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    "A particle moves in a horizontal plane and its position vector at time t is relative to a fixed origin O is given by:

    r = (2 sin t)i + (cos t)j

    Find the values of t in the range "0 <= t <= pi" when the speed of the particle is a maximum.

    ----

    Stuck on this... any help and a walkthrough on what to do?
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    (Original post by 2014_GCSE)
    "A particle moves in a horizontal plane and its position vector at time t is relative to a fixed origin O is given by:

    r = (2 sin t)i + (cos t)j

    Find the values of t in the range "0 <= t <= pi" when the speed of the particle is a maximum.

    ----

    Stuck on this... any help and a walkthrough on what to do?
    We're not allowed to post full solutions but I'll say this: remember that differentiating the position vector gives a velocity vector and differentiating a velocity vector gives an acceleration vector. Also remember that acceleration is zero when the particle is at its maximum speed.
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    (Original post by Hydeman)
    We're not allowed to post full solutions but I'll say this: remember that differentiating the position vector gives a velocity vector and differentiating a velocity vector gives an acceleration vector. Also remember that that acceleration is zero when the particle is at its maximum speed.
    Haha, this is as far as I got!
    But okay, that rule seems fair.

    Here's what I have so far:

    So I differentiated twice and got

    a = (-2 sin t)i + (-cos t)j

    And I'm not quite sure how to set this equal to 0 exactly.
    Do I find the acceleration magnitude and put THAT equal to 0?
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    (Original post by 2014_GCSE)
    Haha, this is as far as I got!
    But okay, that rule seems fair.

    Here's what I have so far:

    So I differentiated twice and got

    a = (-2 sin t)i + (-cos t)j

    And I'm not quite sure how to set this equal to 0 exactly.
    Do I find the acceleration magnitude and put THAT equal to 0?
    Huh. I'm stuck on that too. xD Well, technically the magnitude of the acceleration is zero at the time that you want to work out but I don't understand how you could get further than:

    a = [(-2 sin t)2 + (-cos t)2]0.5 = 0
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    (Original post by Hydeman)
    Huh. I'm stuck on that too. xD Well, technically the magnitude of the acceleration is zero at the time that you want to work out but I don't understand how you could get further than:

    a = [(-2 sin t)2 + (-cos t)2]0.5 = 0
    Expand to: ROOT: 4 sin^2 (t) + cos^2 (t)

    Then use a trig. identity to get ROOT: 3sin^2(t) + 1= 0

    Then you have sin^2(t) = -1/3

    Which has no solutions...
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    (Original post by 2014_GCSE)
    Expand to: ROOT: 4 sin^2 (t) + cos^2 (t)

    Then use a trig. identity to get ROOT: 3sin^2(t) + 1= 0

    Then you have sin^2(t) = -1/3

    Which has no solutions...
    Is this in a book? If so, do you have answers in the back?
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    (Original post by Hydeman)
    Is this in a book? If so, do you have answers in the back?
    t = 0

    t = pi
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    (Original post by 2014_GCSE)
    t = 0

    t = pi
    Huh. That's very strange... :/

    I'll tag a maths teacher I know of and see if she can help you. Sorry I couldn't be of much help.
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    (Original post by 2014_GCSE)
    "A particle moves in a horizontal plane and its position vector at time t is relative to a fixed origin O is given by:

    r = (2 sin t)i + (cos t)j

    Find the values of t in the range "0 <= t <= pi" when the speed of the particle is a maximum.

    ----

    Stuck on this... any help and a walkthrough on what to do?
    Hi - I've just been tagged - I've noted it says 'speed' not velocity.

    I found v^2 and use that trig identity to see when it would be max - try that.
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    (Original post by Muttley79)
    Hi - I've just been tagged - I've noted it says 'speed' not velocity.

    I found v^2 and use that trig identity to see when it would be max - try that.
    I'm not sure why this is? But I did it and I got:

    ROOT: 3 cos^2(t) + 1 = v

    or
    3 cos^2(t) + 1 = v^2
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    (Original post by 2014_GCSE)
    I'm not sure why this is? But I did it and I got:

    ROOT: 3 cos^2(t) + 1 = v

    or
    3 cos^2(t) + 1 = v^2
    When is cos x max? You are nearly there
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    (Original post by Muttley79)
    When is cos x max? You are nearly there
    When cos x = 1

    So you want x to be 0 or 2pi to get max velocity!

    But the answer is t = 0 and t = pi
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    (Original post by 2014_GCSE)
    When cos x = 1

    So you want x to be 0 or 2pi to get max velocity!

    But the answer is t = 0 and t = pi
    The range only goes up to pi in the question.
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    (Original post by Muttley79)
    The range only goes up to pi in the question.
    Yeah but if the answers I got were t = 0 and t = 2pi

    And the range is 0 to pi, then surely the only answer should be t = 0? Where does the t = pi answer come from?
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    (Original post by 2014_GCSE)
    Yeah but if the answers I got were t = 0 and t = 2pi

    And the range is 0 to pi, then surely the only answer should be t = 0? Where does the t = pi answer come from?
    Can you double check the question is copied correctly? I'm busy prepping lessons for tomorrow at the moment but will come back later and check this thread.
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    (Original post by Muttley79)
    Can you double check the question is copied correctly? I'm busy prepping lessons for tomorrow at the moment but will come back later and check this thread.
    Yeah it's definitely copied correctly.
    And no problem, I really appreciate your help!
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    (Original post by 2014_GCSE)
    Yeah it's definitely copied correctly.
    And no problem, I really appreciate your help!
    The other answer is pi because cos x = -1 and we are squaring it ...
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    (Original post by Muttley79)
    The other answer is pi because cos x = -1 and we are squaring it ...
    I'm sorry, but I don't understand this?
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    (Original post by 2014_GCSE)
    I'm sorry, but I don't understand this?
    We are maximising 3cos^2 t + 1, yes?

    Max when cos x = 1 or -1 (because of the squared term)
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    (Original post by Muttley79)
    We are maximising 3cos^2 t + 1, yes?

    Max when cos x = 1 or -1 (because of the squared term)
    Ah, I understand! Okay, thanks!
 
 
 
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