Calculus

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Two people start from rest at the same point. One walks east at 3 mi/h and the other walks northeast at 2 mi/h. How fast is the distance between the people changingg after 15 minutes?

  • Calculus -

    one way, using the law of cosines. Draw a triangle. The angle between the two journeys is 45°. At time t, the distances traveled are 3t and 2t. So, the distance x is

    x^2 = (3t)^2 + (2t)^2 - 2(3t)(2t)*1/√2
    x^2 = 9t^2 + 4t^2 - 6√2 t^2
    x^2 = (13-6√2)t^2

    At t=15, x = 15√(13-6√2)

    2x dx/dt = 2(13-6√2)t
    2*15√(13-6√2) dx/dt = 30(13-6√2)
    dx/dt = √(13-6√2)

    using vectors,
    x = (3t,0) - (√2t,√2t) = ((3-√2)t,√2t)
    dx/dt = (3√2,-√2)
    |dx/dt|^2 = (3-√2)^2 + √2^2 = 9-6√2+2+2 = 13-6√2
    |dx|dt| = √(13-6√2)

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