SOLUTION: Ship A sailing due north at 6 km/h sights ship B sailing northwest 4 km away and sailing due east at 5 km/h. How close will the two ships pass if neither alters course or speed?

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Question 1153729: Ship A sailing due north at 6 km/h sights ship B sailing northwest 4 km away and sailing due east at 5 km/h. How close will the two ships pass if neither alters course or speed?
(We're learning about applications of derivatives and have to use Pythagoras Theorem. Thank you!)

Answer by ikleyn(52786)   (Show Source): You can put this solution on YOUR website!
.

Introduce the coordinate plane (x,y) in a way that ship's A initial location is the origin of the coordinate plane;
x-axis is directed East and y-axis is directed North.


Then the initial location of the ship B is the point (,).


The ship A moves along y-axis, according to equation y = 6*t, where t is the time in hours.

So, the ship A location in time on the coordinate plane is A = (0,6t).


Ship B moves parallel to x-axis, according to equation x =  + 4*t, where t is the time in hours.

So, the ship B location in time on the coordinate plane is B = (,).


Now find the square of distance between the points A and B


    d^2 =  + .


Simplify


    d^2 =  +  = .


Thus you have this quadratic function of "t", and you need to find its MINIMUM.


Use the formulas for the vetex of the quadratic function/parabola


     = " (-b/(2a)) " =  =  =  = 0.5439 hours.


To find the square of the minimal distance, substitute t = 0.5439 into the formula (1). You will get


    d^2 =  = 0.6154.


Hence d =  = 0.7845.


ANSWER.  The minimal distance between the ships will be 0.7845 kilometers.


Notice that I did not use Calculus in my solution.
Algebra is just enough.

But if you want to use  Calculus  to find  ,  you can do it.

The route  (the road)  is open for you now.



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