anonymous
  • anonymous
Convert r=3sin(theta) into a cartesian equation.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[r=\sqrt (x^2+y^2)\] \[\tan^{-1} (y/x)=\theta\]
anonymous
  • anonymous
Put them in place, and try simplifying, the resulting equation will be in terms of x and y
nowhereman
  • nowhereman
No need to make it so complicated for θ. You should know that polar coordinates mean \[\sin θ = y\]

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anonymous
  • anonymous
rsin(theta) = y. I just need to know how to deal with the r being on the left side of the equation basically.
anonymous
  • anonymous
Good point @nowhereman
nowhereman
  • nowhereman
a, right forgot that r
nowhereman
  • nowhereman
you can divide it away then
anonymous
  • anonymous
In r you can put r=(√x^2+y^2)
anonymous
  • anonymous
Ok so you use the equation of a circle? Not sure exactly where that's coming from...
anonymous
  • anonymous
\[x^2+y^2=3y\]
anonymous
  • anonymous
Sorry, you got to square the right side
nowhereman
  • nowhereman
That would just be the definition of polar coordinates. \[r = 3\sinθ ⇔ r = 3ry ⇔ 1 = 3y ∨ r = 0 ⇔ y = \frac 1 3 ∨ x = y = 0\]
anonymous
  • anonymous
Kinda abstract...
nowhereman
  • nowhereman
yeah, that was math is about.
anonymous
  • anonymous
Yeah I understand that but the pathetic American Education System doesn't go into the abstract much. So call me dumb but that's how it is. lol
amistre64
  • amistre64
the graph of this thing is a circle centered on the yaxis stretching from the origin to the y=3 mark...
amistre64
  • amistre64
it should end up as: x^2 + (y -1/2)^2 = 3/2 if i see it right :)
nowhereman
  • nowhereman
mmh, yeah I put the r on the wrong side... guess I'm not awake after all
amistre64
  • amistre64
well (3/2)^2 :)
anonymous
  • anonymous
Amistre, what was your process for getting that answer?
amistre64
  • amistre64
seeing the graph in my head really :) the algebra is this: r = 3sin(t) ; * r r^2 = 3rsin(t) ; r^2 = x^2 + y^2 ; and rsin(t) = y x^2 + y^2 = 3y ; -3y x^2 +y^2 -3y = 0 ; complete the square x^2 +y^2 -3y + (3/2)^2 = (3/2)^2 ; clean up the square.. x^2 +(y-3/2)^2 = (3/2)^2
amistre64
  • amistre64
i was off by a y -1/2 the first time lol....but then again i am not right in the head ;)
anonymous
  • anonymous
Ok I think I'm beginning to see it. Thanks.
amistre64
  • amistre64
any ?s speak up ;)
anonymous
  • anonymous
I may come up with another one in a minute. Just digesting it right now.
amistre64
  • amistre64
at sin(0) we are at the origin; at sin(90) we are at 3; then we swing back to 0 on our way thru and that makes the circle that sits on the origin stretched to y=3 in my head
amistre64
  • amistre64
center at (0,3/2)
amistre64
  • amistre64
but yet i cant seem to get how to turn parametric equations into cartesian equation lol
anonymous
  • anonymous
I think I typed the right answer when I typed \[x^2+y^2=3y\] didn't I?
amistre64
  • amistre64
iam; that was a good start yes :), just needed to turn it into the circle equation
anonymous
  • anonymous
I am not getting you, if we just write it as \[x^2+y^2-3 y = 0 \], then it represents a circle doesn't it?
amistre64
  • amistre64
been wondering how to make money doing this.... got $60 left in the bank.... :/
amistre64
  • amistre64
that form is not the standard for a circle equation.... just needs dressed up a bit ;)
anonymous
  • anonymous
Huh? What do you mean?
anonymous
  • anonymous
"been wondering how to make money doing this.... got $60 left in the bank.... :/"??
amistre64
  • amistre64
just broke, thats all :)
anonymous
  • anonymous
I know some methods, but those are not similar to these
amistre64
  • amistre64
:) I saw something today about a cat.... but it looked used up...
amistre64
  • amistre64
id be great if I could get a job tutoring at the college..
anonymous
  • anonymous
I can help you if you want to tutor people online by getting paid at the same time
amistre64
  • amistre64
thatd be good if it works out :)
amistre64
  • amistre64
i got aloooottt of time, and aloooott of math in me lol
anonymous
  • anonymous
Are you serious about it?
anonymous
  • anonymous
I mean do you really want to do it?
amistre64
  • amistre64
yeah, this experience here has taught me that I am actually good at it :)
anonymous
  • anonymous
You are really good
anonymous
  • anonymous
So if you are interested, I can help you start, but you got to email me when you wish to begin
amistre64
  • amistre64
tony031172@gmail.com is my junkmmail catcher, I check it still, but it keeps my good email from the clutter...
anonymous
  • anonymous
I have your email address, did you forget?
amistre64
  • amistre64
lol.... just wasnt sure ;)
anonymous
  • anonymous
I have your real email address a*******@*****.com
amistre64
  • amistre64
:) ill hit you up tomorrow, my borrowed time i up here at the library :/
anonymous
  • anonymous
Sure, I hope I am disturbing all of them who are sitting here
anonymous
  • anonymous
Bye
amistre64
  • amistre64
Ciao :)

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