Topic: Math - Blog Posts

Why is the Circumference of a Circle the Derivative of its Area?: A Visual Explanation

Circumference = 2πr

Area = πr^2

You may have noticed that the circumference of circle is the derivative of its area with respect to the radius. Similarly, a sphere’s surface area (SA = 4πr^2) is the derivative of its volume (Volume = 4/3πr^3). This isn’t a coincidence! But why? And is there an intuitive way of thinking about it?

Calculus refresher: Finding the derivative of a function is finding its rate of change. For example, consider the function y = x^2. The derivative of this function is 2x, which describes how much, in terms of x, y changes when x changes. Integration is the reverse process of derivation. Finding the integral of a function first considers the function a rate of change. Then, by multiplying it by infinitesimally small increments of x from a lower bound to an upper bound, the process of integration computes the definite integral, a new function whose derivative was the original function. Think of a car moving at a velocity over time. The rate of change of the velocity is the cars acceleration. Additionally, if you multiply the velocity by how much time has passed, you get the total distance traveled by the car. Therefore, acceleration is velocity’s derivative and distance traveled is velocity’s integral.

So what is the rate of change of a circle? Consider a circle with the radius r. If you increase the radius by ∆r, the area of the new circle is πr^2 + the area of the added ring. The ring’s area is 2πr (which is the rings length) * ∆r (the ring’s height). This is indicated by the first gif, in which the new rings have the length ∆r. To find the rate of change, we take the limit as ∆r goes to 0. The limit as ∆r goes to 0 of 2πr∆r is simply 2πr! 

Let’s find the area of a circle with radius r by integrating its 2πr, its circumference. For the lower bound of our integration, think of the smallest circle we can make—a circle with radius 0. The largest circle we can make is a circle with radius r—our upper bound. We draw our smallest circle (radius 0), and then continuously add tiny rings to it by increasing r and drawing another circle, keeping the change of r as tiny as possible. We stop when r has reached our upper bound. As the second gif demonstrates, we are left with what is pretty much a filled circle! We went from 1 dimensional lines, to a 2D figure with an area of πr^2. This a fun way of visualizing the integration of 2πr from 0 to r! 

So, based on this explanation, can you figure out a way to visualize why the surface area of a sphere is the derivative of its volume? Hint: jawbreakers (or onions, alternatively)!