# Thread: Derivative of Areas of Shapes related to their perimter

1. ## Derivative of Areas of Shapes related to their perimter

I'm stumped with a question my boss posed to me today. He explained how since the area of a circle is Pi * r^2, its derivative is 2Pi * r, the same as it's circumference (due to the infinitesimal change in area equaling the circumference).

Now, take a square.
It's area is x^2
derivative = 2x
The perimeter of a square is 4x.
So why doesn't this Area Derivative/Perimeter relationship hold true for squares as it does circles?

2. ## Re: Derivative of Areas of Shapes related to their perimter

Originally Posted by noclist
I'm stumped with a question my boss posed to me today. He explained how since the area of a circle is Pi * r^2, its derivative is 2Pi * r, the same as it's circumference (due to the infinitesimal change in area equaling the circumference).

Now, take a square.
It's area is x^2
derivative = 2x
The perimeter of a square is 4x.
So why doesn't this Area Derivative/Perimeter relationship hold true for squares as it does circles?
Interesting question! The answer lies in the fact that you're expressing your areas with respect to different parameters: in the circle case, you use the radius, and in the square case, you use sidelength. However, there is an analogous construction for the square which does hold!

Consider the lines from the centre of the square to the midpoint of each edge. These clearly have lengths $r=x/2$, and partitions the square into 4 smaller squares, each with sidelength $r$, and area $r^2$. The area of the original square is $A=x^2=4r^2$, and taking the derivative with respect to $r$ shows that $\frac{dA}{dr}=8r=8\left(\frac{x}{2}\right)=4x=P$, as you might expect.

3. ## Re: Derivative of Areas of Shapes related to their perimter

Ah, so the area of a square is expressed with sidelength, which is a parameter analogous to diameter, not radius. Diameter = 2 * radius, which is why the perimeter of a square is 2 times the derivative of the area of a square as opposed to equal to the area. Thanks for the quick reply!