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 Protecting the fleet Part E. (Posted on 2015-12-23)
A fleet of ships is on a straight course at a speed of one knot. It is guarded by a cruiser that travels at two knots.

E. The ships are in the formation of a perfect circle of radius 1 nautical mile. How long does the cruiser take to completely traverse the edge of this circle?

Try to give an exact answer.

 No Solution Yet Submitted by Jer Rating: 4.0000 (1 votes)

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 Solution Comment 3 of 3 |
As with the previous entries, I will assume the fleet is heading north.
In the Part C&D problem I determined that if T is the angle measured clockwise between the fleet north vector and the cruiser's vector relative to the fleet then the relative speed of the cruiser to the fleet is V = -cos T + sqrt[(cos T)^2 + 3]

Because we have a radius of 1 nautical mile then T measured in radians is equal to the distance traveled around the circumference of the fleet.
Then treat the velocity equation as a differential equation, letting x be the amount of time:
dT/dx = Integ{-cos T + sqrt[(cos T)^2 + 3]}

This is a separable differential equation, yielding:
x = (1/3)*Integ{cos T + sqrt[(cos T)^2 + 3]}

Plugging the indefinite integral into Wolfram's online integrator yields:
x = (sin T)/3 + (2/3)*E(T,1/4) + C
Where E(phi,m) is the Incomplete Elliptic Integral of the Second Kind evaluated with angle phi and parameter m (0<m<1).

sin 0 and E(0,1/4) both equal 0, so we can substitute C=0 for our constant of integration.  Since sin (2*pi) = 0 then the time taken is (2/3)*E(2*pi,1/4) = 3.91323 hours, or 3 hours, 54 minutes, 48 seconds.

 Posted by Brian Smith on 2015-12-24 10:27:35

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