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Math Help - GCSE A Level Maths Qns

  1. #1
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    GCSE A Level Maths Qns

    This is a qns from GCSE A level.....Appreciate your help !

    In the diagram, O and A are fixed points 1000m apart on horizontal ground. The point B is vertically above A, and represents a balloon which is ascending at a steady rate of 2m/s. The balloon is being observed from a moving point P on the line OA. At time t=0, the balloon is at A and the observer is at O. The observation point P moves towards A with steady speed 6m/s. At time t, the angle APB is θ radians. Show that dθ/dt = 500/(t^2+(500-3t)^2).
    Last edited by shaojie2k; May 21st 2006 at 01:18 AM.
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  2. #2
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    Quote Originally Posted by shaojie2k
    This is a qns from GCSE A level.....Appreciate your help !

    In the diagram, O and A are fixed points 1000m apart on horizontal ground. The point B is vertically above A, and represents a balloon which is ascending at a steady rate of 2m/s. The balloon is being observed from a moving point P on the line OA. At time t=0, the balloon is at A and the observer is at O. The observation point P moves towards A with steady speed 6m/s. At time t, the angle APB is θ radians. Show that dθ/dt = 500/(t^2+(500-3t)^2).
    Where is the diagram?

    But no need. Can imagine.
    The diagram, after t sec, is a right triangle, with these:
    --right angle is at corner A.
    --hypotenuse = PB = unknown.
    --vertical leg = AB = 2t m.
    --horizontal leg = PA = (1000 -6t) m.
    --angle APB = theta

    tan(theta) = AB/PA
    tan(theta) = 2t/(1000 -6t)
    tan(theta) = t/(500 -3t)
    theta = arctan[t/(500 -3t)] ----(1)

    To get dtheta/dt, we differentiate both sides of (1) with respect to time t.

    Remember,
    d[arctan(u)] = [1/(1 +u^2)](du) ---------(i)
    d[u/v] = (v*du -u*dv)/(v^2)----------------(ii)

    So, applying those to (1),
    dtheta/dt = {1 / [1 +(t/(500-3t))^2]}*[((500-3t)(1) -t(-3)) / (500-3t)^2]
    simplify,
    dtheta/dt = {1 / [1 +(t^2 / (500-3t)^2)]}*[(500-3t +3t) / (500-3t)^2]
    dtheta/dt = {1 / [(1*(500-3t)^2 +t^2)/(500-3t)^2]}*[(500) / (500-3t)^2]
    dtheta/dt = {1*(500-3t)^2 / [(500-3t)^2 +t^2)]}*[(500) / (500-3t)^2]
    dtheta/dt = (500) / [(500-3t)^2 +t^2] -------------answer.

    Shown.
    Last edited by ticbol; May 21st 2006 at 07:17 AM.
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  3. #3
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    Thank You!

    Thank You!
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