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Thread: Point P(t) on unit circle

  1. #1
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    Point P(t) on unit circle

    Hi!

    Show the approximate location on the unit circle of $\displaystyle P(t)$ for the given value of $\displaystyle t$.

    1. $\displaystyle t = -\frac{37\pi}{6} $

    2. $\displaystyle t = -\frac{4\pi}{3}$

    3. $\displaystyle t = -\frac{7\pi}{4}$

    What's the 'process' behind showing where the points are?

    I have tried adding/subtracting $\displaystyle 2\pi$ to obtain a point on the unit circle so I can have a general idea. It wasn't concrete.
    You can use #1. as an example and I'll do the rest and post.
    Last edited by Unreal; May 5th 2013 at 09:22 PM.
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  2. #2
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    Re: Point P(t) on unit circle

    For the first one, I would write:

    $\displaystyle -\frac{37\pi}{6}=-\frac{36\pi+\pi}{6}=-6\pi-\frac{\pi}{6}=3(-2\pi)-\frac{\pi}{6}$

    Now you can see this is 3 complete revolutions in the clockwise (negative) direction plus an additional negative amount...
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  3. #3
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    Re: Point P(t) on unit circle

    Quote Originally Posted by MarkFL View Post
    For the first one, I would write:

    $\displaystyle -\frac{37\pi}{6}=-\frac{36\pi+\pi}{6}=-6\pi-\frac{\pi}{6}=3(-2\pi)-\frac{\pi}{6}$

    Now you can see this is 3 complete revolutions in the clockwise (negative) direction plus an additional negative amount...
    But how do you know where to start, in order to go the 3 revolutions?

    You can't start at $\displaystyle -2\pi$ or $\displaystyle -\frac\pi6$.
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Point P(t) on unit circle

    You start at $\displaystyle t=0$ which is the point (1,0)...
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  5. #5
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    Re: Point P(t) on unit circle

    $\displaystyle t = -\frac{4\pi}{3}$

    $\displaystyle -\frac{4\pi - \pi}{3}$

    $\displaystyle -\frac13\left(4\pi - \pi\right)$

    $\displaystyle 4\left(-\frac\pi3 - \frac{\pi}{3}\right)$

    So, that is 4 clockwise revolutions from $\displaystyle \frac\pi3$, which lands us back on $\displaystyle \frac\pi3$, and from there we move $\displaystyle \frac\pi3$ counterclockwise to get $\displaystyle \frac{2\pi}{3}$ on the unit circle which is the approximate location of $\displaystyle -\frac{4\pi}{3}$

    ASIDE:
    Does this mean $\displaystyle -\frac{36\pi+\pi}{6}= -\left(\frac{36\pi+\pi}{6}\right)$?
    I was unsure because it led to: $\displaystyle \-6\pi-\frac{\pi}{6}$. It appears you factored.
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Point P(t) on unit circle

    Quote Originally Posted by Unreal View Post
    $\displaystyle t = -\frac{4\pi}{3}$

    $\displaystyle -\frac{4\pi - \pi}{3}$

    $\displaystyle -\frac13\left(4\pi - \pi\right)$

    $\displaystyle 4\left(-\frac\pi3 - \frac{\pi}{3}\right)$

    So, that is 4 clockwise revolutions from $\displaystyle \frac\pi3$, which lands us back on $\displaystyle \frac\pi3$, and from there we move $\displaystyle \frac\pi3$ counterclockwise to get $\displaystyle \frac{2\pi}{3}$ on the unit circle which is the approximate location of $\displaystyle -\frac{4\pi}{3}$
    No, this is incorrect...$\displaystyle -\frac{4\pi}{3}$ is less than one complete revolution, so you simply move $\displaystyle \frac{4\pi}{3}$ radians in the negative (clockwise) direction. Since:

    $\displaystyle \frac{\frac{4\pi}{3}}{2\pi}=\frac{2}{3}$, we know we need to make 2/3 of a complete revolution.

    Quote Originally Posted by Unreal View Post
    ASIDE:
    Does this mean $\displaystyle -\frac{36\pi+\pi}{6}= -\left(\frac{36\pi+\pi}{6}\right)$?
    I was unsure because it led to: $\displaystyle -6\pi-\frac{\pi}{6}$. It appears you factored.
    Yes, both are the same...$\displaystyle -\frac{a+b}{c}=-\left(\frac{a+b}{c} \right)$.
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  7. #7
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    Re: Point P(t) on unit circle

    Quote Originally Posted by MarkFL
    Since:

    $\displaystyle \frac{\frac{4\pi}{3}}{2\pi}=\frac{2}{3}$, we know we need to make 2/3 of a complete revolution.
    How do you know where 2/3 lands?
    It's a bit of an awkward number.

    $\displaystyle t = -\frac{7\pi}{4}$

    $\displaystyle \frac{\frac{7\pi}{4}}{2\pi} = \frac78$

    So, we are moving 7/8 of a revolution clockwise, which lands on $\displaystyle \frac\pi4$.
    The circle is divided into 4 parts, so in this case starting from (1,0) and counting clockwise, the third part is 75%. I then used this to approximate where 7/8 is the denominator is 4, so $\displaystyle \frac\pi4$


    Is $\displaystyle t = 0$ always the starting point?
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  8. #8
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    Re: Point P(t) on unit circle

    Quote Originally Posted by Unreal View Post
    How do you know where 2/3 lands?
    It's a bit of an awkward number.
    It may be easier to just think of it in its original form...$\displaystyle -\frac{4\pi}{3}=-\pi-\frac{\pi}{3}$

    So go half way around in the negative direction, then continue a third of another half...

    Quote Originally Posted by Unreal View Post
    $\displaystyle t = -\frac{7\pi}{4}$

    $\displaystyle \frac{\frac{7\pi}{4}}{2\pi} = \frac78$

    So, we are moving 7/8 of a revolution clockwise, which lands on $\displaystyle \frac\pi4$.
    The circle is divided into 4 parts, so in this case starting from (1,0) and counting clockwise, the third part is 75%. I then used this to approximate where 7/8 is the denominator is 4, so $\displaystyle \frac{\pi}{4}$
    You can always add an integral multiple of $\displaystyle 2\pi$ to get to the same place...:

    $\displaystyle -\frac{7\pi}{4}+2\pi=\frac{\pi}{4}$

    Quote Originally Posted by Unreal View Post
    Is $\displaystyle t = 0$ always the starting point?
    Yes, that's where we begin when we measure an angle on the unit circle.
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  9. #9
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    Re: Point P(t) on unit circle

    With due regards I have the following suggestion:
    Whatsoever is

    What so ever is the value of the angle write it in proper fraction e.g.,
    23/4 π=(5+ 3/4 )π=4π+ π+ 3/4 π= π+ 3/4 π [ because even multiples ( positive or negative ) of π we reach OX the starting ray.
    Now suppose it was negative we write it as sum of positive multiples of π and then subtract the angle. For example:
    - 23/4 π=(-6+ 1/4 )π=-6π+ 1/4 π= 1/4 π [ because even multiples ( positive or negative ) of π we reach OX the starting ray.
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