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Derivative of the Area of a Circle

  1. Mar 16, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that the rate of change of the area of a circle with respect to its radius is the same as the circumference of the circle. Can you suggest why?

    2. Relevant equations

    A = [tex]\pi[/tex]r[tex]^{2}[/tex] = f(r)
    L = 2[tex]\pi[/tex]r = g(r)

    3. The attempt at a solution

    I have showed that the derivative of f(r) is equal to g(r).
    But I have no idea why the area and the circumference of the circle are related in such a way. Any suggestions greatly appreciated.
    Thank you.
  2. jcsd
  3. Mar 16, 2009 #2
    Start by thinking about what any derivative of a function is describing in general, and then how it applies here specifically.
  4. Mar 16, 2009 #3
    The derivative describes the slope of a tangent to the circle which is perpendicular to the radius... but I don't seem to go anywhere from here... hmmm
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