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Thread: Ocean Liner Problem

  1. #1
    Sep 2009

    Ocean Liner Problem

    An ocean liner backs out of a dock slip into a narrow channel. The slip is 54 feet wide; the liner is 250 feet long, and we will assume its width may be treated as 0. How wide must the channel be in order to clear the liners stern? Hint: Express everything in terms of the angle between the liner and the dock.

    Thanks for all the help and happy thanksgiving
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  2. #2
    MHF Contributor
    Oct 2008
    Have you drawn a sletch? You want two parallel lines which are the sides of the slip and therefore 54 apart... They stop at the same 'point' - rather, at the same perpendicular. We don't care what happens alot further on up the slip, but the slip stops where it reaches one side of the perpendicular 'channel'. Then a straight line representing the ship (and 250 long) reaches inside the slip so that one end of the ship-line touches one of the slip-side-lines and the end-point of the other slip-side-line touches it.

    Now label...

    z, let's say, is the length of ship inside the slip - this is the hypotenuse of a right triangle, and 54 is the length of the side opposite an angle theta. Call the remaining ('adjacent') side x. Get a trig expression for x and one also for z in terms of theta.

    Now compare with the larger and similar right triangle whose hypotenuse is the whole ship. Let u be the necessary width of channel when the ship is in this position - for the current value of theta. You should be able to get a function (to be maximised with calculus) that gets u from theta. Use the ratios in the similar triangles. Specifically,

    $\displaystyle \frac{u + x}{x} = \frac{250}{z}$

    Make u the subject, and replace x and z with functions of theta. Maximize.

    Here's roughly what I was hoping to describe...

    Hope this helps.
    Last edited by tom@ballooncalculus; Nov 26th 2009 at 01:15 PM. Reason: pic
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