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.
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.
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
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.