calculus
posted by Zumba on .
At noon, ship A is 100 kilometers due east of ship B. Ship A is sailing west at 12 k/h and ship B is sailing S10degrees west at 10 k/h. At what time will the ships be nearest each other and what will this distance be? (hint: You do not have a right triangle, unfortunately)

put B at the origin and A on the positive xaxis
After t hours, draw the paths of ship A and ship B and let A's position be P (to the left of B) and the position of ship B as Q
I see triangle PQB, where angle B=10°
and QB = 10t miles, PB = 12t  100 miles
Let the distance between them be x
by the cosine law:
x^2 = (10t)^2 + (12t100)^2  2(10t)(12t100)cos 10°
= 244t^22400t + 10000  (240cos10°)t^2 + 2000cos10° t
2x dx/dt = 488t  2400  480cos10° t + 2000cos10°
= 0 for a minimum of x
t( 488  480cos10°) = 2400  2000cos10°
t = 28.14
check my arithmetic, I should have written this out first, I just drew the diagram.
So it would be 28.14 hrs after noon (into the next day) 
thank you, this looks correct. But, whouldnt angle B be 80 degrees, not 10 degrees. Because the s10w means 10 degrees to the west side of south.

Yes, I believe you are interpreting
S 10° W in the proper way, I did W10S
so change the values of cos10 to cos80
The steps would remain the same. 
ok, thank you so much for your help. one more question, what happens to the x? does it just drop out when you set it equal to zero?