Posted by Zumba on Monday, July 23, 2012 at 1:34pm.
put B at the origin and A on the positive x-axis
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 + (12t-100)^2 - 2(10t)(12t-100)cos 10°
= 244t^2-2400t + 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?
Related Questions
Calc - At noon, ship A is 100 Kilometers due east of ship B, Ship A is sailing ...
Calc - At noon, ship A is 100 Kilometers due east of ship B, Ship A is sailing ...
Calc - At noon, ship A is 100 Kilometers due east of ship B, Ship A is sailing ...
Calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...
calculus - At noon, ship A is 30 nautical miles due west of ship B. Ship A is ...
calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...
Calculus - At noon, ship A is 180 km west of ship B. Ship A is sailing east at ...
calculus - At noon, ship A is 110 km west of ship B. Ship A is sailing east at ...
Calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...
calculus - At noon, ship A starts sailing due east at the rate of 20km/hr. Ath ...
For Further Reading
