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October 30, 2014

October 30, 2014

Posted by **Darcy ( Please Reply, thank you)** on Monday, December 9, 2013 at 6:05am.

- Differential Calculus -
**Reiny**, Monday, December 9, 2013 at 8:50amplace ship B 65 miles east of ship A and mark them that way.

Let the time passed be t hrs

Draw a line BP , so that P is between A and B showing the distance traveled in those t hours.

Draw a line downwards AQ showing the distance traveled by ship A.

Join PQ to get the right-angled triangle APQ

AP = 65-10t

AQ = 15t

let D be the distance between P andQ

D^2 = (65-10t(^2 = (15t)^2

2D dD/dt = 2(65-10t)(-10) = 2(15t)(15)

for a minimum of D , dD/dt = 0

so 2(10)(65-10t) = 2(15t)(15)

divide by 10

2(65-10t) = 3(15t)

130 - 20t = 45t

65t = 130

t = 130/65 or 2 hrs

They will be closest at 9:00 + 2:00 or 11:00 am

and that distance is ....

D^2 = (65-20)^2 + 30^2

= 2925

D = √2925 = appr 54.1 miles

check my arithmetic

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