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Math Help - Joint variation

  1. #1
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    Joint variation + percent

    The weight (W) that can be supported by a wooden beam varies directly as the square of its diameter (d) and inversely as its length (l).

    a) What percentage increase in the diameter would be necessary for a beam twice as long to support an equivalent weight?


    b)
    What percentage change in the weight would be capable of being supported by a beam
    three times as long with twice the diameter?

    thank you.
    Last edited by alex123; April 23rd 2013 at 12:54 AM.
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  2. #2
    Member agentmulder's Avatar
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    Re: Joint variation

    The setup is

    (1)  W = \frac{kd^2}{l}

    where k is the constant of variation.

    The percentage increase of d is  \frac{x}{100} \cdot d W remains the same but l becomes 2l.


     W = \frac{k(d + \frac{xd}{100})^2}{2l}

    (2)  W = \frac{kd^2(1 + \frac{x}{100})^2}{2l}

    Set (1) = (2) and solve for x

     \frac{kd^2}{l} = \frac{kd^2(1 + \frac{x}{100})^2}{2l}

    We see k and l and d^2 cancel immediately giving

     2 = (1 + \frac{x}{100})^2

    Can you continue from here?

    I'm already estimating x about 41 %...

    Thanks from alex123
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  3. #3
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    Re: Joint variation

    Quote Originally Posted by agentmulder View Post
    The setup is

    (1)  W = \frac{kd^2}{l}

    where k is the constant of variation.

    The percentage increase of d is  \frac{x}{100} \cdot d W remains the same but l becomes 2l.


     W = \frac{k(d + \frac{xd}{100})^2}{2l}

    (2)  W = \frac{kd^2(1 + \frac{x}{100})^2}{2l}

    Set (1) = (2) and solve for x

     \frac{kd^2}{l} = \frac{kd^2(1 + \frac{x}{100})^2}{2l}

    We see k and l and d^2 cancel immediately giving

     2 = (1 + \frac{x}{100})^2

    Can you continue from here?

    I'm already estimating x about 41 %...

    Really appreciate it. Thank you
    Thanks from agentmulder
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  4. #4
    Member agentmulder's Avatar
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    Re: Joint variation

    Quote Originally Posted by alex123 View Post
    Really appreciate it. Thank you
    You're welcome, if you get stuck trying to do part b post again.

    Thanks from alex123
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