# calculus

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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 -

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)

• calculus -

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 -

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 -

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?