At noon ship A is 100 km west of ship B. Ship A travels south at 35km/h and ship B travels North at 25 km/hr. At 4pm how fast does the distance between them change?

I don't quite understand how to do this so a complete explanation would be best.

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- Mar 18th 2010, 03:48 AMNevershDistance of two moving objects.
At noon ship A is 100 km west of ship B. Ship A travels south at 35km/h and ship B travels North at 25 km/hr. At 4pm how fast does the distance between them change?

I don't quite understand how to do this so a complete explanation would be best. - Mar 18th 2010, 04:37 AMZanderist
Haha, I have nearly the same problem.

This has to do with implicit differentiation

Use the formula d = v*t

I would say you could also use a^2+b^2=c^2 but they're going in complete opposite directions.

By the way your problem is featured on youtube.

http://www.youtube.com/watch?v=jv4gTxWqeBE - Mar 18th 2010, 05:41 AMHallsofIvy
Set up a coordinate system so that ship A is at (0, 0) and ship B is at (100, 0). t hours after noon, ship A is at (0, -35t) and ship B is at (100, 25t).

The distance between them, as a function of t, is $\displaystyle \sqrt{100^2+ (25t-(-35t))^2}= \sqrt{100^2+ 60^2t^2}$.

Find the derivative of that at t= 4.