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At noon, ship A is 100 kilometers due east of ship B. Ship A is sailing west at 12 k/h and ship B is sailing S10degrees west at 10 k/h. At what time will the ships be nearest each other and what will this distance be? (hint: You do not have a right triangle, unfortunately)

  • calculus -

    put B at the origin and A on the positive x-axis
    After t hours, draw the paths of ship A and ship B and let A's position be P (to the left of B) and the position of ship B as Q
    I see triangle PQB, where angle B=10°
    and QB = 10t miles, PB = 12t - 100 miles
    Let the distance between them be x

    by the cosine law:
    x^2 = (10t)^2 + (12t-100)^2 - 2(10t)(12t-100)cos 10°
    = 244t^2-2400t + 10000 - (240cos10°)t^2 + 2000cos10° t

    2x dx/dt = 488t - 2400 - 480cos10° t + 2000cos10°
    = 0 for a minimum of x

    t( 488 - 480cos10°) = 2400 - 2000cos10°
    t = 28.14

    check my arithmetic, I should have written this out first, I just drew the diagram.

    So it would be 28.14 hrs after noon (into the next day)

  • calculus -

    thank you, this looks correct. But, whouldnt angle B be 80 degrees, not 10 degrees. Because the s10w means 10 degrees to the west side of south.

  • calculus -

    Yes, I believe you are interpreting
    S 10° W in the proper way, I did W10S
    so change the values of cos10 to cos80
    The steps would remain the same.

  • calculus -

    ok, thank you so much for your help. one more question, what happens to the x? does it just drop out when you set it equal to zero?

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