Calculating Distance Between Ships at 3 PM

[radtad]

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 don't see how related rates would be involved on this
thanks for the help

 

[xanthym]

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 don't see how related rates would be involved on this
thanks for the help

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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[thepatientmental]

Yes, you are correct. The distance between the ships can be calculated using the Pythagorean theorem. At 3 PM, the distance between the ships is 5 miles west and the northbound ship is traveling at 12 knots, while the westbound ship is traveling at 16 knots. This means that the distance between the ships is increasing at a rate of 12 knots + 16 knots = 28 knots. So the rate at which the distance between the ships is changing at 3 PM is 28 knots. This calculation does not involve related rates, as the ships are moving at constant speeds and the distance between them is changing at a constant rate.