1. ## Boats?

A boat leaves a dock at 3:00 P.M. and travels due south at a speed of 20 km/h. Another boat has been heading due east at 15 km/h and reaches the same dock at 4:00 P.M. How many minutes after 3:00 P.M. were the two boats closest together?

2. Originally Posted by sheva2291
A boat leaves a dock at 3:00 P.M. and travels due south at a speed of 20 km/h. Another boat has been heading due east at 15 km/h and reaches the same dock at 4:00 P.M. How many minutes after 3:00 P.M. were the two boats closest together?
Start by setting up an appropriate coordinate system. It will be simplest to set the origin at the dock, the positive x-axis west and the positive y axis south with x and y in km from the dock. Take t in hours with t= 0 at 3:00 P.lM.

The first boat travels due south so its x coordinate is always 0. Its speed is 20 km/h so its distance from the dock, the y coordinate, after t hours is 20t.

The second boat travels due east so its y coordinate is always 0. its speed is -15 (because the positive x-axis points west and the boat is going east) so x= -15t+ b. You can find b by using the fact that the boat "reaches the same dock at 4:00 P.M."

The distance between the two boats is $\sqrt{x^2+ y^2}$. You can find the minimum distance by differentiating with respect to t and setting the derivative equal to 0 or, for this simple problem, by completing the square.

3. Hello, sheva2291!

A boat leaves a dock at 3:00 PM and travels due south at 20 km/hr.
Another boat has been heading due east at 15 km/h and reaches the same dock at 4:00 PM.
How many minutes after 3:00 PM were the two boats closest together?
Code:
      : - - -  15 - - - - :
P  15t  B   15-15t  D
o-------o-----------o
*         |
*       |
x  *     | 20t
*   |
* |
o A

Ship $A$ starts at dock $D$ and sails south at 20 km/hr,
In $t$ hours, it has sailed $20t$ km to point $A$.

Ship $B$ starts at $P$ (15 km west of $D$) and sails east at 15 km/hr.
In $t$ hours, it has sailed $15t$ km to point $B$.
. . Hence: . $BD \:=\:15-15t$

Let $x \:=\:AB.$

In right triangle $BDA\!:\;\;x \;=\;\sqrt{(20t)^2 + (15-15t)^2}$

. . Hence, we have: . $x \;=\;\left(625t^2 - 450t + 225\right)^{\frac{1}{2}}$

And that is the function we must minimize.

I believe the answer is: . $t \;=\;\frac{9}{25}\text{ hours} \;=\;21.6\text{ minutes}$

4. ahhh thank you very much, you explained it very clearly. i changed around the numbers and got it right =)