Transform each polar equation to an equation in rectangular coordinates and identify its shape.
a. θ = -(π / 6)
b. r = (4 / (2cosθ - 3sinθ))

- anonymous

- schrodinger

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- anonymous

- anonymous

I think for A, r = 2 but I'm not sure

- freckles

\[\theta=\frac{-\pi}{6}\]
actually describes a line
|dw:1433651567694:dw|

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## More answers

- freckles

so you know your line will be of the form y=mx

- freckles

now the job is to find m

- anonymous

Okay...

- freckles

\[\text{ recall } \frac{y}{x}=\tan(\theta) \\\]

- freckles

plug in your value of theta

- freckles

you can also write that as y=tan(theta)*x

- anonymous

I am so confused...

- anonymous

I thought we were finding m? Are we going backwards now?

- freckles

m is just a constant value
I showed what m was by replacing m with tan(theta)
and I got m was tan(theta)
by using the equation y/x=tan(theta)

- anonymous

ohhh, so how do I find out what x and y are since I already know theta?

- freckles

I knew my equation would look like this because I knew theta=a number is a line
\[y=mx\]

- freckles

y and x are variables

- freckles

they vary

- freckles

you only need to find m aka tan(theta)

- anonymous

But I already know what theta is for a. I need to know how to find r, right?

- anonymous

- anonymous

- amilapsn

:)

- anonymous

Can you help me?

- amilapsn

You don't need to find r

- anonymous

Then how am I supposed to graph it, if I don't know what r is?

- amilapsn

You just need to find the Cartesian form of the polar equation \(\theta = -\frac{\pi}{6}\)

- anonymous

Okay...how do I do that?

- anonymous

Do I use tan(theta)

- anonymous

or arctan (theta)?

- amilapsn

You know that's a straight line, right?

- anonymous

yes

- amilapsn

Do you know how to get a equation of a line?

- anonymous

no

- amilapsn

you know the Cartesian form of the line is y=mx+c, right?

- anonymous

okay, so how would I write that with -pi/6??

- anonymous

hello...??? @amilapsn

- amilapsn

Calm down... @Shorty1234 take a deep breath.... :)

- amilapsn

yes of course.... You have to find m and c... what would be c, (it's easy)?

- anonymous

-pi/6?

- amilapsn

No... I'll ask again... you know the Cartesian form of the line is y=mx+c, right?
What would m and c represent?

- amilapsn

(Their meaning)

- anonymous

m is the slop and c is the y-intercept

- amilapsn

yes that's it what's the y intercept =c then?

- anonymous

-pi/6

- amilapsn

What do you mean by the y intercept?

- anonymous

is that not what "c" represents ?

- amilapsn

Yes, it is ....

- anonymous

okay...so whats your point

- amilapsn

y- intercept is the y coordinate where straight line cuts the y axis, right?

- anonymous

yes

- freckles

sorry lost internet connection
do you remember the drawing I made
(since you guys are talking about the y-intercept)|dw:1433653769876:dw|
where does that cross the y-axis?

- amilapsn

so?

- anonymous

it doesnt...so is it zero?

- amilapsn

:) @freckles

- amilapsn

you got it..! :)

- freckles

well it does at 0 (0 is a number :))
so that is how I got y=mx earlier

- amilapsn

Now we need to find m right?

- anonymous

so my answer for a would be (0, -pi/6)?

- amilapsn

- freckles

anyways I will let @amilapsn continue since I was gone so long

- amilapsn

No....

- anonymous

.....is there more stuff I have to do??

- amilapsn

Yes you do...

- amilapsn

you do have to find m, right?

- anonymous

ok

- amilapsn

How do you find m?

- amilapsn

can you tell me the relation ship between - pi/6 and slope?

- anonymous

y2 - y1 / x2 - x1

- amilapsn

That's it! So m would be..............?

- anonymous

0?

- freckles

|dw:1433654449002:dw|
I'm sorry
I said I was going to stay away :p
but can you replace -pi/6 and x and y
using any trig ratio?

- freckles

by the way that is a right triangle I made there

- amilapsn

no.... Can you tell me the relationship between m and pi/6

- freckles

relate *not replace

- anonymous

I give up, this makes zero sense

- freckles

do you know tan(theta)=opp/adj?

- anonymous

yes

- freckles

|dw:1433654806812:dw|
so tan(-pi/6)=?

- amilapsn

......in terms of y and x?

- freckles

yes using the right triangle above

- anonymous

I got -0.577350269

- amilapsn

can you express the same in terms of x and y ?

- anonymous

I don't understand???

- amilapsn

using the right triangle drawn by freckles...

- anonymous

but i don't know what x is?

- amilapsn

yes you don't indeed... use just x and y...

- anonymous

so express x and y using x and y? that makes no sense

- freckles

no express tan(-pi/6) in terms of x and y

- freckles

using the right triangle

- freckles

I have even labeled the opposite side with the correct measurement and the adjacent side too for the point (x,y)

- freckles

|dw:1433655304381:dw|

- anonymous

so opp/adj ??

- freckles

where opp is y
and adj is x right?

- freckles

so can you say what tan(-pi/6)=? in terms of x and y now?

- anonymous

tan (-pi/6) = y ???

- freckles

almost

- freckles

you forgot the x on denominator

- anonymous

so tan(-pi/6) / x = y?

- freckles

:(
tan(theta)=y/x
you know what theta is
just plug in your theta

- freckles

you are making this way complication
your Cartesian equation is suppose to be in terms of y and x
you know theta=-pi/6
so you will have an equation in terms of x and y
if you replace the theta with -pi/6

- freckles

lol i'm having trouble with english tonight :p

- anonymous

.....this makes no sense

- freckles

why ?
I thought the visual made it pretty clear

- freckles

|dw:1433655836599:dw|
if this point is (x,y)
then ...

- freckles

|dw:1433655860761:dw|

- freckles

tan(-pi/6)=opp/adj
where opp=y
and adj=x

- anonymous

yeah, I dont get this...

- freckles

let me just do an example I guess
say we have
\[\theta=\frac{\pi}{4}\]
again this is a line going through the origin
this is of the form y=mx
we only need to determine m
we can draw this line
|dw:1433655974295:dw|

- freckles

let's say point (x,y) is somewhere on there
randomly|dw:1433656006691:dw| just choose a spot

- freckles

|dw:1433656020961:dw|

- freckles

well looking at this triangle I can make the observation that
\[\tan(\frac{\pi}{4})=\frac{y}{x} \]
I can be done here

- freckles

or I can simplify

- freckles

tan(pi/4)=1

- freckles

\[1=\frac{y}{x} \\ \text{ I can also multiply } x \text{ on both sides } y=x\]

- freckles

y=x is the Cartesian (aka rectangular form ) of the polar equation theta=pi/4

- freckles

say we have theta=-pi/4
well that would just be tan(-pi/4)=y/x
which is -1=y/x
which is-x=y

- freckles

say we have theta=pi/3
that would just be tan(pi/3)=y/x
or y=tan(pi/3)x
we can evaluate tan(pi/3)
\[\tan(\frac{\pi}{3})=\frac{\sqrt{3}}{1}=\sqrt{3}\]
so theta=pi/3 can be written as the Cartesian equation \[y=\sqrt{3}x\]

- anonymous

So my answer would be y = tan(-pi/6)x ?

- freckles

yes and simplify the tan(-pi/6)

- anonymous

right,I'll do that. So what about the second problem. Hopefully it what take as long to figure out....

- freckles

b. r = (4 / (2cosθ - 3sinθ))
multiply both sides by the denominator on right hand side
you will need the other two equations to help you to convert this in terms of x and y
that you will need rcos(theta)=x
and rsin(theta)=y

- anonymous

ok, thanks

- freckles

by the way these equations are useful when converting from one to the other
\[r^2=x^2+y^2 \\ \tan(\theta)=\frac{y}{x} \\ x=r \cos(\theta) \\ y=r \sin(\theta)\]
these are your tools to work these particular puzzles

- freckles

though it is kinda useful knowing why they are
and that is why I drew the picture above because it doesn't appear you know why these equations are

- freckles

|dw:1433657166077:dw|
all of those equations can be found just by using this picture

- amilapsn

|dw:1433657937284:dw|

- amilapsn

Just a hint: )

- amilapsn

Things may seem complicated... But they aren't..! Math is the way to sort things out!! :)

- amilapsn

Hope is the only thing you should have!!

- amilapsn

Don't give up.. Just try....!

- amilapsn

You'll find the mathematician inside you..

- amilapsn

|dw:1433658575750:dw|

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