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A hot-air balloon is 120 ft above the ground when a motorcycle (traveling in a straight line on a horizontal road) passes directly beneath it going 45 mi hr (66 ft s). If the balloon rises vertically at a rate of 12 ft s, what is the rate of change of the distance between the motorcycle and the balloon 10 seconds later?

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  1. h = 120 + 12 t which is 240 at t = 10
    dh/dt = 12
    x = 66 * 10 = 660 at t = 10
    dx/dt = 66
    hypotenuse^2 = z^2 = h^2 + x^2
    at t = 10 = 240^2 + 660^2 = 492200
    so z= 702 at t = 10
    2 z dz/ dt = 2 h dh/dt + 2 x dx/dt
    2*702*dz/dt = 2 * 240 * 12 + 2 * 660 * 66
    dz/dt = ( 2880 + 43560) / 702 = 66.2

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