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Stuck on couple related rates problems

  1. May 20, 2006 #1
    1. A ship with a long anchor chain is anchored in 11 fathoms of water. The anchor chain is being wound in at a rate of 10 fathoms/minute, causing the ship to move toward the spot directly above the anchor resting on the seabed. The hawsehole ( the point of contact between ship and chain) is located 1 fathom above the water line. At what speed is the ship moving when there are exatly 13 fathoms of chain still out?

    For this problem I started with this drawing.. http://img.photobucket.com/albums/v449/theboywiththegiantmuffin/untitled.jpg

    And then from there, I had no idea where to go... there hawsehole being 1 fathom above the water really gets to me, perhaps making the above drawing void. Another thing I don't understand is that it says it's anchored in 11 fathoms of water.. how could the question be asking what speed the boat would be moving if it were at 13 fathoms?

    2. A ladder 41 feet long was leaning against a vertical wall and begins to slip. Its top slides down the wall whilte its bottom moves along the level ground at a constant speed of 4ft/sec. How fast is the top of the ladder moving when it is 9 feet above the ground?

    For this one.. I didn't even know what to do.. of course I drew a triangle, hypotenuse of 41 and the vertical side of 9 feet.. and then.......?

    Mainly, I think problems such as these are really easy, but I have a really hard time picturing the problem or drawing it out. I don't know which numbers apply to dx/dt and dy/dt..
  2. jcsd
  3. May 20, 2006 #2


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    I might be wrong here, but to me it seems as if you should consider a triangle like the one you drew except that the vertical side will be 12 fathoms long (in other word, you would draw a horizontal line aligned iwth the hawsehole). Do you see what I mean? I might be wrong, however.

    And tehy are asking what is the speed when there is 13 fathoms of *chain* still out, which is the length of the hypothenuse on your triangle. Of course this length will be larger or equal to 12 fathoms (it will be equal to 12 fathom when the boat will be right above the anchor)

    If we call "L" the length of the hypothenuse, then what you want is to write dx/dt in terms of dL/dt (which is the number they give you). All you have to do is to write an expression relating x and L (and other known values), isolate x in terms of those constants and L, and differentiate both dised with respect to t. You will get dx/dt = expression in terms of constant, L and dL/dt.
    Write an expression relating x and y (that is just Pythagora's theorem). Isolate y in terms of x and the length of the ladder. Differentiate both sides with respect to t. You will get dy/dt = expression in terms of the length of the ladder and x and dx/dt. You iwll need to calculate the value of x when y is equal to 9 feet. You plug dx/dt (which they give to you), x and you get your answer.
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