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x=r cosƟ, y=r sinƟ, x^2+y^2=r^2.
I don't get the second one. x^2+y^2=r^2
No, not really. Trig and I don't get along at all
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)
yeah, it's open in my pdf
k and I put inside a circle i called the radius of that circle r
that trangle is a right triangle so we can use Pythagorean them to find what r is in terms of x and y
or like anwar said before x^2+y^2=r^2
yeah, i don't this equation? what do i use it for?
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=.....?
loco, do you know what to fill that blank with?
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=----?
does it equal r?
sorry, if i take too long, trying to figure this out
5r=16 is the polar equation of the cartesian equation 5x^2+5y^2=16
these equations give the same graph using a little different way of graphing
these are equations are equivalent
polar is in the form (r,pheta) Cartesian is in the form (x,y)
what do u mean equivalent? = to what? 16
no to each other
so 5x=5y, is that what u mean?
5r=16 is equivalent to 5x^2+5y^2=16 because they are the same graph
so you want to convert 5x=5y to polar equation?
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
wow... um.. let me try to figure all this out
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
so did you have a question or something?
this might sound arrogant but, how do i set it up and how do i get the answer
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
all not how
write with no x and no y
your goal is to write with r and pheta
x^2+y^2 with r^2*
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
so it would be 5^2+5^2=
where did you get x=5 and y=5?
oh.. crap. i got them from the original equation
hey anwar is going to come back and help you i have to leave k?
ok thank you very much for all ur help though :)
i think you are over thinking this stuff
i tend to do that a lot :P
i hope you get it
i got the whole night, don't worry i'll get it
Just gimme a minute. I'll help you when I finish with another question if you still need help.
I don't know where you have reached. But I'll start by these formula in polar coordinates: \[x=r \cos \theta .... (1)\] \[y=r \sin \theta .... (2)\] \[x^2+y^2=r^2 .... (3)\]
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?
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: \[x=y \implies r \cos \theta=r \sin \theta\]
Is is clear so far?
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: \[x^2+y^2=16 \implies r^2=16\]
is this the answer
Yep. You can simplify it more, but this is OK.