anonymous
  • anonymous
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Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
zepdrix
  • zepdrix
Make sure you remember how to read these brackets. Square bracket means we include the point, round bracket is exclusion.\[\large\rm t\in[0,2\pi)\qquad\to\qquad 0\le t\lt 2\pi\]
zepdrix
  • zepdrix
t=2pi isn't in our interval, ya?

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anonymous
  • anonymous
it is totally in our interval :D
zepdrix
  • zepdrix
Nooo :O
zepdrix
  • zepdrix
Round bracket on the 2pi
anonymous
  • anonymous
Darn! I keep forgetting. It would be like this (2pi)
zepdrix
  • zepdrix
\[\huge\rm t\in[0,2\pi\color{red}{)}\qquad\to\qquad 0\le t\color{red}{\lt} 2\pi\]No 2pi allowed :OOO
anonymous
  • anonymous
so whenever they are asking the distance what exactly do they want?
zepdrix
  • zepdrix
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.
zepdrix
  • zepdrix
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?
anonymous
  • anonymous
Don't know if this is right but maybe pi/2?
zepdrix
  • zepdrix
No. You need to spin a full rotation to find co-terminal angles. full rotation = 2pi, ya?
zepdrix
  • zepdrix
So if we're at 2pi, and we spin backwards an entire rotation, where do we land?
anonymous
  • anonymous
-2pi?
zepdrix
  • zepdrix
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.
zepdrix
  • zepdrix
You're gonna be so mad when you figure out how simple this was LOL
anonymous
  • anonymous
pi/6?
zepdrix
  • zepdrix
pi/6 is not co-terminal to 2pi :c Should I just spill the beans? :3
anonymous
  • anonymous
yes :D I honestly don't know what it is
zepdrix
  • zepdrix
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
anonymous
  • anonymous
o wow. XD
anonymous
  • anonymous
it's really that simple :)
zepdrix
  • zepdrix
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
anonymous
  • anonymous
thank you zepdrix :)

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