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Math Help - Stuck on Related Rate question

  1. #1
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    Stuck on Related Rate question

    At noon, ship A passes the point P that is 8 km west of a harbour, heading due south at 10 km/h. At 9 a.m., ship B had left the harbour, sailing due south at 6 km/h.




    Give an expression for dA, the distance between ship A and the point P, in terms of t, the number of hours after noon.






    I got dA = 10t km




    Give an expression for dB, the distance between ship B and the harbour, in terms of t, the number of hours after noon.
    I got dB= 6(t+3) km
    Give an expression for D, the distance between the two ships, in terms of t, the number of hours after noon.

    I got D = sqrt(136t^2+216t+324)

    by using Pythagoras D^2 = (10t)^2(6t+18)^2

    Somehow this is not coming out correctly? Any suggestions? Can it be simplified further?

    Any help would be wonderful, cheers
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  2. #2
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    Quote Originally Posted by youmuggles View Post
    At noon, ship A passes the point P that is 8 km west of a harbour, heading due south at 10 km/h. At 9 a.m., ship B had left the harbour, sailing due south at 6 km/h.




    Give an expression for dA, the distance between ship A and the point P, in terms of t, the number of hours after noon.






    I got dA = 10t km




    Give an expression for dB, the distance between ship B and the harbour, in terms of t, the number of hours after noon.
    I got dB= 6(t+3) km
    Give an expression for D, the distance between the two ships, in terms of t, the number of hours after noon.

    I got D = sqrt(136t^2+216t+324)

    by using Pythagoras D^2 = (10t)^2(6t+18)^2

    Somehow this is not coming out correctly? Any suggestions? Can it be simplified further?

    Any help would be wonderful, cheers
    east-west distance between the two ships at any time t = 8 km

    north-south distance between the two ships at any time t in km ...

    |10t - 6(t+3)| = |4t - 18|

    D = \sqrt{8^2 + (4t - 18)^2}
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