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Math Help - Relationship rate of change of diagonal and rate of change of area of rectangle

  1. #1
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    Lightbulb Relationship rate of change of diagonal and rate of change of area of rectangle

    Alright i this idea originally came up while i was do a related rates problem for homework in calc class. so the problem was to find how fast the area of a rectangle was increasing, given the rate of change of one of the sides, and the fact that the proportions of the triangle will never change. i understand how to get the answer and know that it is pretty simple.

    My question: is there any known way of relating the rate of change of the diagonal of the rectangle, to the rate of change of the area of the rectangle as a whole.

    my idea is that if you have the rate of change of the diagonal you can relate it to the rate of change of the triangle and then to the rectangle as a whole.

    or i might just be crazy.
    Last edited by HungryForTruth; April 11th 2011 at 06:23 PM. Reason: wordiness
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  2. #2
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    Here's some ideas,

    So you have a rectangle dimensions \displaystyle L and \displaystyle W where \displaystyle A = L \times W

    The length of the diagonal, lets call it \displaystyle d can be found as \displaystyle d = \sqrt{L^2+W^2}

    Now increasing the dimensions by variables \displaystyle x for length and \displaystyle y for width the area becomes \displaystyle A = (L+x) \times (W+y)

    The diagonal is now \displaystyle d = \sqrt{(L+x)^2+(W+y)^2}
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  3. #3
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    so would i take the derivative of the last equation you game me, then just plug in values. if im understanding the equation correctly you said that the diagonal is equal to the area with increased dementions.

    here is the original equaton now i dont want you to solve it the normal way i just want to know if there is anyway to solve it by using the rate of change of the diagonal.

    " a screen saver displays the outline of a 3cm by 2cm rectangle and then expands the rectanle in such a way that the 2cm side is expanding at the rate of 4 cm/sec and the proportions of the rectangle never change. how fast is the area of the rectangle increasing when its dimesions are 12cm by 8cm."
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  4. #4
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    My question: is there any known way of relating the rate of change of the diagonal of the rectangle, to the rate of change of the area of the rectangle as a whole.
    As Pickslides said, for a rectangle of length x and width y, the length of the diagonal is given by L= (x^2+ y^2)^{1/2} and the area is given by A= xy.

    Then, \frac{dL}{dA}= \frac{\partial L}{\partial x}\frac{\partial x}{\partial A}+ \frac{\partial L}{\partial y}\frac{\partial y}{\partial A}

    It is easy to calculate that \frac{dL}{dx}= (1/2)(x^2+ y^2)^{-1/2}(2x)= \frac{x}{(x^2+ y^2)^{1/2}} and that \frac{\partial A}{\partial x}= y so that \frac{\partial x}{\partial A}= \frac{1}{y} and so \frac{\partial L}{\partial x}\frac{\partial A}{\partial x}= \frac{x}{y(x^2+ y^2)^{1/2}}.

    Can you finish that?
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