# Thread: World Problem Involving Derivatives

1. ## World Problem Involving Derivatives

The World Problem:

At 9:30am, one ship was 65 miles due east from a second ship. If the first ship sailed west at 20 miles per hour and the second ship sailed southeast at 30 miles per hour, when were they closest together?

What I Have So Far:

1st ship coordinates: (0,0)
2nd ship coordinates: (-65,0)

x2+x2=(30t)2
x = $\displaystyle 15\sqrt{2}t$

Letting time be 't' in hours since 9:30am :
1st ship coordinates: (-20t,0)
2nd ship coordinates: ($\displaystyle -65 + 15\sqrt{2}t, -15\sqrt{2}t$)

Not sure where to go where to go from here... Can anyone help me out?

2. ## Re: World Problem Involving Derivatives

If the distance between the two boats is $\displaystyle D,$ then

$\displaystyle D^{2} = (-20t+65-15t\sqrt{2})^{2}+(15t\sqrt{2})^{2}.$

For minimum $\displaystyle D,$ differentiate this (no need to take the square root), and equate the result to zero.

3. ## Re: World Problem Involving Derivatives

So then would it be..

$\displaystyle (1300+600\sqrt{2}) t^2 - (2600 + 1950\sqrt{2}) t + 4225$
$\displaystyle 2(1300+600\sqrt{2})t = 2600 + 1950\sqrt{2}$
$\displaystyle t = 1.25$

Which makes it 10:45am

Thank you so much for the guidance!!