Thread: Optimization Problem6

A boat leaves a dock at 2:00PM and travels due south at a speed of 20km/h.
Another boat has been heading due east at 15km/h and reached the same dock at 3PM.
At what time were the two boats closest together.

Originally Posted by camherokid

A boat leaves a dock at 2:00PM and travels due south at a speed of 20km/h.
Another boat has been heading due east at 15km/h and reached the same dock at 3PM.
At what time were the two boats closest together.

Take the position of the dock as the origin, and 2 pm as t=0.

The position of the first boat at time t is (0, -20t), and the position
of the second boat at time t is (-15+15t, 0).

hence the squate distance between them at time t is:

SD = 15^2 (-1+t)^2 + 20^2 t^2.

The distance between the boats is a minimum when the square distance is a
mimimum, that is when d(SD)/dt = 0.

d(SD)/dt = 2 15^2 (-1+t) + 2 20^2 t

Which is zero when t = 9/25 hours, when SD=144, and so the seperation is
12km.

At this point we could use the second derivative test to show this is a
minimum but that is probably not necessary here, as it must be.

RonL

Hello, camherokid!

A slightly different approach . . .

A boat leaves a dock at 2:00PM and travels due south at a speed of 20km/h.
Another boat has been heading due east at 15km/h and reached the same dock at 3PM.
At what time were the two boats closest together?

Code:

      P   15t   B   15-15t    C
      * - - - - * - - - - - - *
                  \           |
                    \         |
                      \       |
                      x \     |20t
                          \   |
                            \ |
                              |
                              * A

Ship #1 starts at C at 2 PM and travels south at 20 km/hr.
In the next t hours, it travels 20t km to point A

Since Ship #2 reaches C at 3 PM at 15 km/hr,
. . it was 15 km west of C at 2 PM at point P.
In the next t hours, it travels 15t km to point B.
. . Hence: . BC = 15 - 15t


Let x = AB, the distance between the ships.

. . Then: .x² .= .(20t)² + (15 - 15t)² .= .625t² - 450t + 225

Differentiate: .2x(dx/dt) .= .1250 t - 450

. . and we have: .dx/dt .= .625t - 225)/x .= .0

. . Hence: .t .= .9/25 .= .0.36 hours .= .21 minutes, 36 seconds


Therefore, the ships are closest at 2:21'36" PM


Since x² .= .625(0.36)² - 450(0.36) + 225 .= .144
. . their minimum distance is: .x = 12 km.