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Expressing area as a function of circumference

  1. Apr 5, 2009 #1
    I don't feel the formatting is right for this question because it's not a problem, it's just algebra I'm not getting.

    A(r)= (pi)(r)^2
    C= 2(pi)r

    r= C/(2pi)

    therefore,

    A(c)= pi[c/(2pi)]^2

    but I don't understand how this is supposed to equal A(c)= C^2/4pi

    my attempt:
    A(c)= pi(c/2pi)(c/2pi)
    = pi(c^2/4pi)
    = c^2pi/4pi

    What am I doing wrong?
     
  2. jcsd
  3. Apr 5, 2009 #2

    symbolipoint

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    You are mostly having trouble with the formatting. Your work up to the function, A(c) looks good. Why are you confused? You are simply forgetting some simple algebra, that's all.

    This, "A(c)= pi[c/(2pi)]^2", is good. You only want to simplify it. Instead of struggling only with simple text in the message body, use of the advanced formatting available on the physicsforums lets us display that as:
    A(c) = [tex]\pi[/tex]([tex]\frac{c}{2\pi}[/tex])2
    Start from there to simplify.

    EDIT: Please excuse the poor arrangment on the parentheses, since I intended them to reach further vertically to enclose the entire rational part of the expression. also, the first pi factor is NOT intended to be superscripted although it is displayed unfortunately in such a superscripted location.
     
    Last edited: Apr 5, 2009
  4. Apr 5, 2009 #3
    Here's what I did.

    A(c) = pi(c/2pi)^2

    now I have to square the brackets first, right?

    therefore, pi(c/2pi)(c/2pi) Is this not what I have to do?

    and then I can multiply the extra pi in there. So I am still confused. I still end up with

    A(c)= pi(c^2/4pi)
     
  5. Apr 5, 2009 #4

    Pengwuino

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    Multiplying by pi will eliminate one pi in the denominator.
     
  6. Apr 5, 2009 #5
    What do you mean?

    if
    A(c)= pi(c/2pi)(c/2pi)

    then I get pi[c*c/2(2pi)] do I not? Because I have to multiply the two brackets. Then I multiply the pi and end up not getting c^2/4pi.

    I am still confused as to what I am doing wrong.
     
  7. Apr 5, 2009 #6

    Pengwuino

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    Gold Member

    When you multiply [tex](\frac{c}{{2\pi }}) \cdot (\frac{c}{{2\pi }})[/tex], you get [tex]\frac{{c^2 }}{{2^2 \pi ^2 }}[/tex]. Multiplying by [tex]\pi[/tex] eliminates 1 [tex]\pi[/tex] in the denominator.
     
  8. Apr 5, 2009 #7
    Thanks for your help Pengwuino and Symboli, I get it now... and I feel like an idiot.:redface:
     
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