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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 30 mi/hr ​(44 ft/s​). If the balloon rises vertically at a rate of 15 ft/s​, what is the rate of change of the distance between the motorcycle and the balloon 6 seconds ​later?

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3 answers

  1. as usual, draw a diagram. If the bike is x ft away, then the distance z is
    z^2 = x^2 + 120^2
    So that means that
    z dz/dt = x dx/dt
    at the moment in question, z^2 = (6*44)^2 + 120^2, so z = 290 ft
    Now to find dz/dt,
    290 dz/dt = (6*44) * 15
    dz/dt = 396/29 ≈ 13.66 ft/s

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  2. it was 43.77

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  3. oops. do you see my mistake? I was considering the balloon at a constant height of 120 ft. So, in reality, at time t,
    z^2 = (44t)^2 + (120+15t)^2 = 2161t^2 + 3600t + 14400
    at t=6, we have z^2 = 113796
    z =337.34 and
    2z dz/dt = 4322t + 3600
    674.68 dz/dt = 29532
    dz/dt = 43.77

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