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anonymous

  • one year ago

The polar coordinates of a point are given. Find the rectangular coordinates of each point. (5,(π / 4)) (-2,(π / 6))

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  1. johnweldon1993
    • one year ago
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    Or instead of going the long way backwards...we can also know that \[\large x = rcos(\theta)\] and \[\large y = rsin(\theta)\]

  2. johnweldon1993
    • one year ago
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    So for example with your first point up there \[\large x = 5cos(\pi/4)\] \[\large y = 5sin(\pi/4)\]

  3. anonymous
    • one year ago
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    And I can do the same for the second equation @johnweldon1993

  4. johnweldon1993
    • one year ago
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    Correct indeed :)

  5. anonymous
    • one year ago
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    so \[x=-2(\cos(\pi/6)\]

  6. johnweldon1993
    • one year ago
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    I just deleted the first post I made so as not to confuse you...why go a long way when there are shortcuts! lol And yes that is correct for your x-coordinate!

  7. anonymous
    • one year ago
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    Thank you so much @johnweldon1993 !

  8. johnweldon1993
    • one year ago
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    Anytime!

  9. anonymous
    • one year ago
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    @Johnweldon1993 is it weird that when I go to convert the first they come out to be the same point?

  10. johnweldon1993
    • one year ago
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    Not weird at all :P Seeing as how \(\large sin(45) = cos(45)\) :)

  11. anonymous
    • one year ago
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    Ah, thank you! All this polar stuff is just alsfjasflkajlfa! @johnweldon1993

  12. johnweldon1993
    • one year ago
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    Lol it will come much easier dont worry :) Actually polar is the easiest of the bunch...wait till you get to cylindrical and spherical THOSE are the best ;) haha

  13. anonymous
    • one year ago
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    So I'm off to go dig my shallow grave.

  14. johnweldon1993
    • one year ago
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    Lol well if you are in DE right now, you wont deal with them any if at all...Calc 3 is where that comes into play :) So nope, stay above ground...for now :P lol no try and stay permanently lol :D

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