Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Calculus - Related rates problem?

Two boats leave the same port at the same time with one boat travelling North at 15 knots/hour and the other boat travelling west at 12 knots/hour. How fast is this distance between the two boats cahnging after 2 hours?

Thanks

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  • Anonymous
    1 decade ago
    Favorite Answer

    First of all, the units that you have "knots/hour" is not an accurate unit. "Knots" is already a unit of speed and this is equivalent to 1.15 miles/hour.

    What you have here is a right triangle configuration, i.e.,

    C^2 = A^2 + B^2

    where

    C = distance between the boats at any time

    A = distance travelled by the north-bound boat

    B = distance travelled by the west-bound boat

    Differentiating the above,

    2C(dC/dt) = 2A(dA/dt) + 2B(dB/dt)

    Modifying the above,

    C(dC/dt) = A(dA/dt) + B(dB/dt)

    After two hours,

    A = 15*2 = 30

    B = 12*2 = 24

    and

    C = sqrt(30^2 + 24^2)

    C = 38.42

    and since dA/dt = 15 and dB/dt = 12, then

    38.42(dC/dt) = 30(15) + 24(12)

    Solving for "dC/dt"

    dC/dt = 19.21 knots

    Hope this helps.

  • 1 decade ago

    Boats A goes north such that dA/dt = 15

    Boats B goes west such that dB/dt = 12

    After 2 hours

    A has travelled 30 knots and boats B has travelled 24 knots

    the distance "r" between them is

    r² = 30² + 24²

    r² = 1476

    r = sqrt1476

    at any point in time

    r² = A² + B²

    so

    2r dr/dt = 2A dA/dt + 2B db/dt

    r dr/dt = A dA/dt + B db/dt

    sqrt1476 dr/dt = (30)(15) + (24)(12)

    sqrt1476 dr/dt = 738

    dr/dt = 738/sqrt1476

    dr/dt = 19.21knots/hour (2dp)

  • CDA
    Lv 6
    1 decade ago

    Simple pythagoras, relative speed is sqrt(15^2 + 12^2)

    knots. This is sqrt(369) knots or 19.21 knots. The boats are receding from each other at 19.21 knots. The "after 2 hours" is irrelevant.

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