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    (Original post by TeeEm)
    it happens ...
    well you learned a bit and I have another nice question to add to my books
    Thank you.
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    (Original post by edothero)
    is |SP| + |TP| = 2a a loci property of an ellipse?

    Was never taught that
    not only you ought to know it but it is examinable !!!
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    (Original post by edothero)
    is |SP| + |TP| = 2a a loci property of an ellipse?

    Was never taught that

    Though to be fair it is quite obvious..
    Yep, it's in the FP3 textbook, went over it a few days ago.
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    (Original post by TeeEm)
    not only you ought to know it but it is examinable !!!
    Thank you anyway.
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    well pleasing ... thanks to the people which nominated me and also voted for me
    http://www.thestudentroom.co.uk/show....php?t=3889039

    6 nominations ... two firsts, two seconds and two thirds ....


    Thank you again

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    (Original post by TeeEm)
    well pleasing ... thanks to the people which nominated me and also voted for me
    http://www.thestudentroom.co.uk/show....php?t=3889039

    6 nominations ... two firsts, two seconds and two thirds ....


    Thank you again

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    Calm tf down, its only a forum site
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    (Original post by GeologyMaths)
    Calm tf down, its only a forum site
    you are clearly very jealous ...
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    Someone help? Found this question on Variable Acceleration in 2 dimensions and don't understand how they get values of maximum and minimum acceleration for part b. I managed to get a=12 using the given rearrangement, but don't know how to get the maximum!

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    (Original post by Hassan55)
    Someone help? Found this question on Variable Acceleration in 2 dimensions and don't understand how they get values of maximum and minimum acceleration for part b. I managed to get a=12 using the given rearrangement, but don't know how to get the maximum!

    Posted from TSR Mobile
    Zacken SeanFM16Characters....


    do you mind please
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    (Original post by Hassan55)
    Someone help? Found this question on Variable Acceleration in 2 dimensions and don't understand how they get values of maximum and minimum acceleration for part b. I managed to get a=12 using the given rearrangement, but don't know how to get the maximum!

    Posted from TSR Mobile
    Let the acceleration be a, then acceleration will achieve minimum and maximum values when \frac{da}{dt} = v =  0, so find the values of t for which this is so, investigate whether they are minimum or maximum values by considering the second derivative and then plug it back into the equation to find the min/max acceleration.

    Edit: ignore this, as per SeanFM, recall that \cos^2 x varies between and 1...
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    (Original post by TeeEm)
    Zacken
    Aye
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    (Original post by Zacken)
    Aye
    thank you
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    (Original post by Hassan55)
    Someone help? Found this question on Variable Acceleration in 2 dimensions and don't understand how they get values of maximum and minimum acceleration for part b. I managed to get a=12 using the given rearrangement, but don't know how to get the maximum!

    Posted from TSR Mobile
    Look at the expression and think the smallest and largest value that it can take.
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    (Original post by Zacken)
    Let the acceleration be a, then acceleration will achieve minimum and maximum values when \frac{da}{dt} = v =  0, so find the values of t for which this is so, investigate whether they are minimum or maximum values by considering the second derivative and then plug it back into the equation to find the min/max acceleration.

    Edit: ignore this, as per SeanFM, recall that \cos^2 x varies between and 1...
    Why do you put v=0,can't you put a=0 and instead of using the suggested expression, can't you use dv/da=a=0 and solve for t in both i and j. Is there any other approach,ie with the expression they give, to solve this?
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    (Original post by Hassan55)
    Why do you put v=0,can't you put a=0 and instead of using the suggested expression, can't you use dv/da=a=0 and solve for t in both i and j. Is there any other approach,ie with the expression they give, to solve this?
    I don't know what I was thinking, but da/dt is not v... , themethod you propose would find the maximum/minimum velocity, not acceleration. You could differentitate the position vector they've given you *four* times and set thag equal to 0, but it's a lot of work when you can just look at how cos x varies from the dervived expression!
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    My proposed method seemed to have worked surprisingly, where I set the i and j components of a to 0 and worked out values for t to get 12 and 14. I used the method similar to this mark scheme.. you're right saying a=0 will give max/min velocity, but how come they do this?

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    Also I differentiate 4 times (only had to do it twice more because of part a) and got tan(2t)=-4/3... And by getting t I didn't get 12 or 14 :/

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    (Original post by Hassan55)
    Also I differentiate 4 times (only had to do it twice more because of part a) and got tan(2t)=-4/3... And by getting t I didn't get 12 or 14 :/

    Posted from TSR Mobile
    Oops, did I say 4 times? My mistake, it was really late at night and I was in bed. Differentitate once to get velocity, second time to get acceleration, third time to get jerk and then that =0. So differentiate thrice, not four times.
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    Do you differentiate a naturally as you keep differentiating from r or do you differentiate the expression the give you?

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    I got 32sin(2t)i-24cos(2t)j=0 after differentiating 3 times. Would I then 32sin(2t)=0 and -24cos(2t)j=0 to get values of t in range 0<t<2Pi and sub in to given expression of a to obtain max/min?

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