Hi, Thanks for replying... The problem does ask for the time that the rate of separation is equal to 18.86 mph. The problem is from a calculus text from 1972. The answer from the back of the text, is 11:00 am , but I have no idea how this is calculated. The portion of the distance ship A has as a head start is 6 miles and this becomes proportionally smaller as time passes. If you work backwards and put the relative values in using the answer of time equallying 1 hour dS/dt is indeed 18.86,
S dS/dt= A dA/dt + B dB/dt
S*18.86= 12(12) + 8(16). And S equals 14.422
I am sure the problem is do-able, just tricky, probably really simple too
but finding the relevant equational relationship to calculate this time given only the rates has proven to be exceedingly frustrating. I am sure I am overlooking something. I hoped someone would see a way to solve it.
The way it is asked, this problem is relegating time to a dependant variable status, in most other problems, time is an independant variable, so solving it might involve differentiating (implicitly) with respect to something other than time. I have attempted solutions with this in mind but have had no luck.