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anonymous

  • one year ago

Convert the polar equation r=4sinθ to a Cartesian equation. Answer choices: x^2 + y^2 = 4y x^2 + y^2 = 4x

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  1. jtvatsim
    • one year ago
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    Hello again! We're you able to make sense of that other document that had been given? :)

  2. anonymous
    • one year ago
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    Kind of, but I was still a little confused.

  3. jtvatsim
    • one year ago
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    OK. Well let me give you a summary of key things you need to know to solve these sort of problems.

  4. anonymous
    • one year ago
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    Ok!

  5. jtvatsim
    • one year ago
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    \[r = \sqrt{x^2+y^2}\] \[\cos(\theta) = \frac{x}{r}\] \[\sin(\theta) = \frac{y}{r}\]

  6. jtvatsim
    • one year ago
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    Those are the three most important facts you need to know. The question is... how to use them?

  7. anonymous
    • one year ago
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    ok

  8. jtvatsim
    • one year ago
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    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.

  9. jtvatsim
    • one year ago
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    Let's write down the change as: \[r = 4 \sin \theta\] becomes \[r = 4 \cdot (\frac{y}{r})\] Does that make sense so far?

  10. anonymous
    • one year ago
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    yes it does

  11. jtvatsim
    • one year ago
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    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.

  12. jtvatsim
    • one year ago
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    \[r = 4 \cdot (\frac{y}{r})\] becomes \[r^2 = 4y\]

  13. anonymous
    • one year ago
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    ok

  14. jtvatsim
    • one year ago
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    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...

  15. jtvatsim
    • one year ago
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    \[r^2 = 4y\] becomes \[(\sqrt{x^2+y^2})^2 = 4y\] which is easy \[x^2 + y^2 = 4y\]

  16. anonymous
    • one year ago
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    Thank you so much it makes a lot more sense now! :)

  17. jtvatsim
    • one year ago
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    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! :)

  18. anonymous
    • one year ago
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    Thanks!

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