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anonymous
 one year ago
medal award!
anonymous
 one year ago
medal award!

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zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Make sure you remember how to read these brackets. Square bracket means we include the point, round bracket is exclusion.\[\large\rm t\in[0,2\pi)\qquad\to\qquad 0\le t\lt 2\pi\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1t=2pi isn't in our interval, ya?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it is totally in our interval :D

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Round bracket on the 2pi

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Darn! I keep forgetting. It would be like this (2pi)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\huge\rm t\in[0,2\pi\color{red}{)}\qquad\to\qquad 0\le t\color{red}{\lt} 2\pi\]No 2pi allowed :OOO

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so whenever they are asking the distance what exactly do they want?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Well... you're right. If we spun 2pi, we'd get to the point they labeled. So we need a value that is `coterminal with 2pi` and inside our interval.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1If 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Don't know if this is right but maybe pi/2?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1No. You need to spin a full rotation to find coterminal angles. full rotation = 2pi, ya?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So if we're at 2pi, and we spin backwards an entire rotation, where do we land?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Hmm that certainly is coterminal with 2pi :) But from 2pi, you spun around the circle TWICE to get to 2pi. Too far. That's outside of our interval.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1You're gonna be so mad when you figure out how simple this was LOL

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1pi/6 is not coterminal to 2pi :c Should I just spill the beans? :3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes :D I honestly don't know what it is

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1We'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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it's really that simple :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Ya, 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you zepdrix :)
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