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Math Help - Inverse variation problem

  1. #1
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    Inverse variation problem

    A magnet attracts a nail with a force that varies inversely with the square of the distance between them. The nail is 12 inches from the magnet. How many inches from the magnet would the nail need to be to have twice as much attractive force on it as it current has?

    I can't seem to get 6*sqrt(2) even though I think I'm setting up right. F=k/d^2 right?
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  2. #2
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    Re: Inverse variation problem

    "inverse variation" means that "x times y" is a constant: xy= k. If we let y be the magnetic force at x= 12 inches. Then 12y= k. If we let z be the distance the nail needs to be in order to have twice the force, 2y, then we must have z(2y)= k. Since they are both equal to k, 12y= 2zy. Dividing both sides by 2y, z= 6. In other words, because this is "inverse variation", you double the force by halving the distance.
    Thanks from dannyc
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  3. #3
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    Re: Inverse variation problem

    Quote Originally Posted by HallsofIvy View Post
    "inverse variation" means that "x times y" is a constant: xy= k.
    But the problem says that the force varies inversely with the square of the distance.

    Quote Originally Posted by dannyc View Post
    F=k/d^2 right?
    Yes. To solve this completely mechanically, you substitute 12 and get F = k/144. Therefore, 2F = k/72 = k/d'^2 where d' is the new distance. From there, d'^2 = 72 = 6^2\cdot 2, so d' = 6\sqrt{2}.

    Intuitively, if the force were inversely proportional to the distance, then doubling the force would mean halving the distance, as was said in post #2. If the force is inversely proportional to the square of the distance, then squaring the change in the distance and taking the reciprocal would give the change in the force. If the distance is multiplied by 1/\sqrt{2}, then squaring this gives 1/2, and taking the reciprocal gives 2, which is the required change in the force. Note that this reasoning works only when you multiply the force or the distance, not when you add something to them.
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