anonymous
  • anonymous
Convert the polar equation r=4sinθ to a Cartesian equation. Answer choices: x^2 + y^2 = 4y x^2 + y^2 = 4x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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jtvatsim
  • jtvatsim
Hello again! We're you able to make sense of that other document that had been given? :)
anonymous
  • anonymous
Kind of, but I was still a little confused.
jtvatsim
  • jtvatsim
OK. Well let me give you a summary of key things you need to know to solve these sort of problems.

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anonymous
  • anonymous
Ok!
jtvatsim
  • jtvatsim
\[r = \sqrt{x^2+y^2}\] \[\cos(\theta) = \frac{x}{r}\] \[\sin(\theta) = \frac{y}{r}\]
jtvatsim
  • jtvatsim
Those are the three most important facts you need to know. The question is... how to use them?
anonymous
  • anonymous
ok
jtvatsim
  • jtvatsim
It's a good idea to convert the trig functions first. They are kind of annoying anyways. Let's start by changing the sin into something else.
jtvatsim
  • jtvatsim
Let's write down the change as: \[r = 4 \sin \theta\] becomes \[r = 4 \cdot (\frac{y}{r})\] Does that make sense so far?
anonymous
  • anonymous
yes it does
jtvatsim
  • jtvatsim
Perfect! Now, if you wanted you could switch out the r's now. You would get a bit of a mess though. I'm going to be a little tricky and multiply both sides by r, to get the r's by themself.
jtvatsim
  • jtvatsim
\[r = 4 \cdot (\frac{y}{r})\] becomes \[r^2 = 4y\]
anonymous
  • anonymous
ok
jtvatsim
  • jtvatsim
Now this is a wonderful equation to have! Remember that r is a square root of x^2 and y^2. But since we now have r^2 the square root goes away! Like this...
jtvatsim
  • jtvatsim
\[r^2 = 4y\] becomes \[(\sqrt{x^2+y^2})^2 = 4y\] which is easy \[x^2 + y^2 = 4y\]
anonymous
  • anonymous
Thank you so much it makes a lot more sense now! :)
jtvatsim
  • jtvatsim
Glad to help! Just remember, convert trig first, then see if you can make an r^2 (they are friendly), then convert r and you are done! :)
anonymous
  • anonymous
Thanks!

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