Posted by Allison on .
At 2:00PM boat X is heading due west at 15mph towards a dock that is exactly 15 miles away. At the same time boat Y leaves the dock traveling due north at a speed of 20mph. At what time will the boats be closest together?

Calculus 
Reiny,
Let the time passed since the northbound ship left the dock be t hrs
Make a sketch
Distance traveled by northbound boat is 20t
distance from dock of westbound boat is 1515t.
let the distance between them be D
D^2 = (20t)^2 + (1515t)^2
= 400t^2 + 225  450t + 225t^2
= 625t^2  450t + 225
2D dD/dt = 1250t  450
dD/dt = (625t  225)/d
= 0 for a max/min of D
625t = 225
t = 225/625 = .36
so the time is .36 hours past 2:00 pm
.36 hrs = 21.6 minutes or 21 minutes 36 seconds
time is 2:21:36 PM 
Calculus 
Allison,
THANK YOU! I get it now!