This is a problem in an old calculus text book from the 1970's.
The text gives the answer to the problem but not the way in which it was calculated.
Two ships leave the same point going in different directions and leaving at different times.
Ship A leaves at 10 am going north at 12 mph
Ship B leaves at 10:30 am going west at 16 mph
Find the Time when the rate of change of the distance between the ships equals 18.86 mph
The distance between the ships (S) is the hypotenuse ...we want to find the time when dS/dt =18.86
Here's what I did to try and solve it:
Put everything in terms of t
Distance ship A goes is 12 t + 6
Distance ship B goes is 16 t
dA/dt=12 mph. dB/dt=16 mph and dS/dt=18.86
When A= 12t, B=16t-8
When A=12t+6, B =16 t
Getting S in terms of t Similar triangles: 12t + 6/S = 6/10 (10 is from Pythagoras with 6 vertical and 8 horizontal)
(120t + 6)10 = 6S
Substituting into S dS/ dt = A dA/dt + B dB/dt
(20t + 10) 18.86 = (12t + 6)12 + 16t* 16
377.2t+ 188.6 =144t+72+256t
377.2t + 188.6 = 400t + 72
And this is the wrong answer. Can anyone tell me how this problem is Solved?