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Calculus/related rate


A railroad bridge is 20 feet above a river. A baby is driving a train traveling 60mi/hr passes over the middle of the railroad bridge at the same instant that a puppy in a boat traveling 20 mi/hr passes under the middle of the bridge. Looking from the view of a satellite how fast are the baby and puppy moving away from each other 10 sec later?

Since a mile per hour is the same as 88 feet per minute,
divide by 60 again and get 22/15 feet per second.

This make 60 mi/hr into 88 f/s and 20 mi/hr into 44/5 ft/s.

After 10 sec, the train is 880 feet away from the bridge
and the ship is 44/3 feet away from the bridge.

If the train is t feet away and the boat is b feet away,
the distance d between them is found using tē + bē = dē.

Differentiating gives 2tt' + 2bb' = 2dd'.
Dividing by 2d gives (tt' + bb')/d = d'.

Using what I have said, t and b are the distances away, which were found at 10 sec,
t' and b' are the speeds, which were given, and
d is the distance, which can be found to be sqrt(tē + bē).


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