Asked by gh
A balloon is at a height of 30 meters, and is rising at the constant rate of 5 m/sec. A bicyclist passes beneath it, traveling in a straight line at the constant speed of 10 m/sec. How fast is the distance between the bicyclist and the balloon increasing 2 seconds later?
Answers
Answered by
oobleck
after t seconds, the balloon is at height 30+5t
the cyclist is at a distance of 10t
so, the distance from cyclist to balloon is
z^2 = (30+5t)^2 + (10t)^2 = 125t^2 + 300t + 900
at t= 2, z = 20√5
2z dz/dt = 2(30+5t)(5) + 2(10t)(10) = 250t+300
so, at t=2,
40√5 dz/dt = 800
dz/dt = 4√5 m/s
the cyclist is at a distance of 10t
so, the distance from cyclist to balloon is
z^2 = (30+5t)^2 + (10t)^2 = 125t^2 + 300t + 900
at t= 2, z = 20√5
2z dz/dt = 2(30+5t)(5) + 2(10t)(10) = 250t+300
so, at t=2,
40√5 dz/dt = 800
dz/dt = 4√5 m/s
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