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There's this question in my calculus book that goes something like this:

The derivative of the area of a circle with respect to its radius is equal to the circle's circumference ($dA/dr = 2 \pi r$). Give a geometric explanation of why this is the case.

To me this is really obvious, but I find it hard to put into words. If you increased the radius of the circle by putting your finger inside it and pushing the edge outward, then you would have to push it around the whole circumference of the circle. Heh. I don't know.

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Related: math.stackexchange.com/questions/625/… –  Henry May 21 '11 at 23:49
    
This kind of argument also works for a square: if $h$ is half the side then the circumference is $8s$ and $dA/dh = 8s$. –  Henry May 21 '11 at 23:59
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@Henry: I'm assuming you mean that "if $h$ is half [the length of $s$, the length of] the side [of a square], then the circumference [of the square] is $4s = 8h$, and its area $A = (2h)^2 = 4h^2$, so $dA/dh = 8h$?" –  amWhy May 22 '11 at 0:32
    
@Amy: Yes - thank you for spotting the typo. I intended ending with "the circumference is 8h and $dA/dh=8h$". –  Henry May 22 '11 at 0:54
    
@Henry: that's what I figured (I knew you knew what you were doing!)...I just thought I'd clarify your comment, since, unfortunately, the "window" in which to edit comments expires! –  amWhy May 22 '11 at 0:58
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3 Answers 3

up vote 8 down vote accepted

It's because the area of a circle is really the sum of infinitely many circles' circumferences, which is the integral: $$ \int^r_0 \! 2\pi{}t \, dt $$

See http://betterexplained.com/articles/a-gentle-introduction-to-learning-calculus/ for a nice explanation

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Awesome! Thank you! –  Matt Gregory May 21 '11 at 23:49
    
Wow, that site is really awesome! "Calculus does to algebra what algebra did to arithmetic." I've never thought of it that way! –  Matt Gregory May 21 '11 at 23:58
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That's not too bad. Try this one:

If you want a larger circle with the same centre which has radius $r+\Delta r$ then the circumference of the bigger circle is $2\pi (r+\Delta r)$. The edge is a constant $\Delta r$ from the edge of the original circle all the way round.

The area between the two circles is somewhere between that of a $\Delta r \times 2\pi r$ rectangle and a $\Delta r \times 2\pi (r+\Delta r)$ rectangle. So the rate of change of this area as $\Delta r$ changes is between $2\pi r$ and $2\pi (r+\Delta r)$; take the limit as $\Delta r$ tends to $0$ to get the result.

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Draw a circle with area $A$ and radius $r$, then draw around it a larger circle. The distance from your first circle to the second is $dr$, and the difference of the areas is $dA$. As $dr \to 0$, $dA$ goes to the circumference of the interior circle, hence $dA \to 2\pi r\:dr$. I hope this is helpful, I apologize as I don't know how to use TeX on this site =S
No worries; the LaTeX formatting has been provided.

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