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December 18, 2014

December 18, 2014

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

- calculus -
**Reiny**, Monday, July 23, 2012 at 2:05pmput 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 -
**Zumba**, Monday, July 23, 2012 at 2:14pmthank 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 2:18pmYes, 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 2:41pmok, 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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