Convert the equation to polar form.
5x=5y
AND
x^2+y^2=16

- anonymous

Convert the equation to polar form.
5x=5y
AND
x^2+y^2=16

- Stacey Warren - Expert brainly.com

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- chestercat

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- anonymous

x=r cosƟ, y=r sinƟ, x^2+y^2=r^2.

- anonymous

I don't get the second one. x^2+y^2=r^2

- anonymous

No, not really. Trig and I don't get along at all

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## More answers

- myininaya

ok loco do you have that picture i drawed you where I graph (x,y) and called the angle between the initial ray and I guess we can call it the vector with endpoints at the origin and at the point (x,y)

- anonymous

yeah, it's open in my pdf

- myininaya

k and I put inside a circle i called the radius of that circle r

- myininaya

that trangle is a right triangle so we can use Pythagorean them to find what r is in terms of x and y

- myininaya

r=sqrt(x^2+y^2)

- myininaya

or like anwar said before x^2+y^2=r^2

- anonymous

yeah, i don't this equation? what do i use it for?

- myininaya

we want to convert x^2+y^2=16 to a polar equation so that means we want it in terms of r and pheta
but what is x^2+y^2=.....?

- myininaya

loco, do you know what to fill that blank with?

- anonymous

5x^2+5y^2=16

- myininaya

if we are doing 5x^2+5y^2=16
then we can factor out 5 so we have 5(x^2+y^2)=16
but what does x^2+y^2=----?

- anonymous

does it equal r?

- myininaya

right!

- anonymous

sorry, if i take too long, trying to figure this out

- myininaya

5r=16 is the polar equation of the cartesian equation 5x^2+5y^2=16

- myininaya

these equations give the same graph using a little different way of graphing

- myininaya

these are equations are equivalent

- myininaya

polar is in the form (r,pheta)
Cartesian is in the form (x,y)

- anonymous

what do u mean equivalent? = to what? 16

- myininaya

no to each other

- anonymous

so 5x=5y, is that what u mean?

- myininaya

5r=16 is equivalent to 5x^2+5y^2=16 because they are the same graph

- anonymous

oh ok...

- myininaya

so you want to convert 5x=5y to polar equation?

- anonymous

yes

- myininaya

so remember polar has (r,pheta) form
so x=rcos(pheta) and y=rsin(pheta)
5rcos(pheta)=5rsin(pheta) we could divide both sides by 5
so we have
rcos(pheta)=rsin(pheta)
rcos(pheta)-rsin(pheta)=0
r(cos(pheta)-sin(pheta))=0
so we have r=0 and cos(pheta)-sin(pheta)=0

- anonymous

wow... um.. let me try to figure all this out

- anonymous

ok

- anonymous

There?

- myininaya

hey i was talking to my hubby. hes gonna be home soon so I'm going to leave soon but we were done with writing 5x=5y to a polar equation

- anonymous

yes

- myininaya

so did you have a question or something?

- anonymous

this might sound arrogant but, how do i set it up and how do i get the answer

- myininaya

how you have to do is place x^2+y^2 with r and replace x with rcos(pheta) and y with rsin(pheta) and usually I tried to simplified like I did above

- myininaya

all not how

- myininaya

write with no x and no y

- myininaya

your goal is to write with r and pheta

- anonymous

ok

- myininaya

x^2+y^2 with r^2*

- myininaya

do you see my mistake way up there?
i forgot r^2 when i wrote 5r^2=16 is equivalent to 5x^2+5y^2=16

- anonymous

so it would be 5^2+5^2=

- myininaya

?

- myininaya

where did you get x=5 and y=5?

- anonymous

oh.. crap. i got them from the original equation

- myininaya

hey anwar is going to come back and help you i have to leave k?

- anonymous

ok thank you very much for all ur help though :)

- myininaya

i think you are over thinking this stuff

- anonymous

i tend to do that a lot :P

- myininaya

i hope you get it

- anonymous

thank you

- anonymous

i got the whole night, don't worry i'll get it

- anonymous

Just gimme a minute. I'll help you when I finish with another question if you still need help.

- anonymous

ok

- anonymous

I don't know where you have reached. But I'll start by these formula in polar coordinates:
\[x=r \cos \theta .... (1)\]
\[y=r \sin \theta .... (2)\]
\[x^2+y^2=r^2 .... (3)\]

- anonymous

We're trying to convert the given equations that are in terms of x and y to an a form in terms of r and theta, which is called polar coordinates. Is that clear so far?

- anonymous

Good, Now consider the first equation: 5x=5y. Divide both sides by 5 you get:
\[x=y\] But we can substitute fot x with equation (1) and for y with equation (2). That's:
\[x=y \implies r \cos \theta=r \sin \theta\]

- anonymous

ok

- anonymous

Is is clear so far?

- anonymous

Is it*

- anonymous

For the 2nd equation:
\[x^2+y^2=16\]
we will just apply directly the third formula, since x^2+y^2 is just r^2. Then:
\[x^2+y^2=16 \implies r^2=16\]

- anonymous

is this the answer

- anonymous

Yep. You can simplify it more, but this is OK.

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