Maths

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Two pistons A & B move backwards and forwards in a cyclinder. Distance x cm of the right hand end of piston A from point 0 at time t seconds is x=3sin(2t)+3 and the distance y cm of the left hand end of piston B by formula y=2sin(3t-pi/4)+8.5
The pistons are set in motion at t=0
How do I work out the minimun distance between the pistons and the time it occurs? I am really lost on this, can you please help.
Thanks

  • Maths -

    Looks like a rather nasty question
    I see the distance (d) as
    d = 3sin(2t)+3 + 2sin(3t-pi/4)+8.5
    = 3sin(2t)+ 2sin(3t-pi/4)+ 11.5

    I first ran this through a primitive graphing program I have, and it showed the above to have a period of 2pi, and a minimum value roughly between t = 1.5 and t = 1.7

    to get exact answers is a messy Calculus problem
    dd/dt = 6cos(2t) - 6cos(3t-pi/4)
    = 0 for a max/min of d

    So we have to solve
    6cos(2t) - 6cos(3t-pi/4) = 0 or
    cos(2t) - cos(3t-pi/4) = 0
    the trouble is that we need to have our trig functions contain the same period.
    we know cos 2t = 2cos^2(t) - 1
    and cos(3t - pi/4) = cos3tcospi/4 + sin3tsinpi/4
    = (1/√2)(cos3t + sin3t)

    now there are expansion formulas for
    cos 3x and sin 3x in terms of cos x, but I have a feeling that the question couldn't possibly be that messy.

    Are you using graphig calculators?
    What level is this?

  • Math experts needed here. -

    Can somebody see through the above mess?
    Am I on the wrong track?

  • Maths -

    I see the distance between the pistons as y-x. It is so hard to see without a diagram.

    d= 2sin(3t-pi/4)+8.5 -3sin(2t)+3
    Then proceed as in Dr Reiny's solution.

    In past days, we would have put this on the analogue computer and ran a graph for few seconds.

  • Maths -

    I think I have made some headway.

    I started from Reiny's result
    cos(2t) - cos(3t-pi/4) = 0
    which I validated as correct.

    By equating the two cosines, I conclude that:
    2t + 2kπ = 3t-π/4
    after adding 2kπ for generality, which gives
    t = π/4 + 2kπ
    Numerical solutions are: ... -5.5, 0.78, 7.07, ...
    Since cosine is an even function, one of the arguments of cosine could be negative, therefore
    -2t + 2kπ = 3t - %pi;/4
    which gives
    t=(1/5)(π/4+2kπ)
    Numerical solutions are: ... -6.12, -4.86, -3.61, -1.1, 0.157, 1.41, 2.67, 3.92, 5.18, 6.44 ...

    These results can be verified in the graphed links below:
    for the function:
    http://img62.imageshack.us/i/1255515491func.png/
    for the derivative:
    http://img379.imageshack.us/i/1255515491deriv.png/

  • To MathMate -

    Of course!
    How simple was that step, eh?

    And here I was about to enter the mathematical hinterland.

    Thank you!

  • To Reiny -

    That's what happens when you know a lot!

  • Maths -

    Nice work.

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