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Rate of change of area of a square with respect to its perimeter

  1. May 27, 2013 #1
    oops I meant "Rate of change of area of a square with respect to its side length"

    Ok I have to use this annoying Stewart textbook for my Calc class in college. Most of the questions require what I like to call "Monkey Math," where you just memorize a set of steps and then follow them rigidly for each and every problem.

    However, this problem I found has me really thinking, here it is:

    Show that the rate of change of the area of a square with respect to its side length is half its perimeter. Try to explain geometrically why this is true.

    So I can easily "show" why this is true.



    so (A'(x))/P(x)=1/2

    But I am drawing this out on my white board and I cant conceptually understand why the rate of change of the area would be 1/2 the perimeter at the exact same moment...

    I can understand why this is true algebraically, but I guess I can't visualize what the derivative of a geometric shape's area is.

    Can anyone shed some light on this for me?
    Last edited: May 27, 2013
  2. jcsd
  3. May 27, 2013 #2


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    Draw a square

    Draw a slightly larger square with the smaller square nestled in one corner.

    The change in area is the two rectangles above and next to the smaller square, plus the reeeaaally tiny square nestled in the opposite corner.

    If the larger square has side length [itex] x + \Delta x[/itex], find the area of those rectangles and squares in terms of x and [itex] \Delta x[/itex]
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