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October 20, 2014

October 20, 2014

Posted by **Anonymous** on Sunday, November 4, 2012 at 11:26pm.

- calculus -
**Reiny**, Monday, November 5, 2012 at 12:44amAt a time of t hours after noon

Put the ships on a grid, A along the x-axis and B along the y-axis

In my diagram I have a right-angled triangle with AB the hypotenuse, AO is 15t + 10 and BO = 15t

AB^2 = (15t)^2 + (15t+10)^2 = 450t^2 + 300t + 100

2 AB d(AB)/dt = 900t + 300

when t = 3 (3:00 pm)

AB^2 = 450(9) + 300(3) + 100 = 5050

AB=√5050

d(AB)/dt = (900t+300)/(2AB) = (450t + 150)?AB

= (450(3) + 150)/√5050 = appr 21.11 knots

- calculus -
**Steve**, Monday, November 5, 2012 at 12:50amafter x hours, the distance y between the ships is

y^2 = (10+15x)^2 + (15x)^2

at 3 pm, x=3, so

y^2 = 55^2 + 45^2

y = √5050 = 71.06

2y dy/dt = 2(10+15x)(15) + 2(15x)(15) = 300(3x+1)

when x=3,

2(71.06) dy/dt = 3000

dy/dt = 3000/142.12 = 21.11 knots

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