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Calc

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At noon, ship A is 30 nautical miles due west of ship B. Ship A is sailing west at 16 knots and ship B is sailing north at 15 knots. How fast (in knots) is the distance between the ships changing at 7 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)

  • Calc -

    First draw a right triangle

    Let side y be the distance by B in the time after noon this will be 15*7
    Let side x be the distance traveled in total by A this will be 30+ (16*7)
    so you have y=105 and x=142

    Ship B is sailing at 15 knots, this is the rate it is changing, so it is also written as dy/dt=15

    Ship A is sailing at 16 knots so this is written as dx/dt=16

    you need the distance formula, write is as
    h^2=x^2 + y^2
    h is the hypotenuse of your triangle, or the distance between the two points

    At this point there are two different approaches


    APPROACH 1

    h^2=x^2 + y^2 is the same as
    h=(x^2+ y^2)^(1/2)

    Find dh/dt

    dh/dt=(1/2)(x^2+ y^2)^(-1/2) (2x dx/dt + 2y dy/dt)
    dh/dt=(2x dx/dt + 2y dy/dt)/ [2(x^2+ y^2)^(1/2)]

    plug in:
    y=105
    x=142
    dx/dt=16
    dy/dt=15

    dh/dt=___

    ------------------------------------------------------
    APPROACH 2

    h^2=x^2 + y^2

    plug in your x and y and solve for h
    h=(x^2+ y^2)^(1/2)
    h=[(142)^2 + (105)^2]^(1/2)
    h=(31189)^(1/2)

    using h^2=x^2 + y^2 find dh/dt
    2h dh/dt= 2x dx/dt + 2y dy/dt

    solve for dh/dt by plugging in:
    h=(31189)^(1/2)
    y=105
    x=142
    dx/dt=16
    dy/dt=15

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