Posted by Zumba on Monday, July 23, 2012 at 1:34pm.
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)

calculus  Reiny, Monday, July 23, 2012 at 2:05pm
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)

calculus  Zumba, Monday, July 23, 2012 at 2:14pm
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.

calculus  Reiny, Monday, July 23, 2012 at 2:18pm
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.

calculus  Zumba, Monday, July 23, 2012 at 2:41pm
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?
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