Tuesday

July 29, 2014

July 29, 2014

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

- Calculus -
**Reiny**, Tuesday, July 26, 2011 at 8:28pmLet 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 11:32amTHANK YOU! I get it now!

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