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 $\displaystyle \sqrt{x^2+ y^2}$. 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.
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 $\displaystyle A$ starts at dock $\displaystyle D$ and sails south at 20 km/hr,
In $\displaystyle t$ hours, it has sailed $\displaystyle 20t$ km to point $\displaystyle A$.
Ship $\displaystyle B$ starts at $\displaystyle P$ (15 km west of $\displaystyle D$) and sails east at 15 km/hr.
In $\displaystyle t$ hours, it has sailed $\displaystyle 15t$ km to point $\displaystyle B$.
. . Hence: .$\displaystyle BD \:=\:15-15t$
Let $\displaystyle x \:=\:AB.$
In right triangle $\displaystyle BDA\!:\;\;x \;=\;\sqrt{(20t)^2 + (15-15t)^2} $
. . Hence, we have: .$\displaystyle x \;=\;\left(625t^2 - 450t + 225\right)^{\frac{1}{2}}$
And that is the function we must minimize.
I believe the answer is: .$\displaystyle t \;=\;\frac{9}{25}\text{ hours} \;=\;21.6\text{ minutes}$