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Math Help - Related Rates

  1. #1
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    Exclamation Related Rates

    Can someone please help me with the questions in the attachment. Thanks
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  2. #2
    Grand Panjandrum
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    Clearly we would be doing you no favours in just solving them
    all for you, so I will do one example and if you have specific
    problems with the others you can post the question, what you have
    done and what the difficulty is.

    <usual warning about checking the working yourself>

    <another sugestion-it might be worthwhile converting the units to
    metres (m), kilograms (kg) and seconds (s)>


    2.An airplane of mass 10^5 kg flies in a straight line while ice builds up
    on the leading edges of its wings at a constant rate of 30kg/h.

    a) At what rate is the planesís momentum changing if it
    is flying at a constant rate of 800 km/h?
    (Momentum (p) is the product of an objectís mass and
    velocity: p = mv)

    b) At what rate is the momentum of the plane changing
    at t = 1h if at that instant its velocity is 750 km/h and
    is increasing at a rate of 20 km/h/h?
    The planes momentum is:

    p=m.v,

    so the rate of change of momentum (using the product rule) is:

    \frac{dp}{dt}=\frac{dm}{dt}.v + m. \frac{dv}{dt}

    Part (a) we have the speed is constant:

    \frac{dv}{dt}=0,

    so:

    \frac{dp}{dt}= \frac{dm}{dt}.v.

    We are told: \frac{dm}{dt}=30\  \mbox{kg/hr}, and  v=800\  \mbox{km/hr},
    plugging these in (and leaving the units as they are):

    \frac{dp}{dt}= 30.800=24000\ \mbox{km.kg.hr^{-2}}

    Part (b) at 1hr we have the mass of the aircraft and ice is 100,030 \mbox{kg}, the speed is 750 \mbox{km/hr},
    \frac{dm}{dt}=30\  \mbox{kg/hr}, and  \frac{dv}{dt}=20 \mbox{km/hr^2} , so:

    \frac{dp}{dt}= 30.750+100030.20=2023100\ \mbox{km.kg.hr^{-2}}

    RonL
    Last edited by CaptainBlack; December 18th 2005 at 02:20 PM.
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