How fast is the distance s(t) between the bicycle and balloon increasing 3 sec later?

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Fey L

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A balloon is rising vertically above a level, straight road at a constant rate of 1ft/sec. Just when the balloon is 65 ft above the ground, a bicycle moving at a constant rate of 17 ft/sec passes under it. How fast is the distance s(t) between the bicycle and balloon increasing 3 sec later?
 
rate = distance / time
distance = sqrt [ (65 + t)^2 + (0 + 17t)^2 ]

That is distance in terms of time
We have to find the instantaneous rate of change when t = 3.

dy/dt = (130 + 2t + 578t) / (2sqrt ((65 + t)^2 + (17t)^2)
dy/dt = (130 + 580t) / (2sqrt ((65+t)^2 + 289t^2))

y'(3) = (130 + 1740) / (2sqrt (4624 + 2601))
y'(3) = 1870 / (2sqrt (7225))
y'(3) = 1870 / (2(85))
y'(3) = 1870 / 170
y'(3) = 11

11ft / s
 
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