Understanding intuitively Why the rate of change of area of a circle with radius is the circumference of the circle

I don't know whether this a rigorous mathematical question. But I was trying to figure out the intuition behind the answer . I have through out my high school used the fact that d/dr (πr^2)= 2πr . It is obvious from the formula. My question is how should I understand the answer intuitively or geometrically ?

We all know that the rate of change of displacement with time is the instantaneous velocity. The answer seems obvious in the first reading of any basic calculus text.

Derivatives of certain other functions also seem obvious (or atleast not as unobvious as the above circle one ) at the first go , or if they don't seem so obvious we can understand them after drawing a picture and analysing it geometrically.

But how should I intuitively understand the rate of change of area of circle with radius being equal to the circumference of the circle ? Why is it so ? I need to understand it geometrically too. How can the rate of change of area of circle with radius give the circumference of the circle ( intuitively ) ?

• The onion proof works for me. Also here – leonbloy Sep 12 '16 at 20:22
• Concider a disk of radius r and make tha radius a bit larger, that is, r + dr. Then the area will increase approximately by dr times the outer boundary.. just draw a picture. – Peter Franek Sep 12 '16 at 20:23

Imagine we start with a disk of radius $r$, and we are going to add a ring of width $\Delta r$ to the disk. If $\Delta r \ll 1$, then $\Delta r^2 \approx 0$, a representation of the fact that a small number squared is less than a small number to the first power. The area of the ring is the area of an annulus of outer radius $r + \Delta r$ and inner radius $r$: $$A_\text{ring} = \pi[(r + \Delta r)^2 - r^2] = \pi(2r\,\Delta r + \Delta r^2) \approx 2\pi r\,\Delta r$$ which is the product of the circumference of the ring with its width $\Delta r$.

The differential area element $dA = 2\pi r\,dr$ represents a small change in area for a small amount of radius added to the disk, and is the limit of the finite expression $\Delta A \approx 2\pi r\,\Delta r$. In the language of the derivative, the rate of change of the area $A$ with respect to its radius is $dA/dr = 2\pi r$.

To see why this makes sense intuitively, suppose that $r \approx 0$. Then, when we change the area of the circle by a very small amount $\Delta A$, we should expect that $\Delta A \approx 0$ as well since the ring will have very little area, almost like adding on a "thick point" to the circle, and points have zero area. On the other hand, suppose $r$ was very large. Then, when we add on a small amount of area $\Delta A$, this should be in proportion to the larger radius of the circle. Hence, the amount of area we add with rings gets larger with larger values of $r$, sort of how the amount of fence we need increases the larger an area we want to surround.

In other words, we intuitively expect that $$\frac{dA}{dr} \propto r$$ based on the heuristic argument we have given here. The factor of $2\pi$ comes from the argument we gave in the finite case for computing the area of a thin annulus that we added to the circle.

• Thanks , I got it. I was thinking on the same lines as well – Shashaank Sep 12 '16 at 21:19

I like the integral formulation better. The main idea is that the disk of radius $s$ is made up of the circles of radius $r$ for each $0 \leq r \leq s$, and to argue that

$$\text{Area}(\bullet) = \int_0^s \text{Perimeter}(\circ) \, \mathrm{d} r$$

where $\bullet$ denotes the disk of radius $s$ and $\circ$ the circle of radius $r$.

This is done by looking at the integral formula for area:

$$\text{Area}(\bullet) = \iint_\bullet 1 \, \mathrm{d}A$$

where $\mathrm{d}A$ is the area form; e.g. $\mathrm{d}A = \mathrm{d}x \mathrm{d}y$.

My first goal is to find another coordinate so that I can write $\mathrm{d}A = \text{something} \, \mathrm{d}r$. Polar coordinates are familiar, so I'll use those. We know or can argue or can calculate that $\mathrm{d}A = r\, \mathrm{d}r \mathrm{d} \theta$.

So, we can break the double integral up into an iterated integral, and recognize $r \, \mathrm{d} \theta$ measures arclength, to get

$$\iint_\bullet 1 \, \mathrm{d}A = \int_0^s \left( \int_\circ r \, \mathrm{d} \theta \right) \mathrm{d} r = \int_0^s \text{Perimeter}(\circ) \, \mathrm{d} r$$

as desired.

The reason I prefer this approach is that it generalizes better. In this specific example, it's not too hard to rationalize that $r$ varies at unit rate and perpendicularly to each circle, so that $\mathrm{d}A = \mathrm{d}r \mathrm{d}(\text{arclength})$, but in more complicated examples, it's often easier to just compute the right formula and compute the resulting integrals rather than guess at a 'nicer' set of variables to use that breaks the area form apart in a nice manner.