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

Question

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?

Answer 1

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 $2\pi(r+dr)$. As 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 $2\pi r$. Please note that this is just an informative, intuitive explanation as opposed to a formal proof. The same reasoning works with a sphere, we just flatten it out to a rectangular prism instead.

Answer 2

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.

Answer 3

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:

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

Answer 4

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?"”

Here is the link:

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

Answer 5

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.