# Why is the derivative of a circle's area its perimeter (and similarly for spheres)?

When differentiated with respect to $r$, the derivative of $\pi r^2$ is $2 \pi r$, which is the circumference of a circle.

Similarly, when the formula for a sphere's volume $\frac{4}{3} \pi r^3$ is differentiated with respect to $r$, we get $4 \pi r^2$.

Is this just a coincidence, or is there some deep explanation for why we should expect this?

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(I realise that it might not be clear what the $n$-dimensional generalisation is of this, but perhaps this would happen even in different geometries or metric spaces?). –  bryn Jul 24 '10 at 3:01
Its deep. Look at the most general version of the fundamental theorem of calculus en.wikipedia.org/wiki/… –  Jonathan Fischoff Jul 24 '10 at 3:23
And next explain why it fails for the square... or the ellipse... –  GEdgar Dec 6 '11 at 14:22
You mentioned that it's true for the 2-sphere, and for the 3-sphere, but it should be noted that it is also true for the 1-sphere, which is the interval from -r to r, which has 1-volume of 2r. The derivative of 2r wrt r is 2, which is the measure of its "surface", measure for 0-dimensional items being the same as cardinality. –  Hexagon Tiling Jan 7 '12 at 11:47
@GEdgar : I make out the area of a square of 'radius' $r$ as $4r^2$ and the perimeter as $8r$; the idea continues to work there (for essentially the same uniformity reason that it does on the sphere). Of course, it doesn't work on rectangles for the same reason it doesn't work on ellipses... –  Steven Stadnicki Apr 8 '12 at 18:11

Consider increasing the radius of a circle by an infinitesimally small amount, $dr$. This increases the area by an annulus (or ring) with inner radius $2 \pi r$ and outer radius $2\pi(r+dr)$. At this ring is extremely thin, we can imagine cutting the ring and then flattening it out to form a rectangle with width $2\pi r$ and height $dr$ (the side of length $2\pi(r+dr)$ is close enough to $2\pi r$ that we can ignore that). So the area gain is $2\pi r\cdot dr$ and to determine the rate of change with respect to $r$, we divide by $dr$ and so we get $2\pi r$. This is of course, just an informative, intuitive explanation and not a formal proof. The same reasoning works with a sphere, we just flatten it out to a rectangular prism instead.

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There is an article on the web that deals, in depth, with this question. Here is a quote from it:

“We were intrigued by the students' work, and this paper is the result of our attempt to answer the question, “When is surface area equal to the derivative of volume?"”

www.math.byu.edu/~mdorff/docs/DorffPaper07.pdf

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The explanation is very simple. Take a sphere of radius $r$, volume $V$, and surface area $A$. Now paint it, with a layer of thickness $\delta r$. The volume of paint required is (to first order in $\delta r$) $A\delta r$, which gives you straight away: $$\delta V = A \delta r$$ Hence, in the limit:

$$\frac{dV}{dr} = A$$

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The circle (and sphere) is not really that special. It also works for the square if you measure it using not the side length $s$, but half that, $h=s/2$. Then its area is $A=(2h)^2=4h^2$ with derivative $dA/dh=8h$ which is its perimeter.

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It works for the cube as well. –  Lucian yesterday

The size of the boundary times the rate at which the boundary moves equals the rate at which the size of the bounded region changes.

There appears to be no conventional name for this fact. I've called it the boundary rule sometimes.

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How does one set up the integral to find the area of a circle? An area was defined for a square or rectangle to be the width times the length. It is the equivalent for all geometries. For a circle working in polar coordinates the differential area equivalent is $dr$ while the differential width would be $r \,d\theta$.

So... $$dA = r\, d \theta\, dr.$$ Here $r \,d\theta$ is the differential arc (width) times the differential length $dr$. You can see that by inspecting the form of this differential equation the fundamental form for finding the area of a circle is in the form of what we know to be the circumference of a circle. If we divide through by $dr$. So the connection is implicit in the basic geometry. Because we are working in a polar system.

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You can use LaTeX pretty much as usual. Just enclose your formulas in dollar signs. For example, $\theta$ gives $\theta$. –  t.b. Dec 6 '11 at 14:52

I thought about this recently. This is what I've come up with. It relies on what I think has been called dimensional analysis. But I'm not sure about that name.

Suppose you have a polygon and $A(x)$ is a formula that takes as input something in terms of meters (like a side length or a circumference, or half a side length etc), then $A(x)$ must be of the form $A(x)=k_1x^2$ for some $k_1 \in \mathbb{R}$ since the answer that it outputs is in terms of meters squared.

Similarly, a formula $P(x)$ for perimeter which takes input a one dimensional'' variable (that is, in terms of meters), will be of the form $P(x)=k_2x$ for $k_2 \in \mathbb{R}$.

Note that the input need only be one dimensional in terms of length, as mentioned above, the square case works when you take the area formula to be $A_{square}(y)=(2y)^2$ where $y$ is HALF of the side length. Note that we just rewrote our formula in terms of some scalar multiple of the side length.

My claim is that we can always take some scalar multiple and have this desirable property of the derivative of the area formula equaling the perimeter formula. For example, take $k_3y=x$ (in the square case we let $k_3=2$. Then $A(y)=k_1(k_3x)^2=k_1k_3^2y^2$ and $A'(y)=2k_1k_3^2y=2k_1k_3x$, if we want this to be equal to the perimeter formula $P(x)=k_2x$ we just solve $2k_1k_3x=k_2x$ for $x$, that is, take $k_3=\frac{k_2}{2k_1}$.

To illustrate this, note that we can do this for side length too! Suppose I want the derivative of the area formula for a square to equal the formula for the sidelength of a square. That is, I want $A(x)'=S(x)$. Well, $A(x)=x^2$ and $S(x)=x$ if $x$ is side length. So choose $k_3=\frac 1 2$ that is, I want the input of my functions to be TWICE the side length. Then $A(y)=(\frac 1 2 y)^2=\frac 1 4 y^2$ so $A'(y)= \frac 1 2 y= x = S(x)$.

The very curious part of this however (to me at least), is how what makes this work for the circle is using the RADIUS as input, and what makes it work for the square is choosing half the side length, which seems very analogous to a radius right?

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Because you use the integral (read: anti-derivative) to find the area under the curve - even a curve in polar coordinates.

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This doesn't explain why the coefficients match up. –  Ben Alpert Jul 24 '10 at 3:35
@Ben: Yes it does. Try reading 'integral' as 'anti-derivative'. –  BlueRaja - Danny Pflughoeft Jul 24 '10 at 4:06
This answer is right, but I understand Ben Alpert's confusion. It's slightly unclear when I read it over again. –  Justin L. Jul 24 '10 at 11:10
I hadn't thought of this. Does this also explain why it works in 3 dimensions? –  bryn Jul 27 '10 at 4:59
I agree it doesn't really explain why the coefficients match up. If this answers the question, then how does this work with a square? Is the derivative of $x^2$ equal to $4x$? Your explanation works for one specific example. How about an explanation that really works in general? At the least, you need more details to explain it. –  Graphth Dec 6 '11 at 13:10