Differential Calculus
posted by Darcy ( Please Reply, thank you) .
At 9am ship B is 65 miles due east of another ship A. Ship B is then sailing due west at 10mi/h and A is sailing due south at 15 mi/hr if they continue in their respective course when will they be nearest to one another? and how near?

place ship B 65 miles east of ship A and mark them that way.
Let the time passed be t hrs
Draw a line BP , so that P is between A and B showing the distance traveled in those t hours.
Draw a line downwards AQ showing the distance traveled by ship A.
Join PQ to get the rightangled triangle APQ
AP = 6510t
AQ = 15t
let D be the distance between P andQ
D^2 = (6510t(^2 = (15t)^2
2D dD/dt = 2(6510t)(10) = 2(15t)(15)
for a minimum of D , dD/dt = 0
so 2(10)(6510t) = 2(15t)(15)
divide by 10
2(6510t) = 3(15t)
130  20t = 45t
65t = 130
t = 130/65 or 2 hrs
They will be closest at 9:00 + 2:00 or 11:00 am
and that distance is ....
D^2 = (6520)^2 + 30^2
= 2925
D = √2925 = appr 54.1 miles
check my arithmetic