

## Converting from polar coordinates to rectangular coordinates

To convert polar coordinates $\text{\hspace{0.17em}}\left(r,\text{\hspace{0.17em}}\theta \right)\text{\hspace{0.17em}}$ to rectangular coordinates $\text{\hspace{0.17em}}\left(x,\text{\hspace{0.17em}}y\right),$ let

$\mathrm{cos}\text{\hspace{0.17em}}\theta =\frac{x}{r}\to x=r\mathrm{cos}\text{\hspace{0.17em}}\theta$
$\mathrm{sin}\text{\hspace{0.17em}}\theta =\frac{y}{r}\to y=r\mathrm{sin}\text{\hspace{0.17em}}\theta$

Given polar coordinates, convert to rectangular coordinates.

1. Given the polar coordinate $\text{\hspace{0.17em}}\left(r,\theta \right),$ write $\text{\hspace{0.17em}}x=r\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y=r\mathrm{sin}\text{\hspace{0.17em}}\theta .$
2. Evaluate $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta .$
3. Multiply $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ to find the x- coordinate of the rectangular form.
4. Multiply $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ by $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ to find the y- coordinate of the rectangular form.

## Writing polar coordinates as rectangular coordinates

Write the polar coordinates $\text{\hspace{0.17em}}\left(3,\frac{\pi }{2}\right)\text{\hspace{0.17em}}$ as rectangular coordinates.

Use the equivalent relationships.

$\begin{array}{l}\begin{array}{l}\\ x=r\mathrm{cos}\text{\hspace{0.17em}}\theta \end{array}\hfill \\ x=3\mathrm{cos}\text{\hspace{0.17em}}\frac{\pi }{2}=0\hfill \\ y=r\mathrm{sin}\text{\hspace{0.17em}}\theta \hfill \\ y=3\mathrm{sin}\text{\hspace{0.17em}}\frac{\pi }{2}=3\hfill \end{array}$

The rectangular coordinates are $\text{\hspace{0.17em}}\left(0,3\right).\text{\hspace{0.17em}}$ See [link] .

## Writing polar coordinates as rectangular coordinates

Write the polar coordinates $\text{\hspace{0.17em}}\left(-2,0\right)\text{\hspace{0.17em}}$ as rectangular coordinates.

See [link] . Writing the polar coordinates as rectangular, we have

$\begin{array}{l}x=r\mathrm{cos}\text{\hspace{0.17em}}\theta \hfill \\ x=-2\mathrm{cos}\left(0\right)=-2\hfill \\ \hfill \\ y=r\mathrm{sin}\text{\hspace{0.17em}}\theta \hfill \\ y=-2\mathrm{sin}\left(0\right)=0\hfill \end{array}$

The rectangular coordinates are also $\text{\hspace{0.17em}}\left(-2,0\right).$

Write the polar coordinates $\text{\hspace{0.17em}}\left(-1,\frac{2\pi }{3}\right)\text{\hspace{0.17em}}$ as rectangular coordinates.

$\left(x,y\right)=\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$

## Converting from rectangular coordinates to polar coordinates

To convert rectangular coordinates to polar coordinates    , we will use two other familiar relationships. With this conversion, however, we need to be aware that a set of rectangular coordinates will yield more than one polar point.

## Converting from rectangular coordinates to polar coordinates

Converting from rectangular coordinates to polar coordinates requires the use of one or more of the relationships illustrated in [link] .

## Writing rectangular coordinates as polar coordinates

Convert the rectangular coordinates $\text{\hspace{0.17em}}\left(3,3\right)\text{\hspace{0.17em}}$ to polar coordinates.

We see that the original point $\text{\hspace{0.17em}}\left(3,3\right)\text{\hspace{0.17em}}$ is in the first quadrant. To find $\text{\hspace{0.17em}}\theta ,\text{\hspace{0.17em}}$ use the formula $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{y}{x}.\text{\hspace{0.17em}}$ This gives

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{3}{3}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =1\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\mathrm{tan}}^{-1}\left(1\right)=\frac{\pi }{4}\hfill \end{array}$

To find $\text{\hspace{0.17em}}r,\text{\hspace{0.17em}}$ we substitute the values for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ into the formula $\text{\hspace{0.17em}}r=\sqrt{{x}^{2}+{y}^{2}}.\text{\hspace{0.17em}}$ We know that $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ must be positive, as $\text{\hspace{0.17em}}\frac{\pi }{4}\text{\hspace{0.17em}}$ is in the first quadrant. Thus

$\begin{array}{l}\begin{array}{l}\\ r=\sqrt{{3}^{2}+{3}^{2}}\end{array}\hfill \\ r=\sqrt{9+9}\hfill \\ r=\sqrt{18}=3\sqrt{2}\hfill \end{array}$

So, $\text{\hspace{0.17em}}r=3\sqrt{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\theta \text{=}\frac{\pi }{4},\text{\hspace{0.17em}}$ giving us the polar point $\text{\hspace{0.17em}}\left(3\sqrt{2},\frac{\pi }{4}\right).\text{\hspace{0.17em}}$ See [link] .

## Transforming equations between polar and rectangular forms

We can now convert coordinates between polar and rectangular form. Converting equations can be more difficult, but it can be beneficial to be able to convert between the two forms. Since there are a number of polar equations that cannot be expressed clearly in Cartesian form, and vice versa, we can use the same procedures we used to convert points between the coordinate systems. We can then use a graphing calculator to graph either the rectangular form or the polar form of the equation.

Given an equation in polar form, graph it using a graphing calculator.

1. Change the MODE to POL , representing polar form.
2. Press the Y= button to bring up a screen allowing the input of six equations: $\text{\hspace{0.17em}}{r}_{1},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{r}_{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\text{\hspace{0.17em}}.\text{\hspace{0.17em}}\text{\hspace{0.17em}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{r}_{6}.$
3. Enter the polar equation, set equal to $\text{\hspace{0.17em}}r.$
4. Press GRAPH .

#### Questions & Answers

The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
Rima Reply
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.
Brittany Reply
how do you find the period of a sine graph
Imani Reply
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
Jhon Reply
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
Baptiste Reply
the sum of any two linear polynomial is what
Esther Reply
divide simplify each answer 3/2÷5/4
Momo Reply
divide simplify each answer 25/3÷5/12
Momo
how can are find the domain and range of a relations
austin Reply
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?
Diddy Reply
6000
Robert
more than 6000
Robert
can I see the picture
Zairen Reply
How would you find if a radical function is one to one?
Peighton Reply
how to understand calculus?
Jenica Reply
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
rachel Reply
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
Reena Reply
what is foci?
Reena Reply
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris

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