Homework Help: Calculus

Posted by Allison on Tuesday, July 26, 2011 at 6:21pm.

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, Tuesday, July 26, 2011 at 8:28pm
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 15-15t.
let the distance between them be D
D^2 = (20t)^2 + (15-15t)^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, Wednesday, July 27, 2011 at 11:32am
THANK YOU! I get it now!

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