The Sub and the Destroyer (Problem 44, Section 2.2) Since there is no answer at the back of the book for this one, I will post this solution. [I will not usually do this.] There is more than one way for the destroyer to catch the submarine, so do not assume that you have it wrong if your solution is not exactly this one. At the start the destroyer can just sit still until the time that the sub would reach it if the sub was headed straight toward it. If the sub does not appear at the time it should, the captain of the destroyer resorts to calculus. Putting the origin of polar coordinates at the point where the sub was sighted, the position of the destroyer at time t will be ( r ( t ) ,θ ( t )) in polar coordinates or ( x ( t ) ,y ( t )) = ( r ( t )cos θ ( t ) ,r ( t )sin θ ( t )) in cartesian coordinates. I will assume that t = 0 corre-sponds to the time when the captain realizes he has to use calculus, and that the straight line from the point where the sub was sighted to the destroyer (before it
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