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Two roads intersect at right angles. At a certain moment, one bicyclist is 8 miles due NORTH of the intersection traveling TOWARDS the intersection at a rate of 16 miles/hour. A second bicyclist is 10 miles due WEST of the intersection traveling AWAY from the intersection at 12 miles/hour.
(a) How fast is distance between the two bicyclists changing at this moment?

  • AP CALC -

    make a diagram, let the distance the north biker is from the intersection be y miles, let the distance the west biker is from the intersection be x miles
    Let the distance between them be D
    D^2 = x^2 + y^2
    2D dD/dt = 2x dx/dt + 2y dy/dt

    dx/dt = 12 mph
    dy/dt = -16 mph (y is decreasing)
    find dD/dt when x=10 and y=8
    then D^2 = 100+64
    D = √164

    dD/dt =(2(10)(12) + 2(8)(-16))/(2√164) = .....

    you finish the arithmetic

  • AP CALC -

    My final answer was -8/square root of 164. Is that correct?

  • AP CALC -

    I believe the correct answer is 0.635 but i got 0.625.

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