Why is it the differentiation of the area of a circle is equal to the perimeter of the circumfrence?

[ChocolateDrop]

It's been bugging me for a while but, why is it when you differentiate the area of a circle: πr² is the perimeter of a circle: 2πr?

How come this pattern is not the same in other shapes like squares? When you differentiate the area square: length* width, you do not get the perimeter 2length + 2width.

Any suggestion to why this is?

My maths teachers, (and other maths teachers) are baffled as well.

Thanks in advance

Edited October 19, 2012 by ChocolateDrop

[Drake Glau]

A square does not have equal distance from any two points through its center. This causes the perimeter and area to be quite different: 4W (or 4L) and LW. When Differentiating the area in this case would require you to pick one variable and force the other to be constant so you would have just a constant of L or W but you need 4 of them, whereas the area of a circle and perimeter of a circle are both equations with the single variable r and are all multiplication (forcing the derivative of all of them to be multiplication).

This is all I have at 8am and I'm not even sure if it's sufficient enough, but it made sense in my head =/

Edited October 19, 2012 by Drake Glau

[aldld]

Try using integration (if your class has covered that) to derive the formula for the area of the circle from the circumference. Intuitively, suppose we have a circle of radius r and let 0 < R < r. Think of dR as a very small change (again just for intuition, not for rigour) in R. At a particular value of R, 2πr dR approximates the area of a thin ring for small enough values of dR.

Now a circle can be though of as being approximated by many of these thin concentric "rings" and indeed as dR becomes very small the number of rings increases and can more closely approximate the actual circle (well, disk technically since it contains its interior). Now, imaging "unrolling" each of these rings into a thin rectangle and arrange them in increasing order by height. As the width of the rectangles becomes finer (which can be described more rigorously in terms of partitions) in the limit this figure approaches the graph of a line passing through the origin with slope 2π from R = 0 to r. The area of the region under this line is equal to the area of the circle.

You could just calculate the area directly, but using integration shows where the derivative comes in. Using the fundamental theorem of calculus, A = integral over [0, R] of 2πR dR = πr²

It's a bit of a long explanation, which could be shortened with the help of a picture

Edit: LaTeX equations never work right on IBS.

Edited October 19, 2012 by aldld

[ChocolateDrop]

A square does not have equal distance from any two points through its center. This causes the perimeter and area to be quite different: 4W (or 4L) and LW. When Differentiating the area in this case would require you to pick one variable and force the other to be constant so you would have just a constant of L or W but you need 4 of them, whereas the area of a circle and perimeter of a circle are both equations with the single variable r and are all multiplication (forcing the derivative of all of them to be multiplication).

This is all I have at 8am and I'm not even sure if it's sufficient enough, but it made sense in my head =/

Interesting food for thought. So a square with length r would have an area of r² and the reason that the first differential of the square 2r is not equal to the perimeter of the square which in actuality would be 4r is because it does not have any two points which are equidistant from its center? Why does the fact that it doesn't have any two points equal distance from the center matter?

[macrofire]

To add on to aldld's point, you might want to see this proof without words. This will explain how the area of a circle and its circumference are related:

[aldld]

A square does not have equal distance from any two points through its center. This causes the perimeter and area to be quite different: 4W (or 4L) and LW. When Differentiating the area in this case would require you to pick one variable and force the other to be constant so you would have just a constant of L or W but you need 4 of them, whereas the area of a circle and perimeter of a circle are both equations with the single variable r and are all multiplication (forcing the derivative of all of them to be multiplication).

This is all I have at 8am and I'm not even sure if it's sufficient enough, but it made sense in my head =/

Interesting food for thought. So a square with length r would have an area of r² and the reason that the first differential of the square 2r is not equal to the perimeter of the square which in actuality would be 4r is because it does not have any two points which are equidistant from its center? Why does the fact that it doesn't have any two points equal distance from the center matter?

Actually, with a square, you can think of a square in the same way as a circle and find that the same principle does in fact apply. But instead of using the side length as the variable, consider the distance r from the centre to the closest point on the edge (Or r would be the radius of a circle inscribed in the square.)

This time, the perimeter is given by 8R, and integrating these perimeters gives A = integral 0 to r of 8R dR = 4r^2, which is indeed the area of the square.

[Peanut Butter Jelly]

First, use the definition of the derivative to get an expression. (Use the definition for h-->0.)

You can then represent this geometrically. In other words, you are finding the difference of the areas between a circle with radius (r+h) and a circle with radius r. As h approaches 0, it gets infinitely close to r. So, in other words, the difference between the areas of a circle with radius (r+h) and r is simply the circumference, or 2r(pi).