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anonymous

  • one year ago

medal award!

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  1. anonymous
    • one year ago
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  2. zepdrix
    • one year ago
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    Make sure you remember how to read these brackets. Square bracket means we include the point, round bracket is exclusion.\[\large\rm t\in[0,2\pi)\qquad\to\qquad 0\le t\lt 2\pi\]

  3. zepdrix
    • one year ago
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    t=2pi isn't in our interval, ya?

  4. anonymous
    • one year ago
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    it is totally in our interval :D

  5. zepdrix
    • one year ago
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    Nooo :O

  6. zepdrix
    • one year ago
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    Round bracket on the 2pi

  7. anonymous
    • one year ago
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    Darn! I keep forgetting. It would be like this (2pi)

  8. zepdrix
    • one year ago
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    \[\huge\rm t\in[0,2\pi\color{red}{)}\qquad\to\qquad 0\le t\color{red}{\lt} 2\pi\]No 2pi allowed :OOO

  9. anonymous
    • one year ago
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    so whenever they are asking the distance what exactly do they want?

  10. zepdrix
    • one year ago
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    Well... you're right. If we spun 2pi, we'd get to the point they labeled. So we need a value that is `co-terminal with 2pi` and inside our interval.

  11. zepdrix
    • one year ago
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    If we spin around 2pi, we get to 4pi. That's way way outside of our interval though. Can we go backwards and get to a value in our interval maybe?

  12. anonymous
    • one year ago
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    Don't know if this is right but maybe pi/2?

  13. zepdrix
    • one year ago
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    No. You need to spin a full rotation to find co-terminal angles. full rotation = 2pi, ya?

  14. zepdrix
    • one year ago
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    So if we're at 2pi, and we spin backwards an entire rotation, where do we land?

  15. anonymous
    • one year ago
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    -2pi?

  16. zepdrix
    • one year ago
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    Hmm that certainly is co-terminal with 2pi :) But from 2pi, you spun around the circle TWICE to get to -2pi. Too far. That's outside of our interval.

  17. zepdrix
    • one year ago
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    You're gonna be so mad when you figure out how simple this was LOL

  18. anonymous
    • one year ago
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    pi/6?

  19. zepdrix
    • one year ago
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    pi/6 is not co-terminal to 2pi :c Should I just spill the beans? :3

  20. anonymous
    • one year ago
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    yes :D I honestly don't know what it is

  21. zepdrix
    • one year ago
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    We're starting at (1,0) and ending at (1,0). In order for this to happen, we have to spin multiples of 2pi. If we're between 0 and 2pi (excluding the value 2pi), then we have to spin how many 2pi's around to land in the same spot? 0 of them. t=0

  22. anonymous
    • one year ago
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    o wow. XD

  23. anonymous
    • one year ago
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    it's really that simple :)

  24. zepdrix
    • one year ago
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    Ya, kind of a trick question :c The point to point tells us that we have to move 2pi's. But the interval tells us that we're not able to. So you can't move at all :p

  25. anonymous
    • one year ago
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    thank you zepdrix :)

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