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April 20, 2014

April 20, 2014

Posted by **LALA** on Monday, November 14, 2011 at 5:51pm.

- calculus -
**Reiny**, Monday, November 14, 2011 at 7:14pmmake a diagram.

It took the eastbound ship 1 hr to reach the dock, so when the southbound ship was at the dock, the eastbound ship was 15 km from the dock

So after t hrs, the southbound ship went 20t

and the eastbound ship went 15t

I see a right-angled triangle with a vertical of 20t and a horizontal of 15-15t

let d be the distance between them

d^2 = (20t)^2 + (15-15t)^2

2d dd/dt = 2(20t)(20) + 2(15-15t)(-15)

at a minimum of d, dd/dt = 0

800t - 30(15-15t) = 0

800t - 450 + 450t = 0

1250t = 450

t = .36 hrs or 21.6 minutes

so they were closest at 2:21:36 pm

check:

when t = .36 , d^2 = 51.84 + 92.16 = 144, d = 12

take a value slightly higher and smaller

t = .37 , d^2 = 54.76 + 89.3025 = 144.0625 , d = 12.0026 , slightly farther

t = .35 , d^2 = 49 + 95.0625 = 144.0625 , d = 12.0026 , again slightly farther

my answer is correct

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