# Homework Help: Help with calculus

1. Mar 16, 2005

At 3 PM a ship which is sailing due north at 12 knots(nautical miles/hour) is 5 miles west of a westbound ship which is making 16 knots. a) At what rate is the distance between the ships changing at 3 PM

is it just simply pythagorean theorem of the rates of both ships?
i got the answer as 20 knots

i dont see how related rates would be involved on this
thanks for the help

Last edited: Mar 16, 2005
2. Mar 16, 2005

### xanthym

Let time t=(0) correspond to 3 PM. Then:
{North} = (+y Direction)
{West} = (-x Direction)
{Northbound Boat Position} = {x=(-5 miles), y=(12*t)}
{Westbound Boat Position} = {x=(-16*t), y=(0)}

{Distance Between Boats} = D = sqrt{(Δx)2 + (Δy)2} =
= sqrt{((-5) - (-16*t))2 + ((12*t) - (0))2} =
= sqrt{(16*t - 5)2 + (12*t)2} =
= sqrt{256*t2 - 160*t + 25 + 144*t2} =
= sqrt{400*t2 - 160*t + 25}

The time rate of separation distance change can be determined from:
(dD/dt) = (1/2)*{400*t2 - 160*t + 25}(-1/2)*(800*t - 160)
Thus, at 3 PM {or t=(0)}:
(dD/dt) = (1/2)*{0 - 0 + 25}(-1/2)*(0 - 160) = (1/2)*(25)(-1/2)*(-160)
(dD/dt) = (1/2)*(-160)/5 = (-16 knots)

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Last edited: Mar 17, 2005