## derivative

An enlightening discussion about pi and tau.

### derivative

Any even notice the first derivative of Area with respect to the radius is the circumference?

A = pi R^2

DA = 2 pi R DR

c = 2 pi R

That not as pretty if using Tau.
Kindergarten

Posts: 1
Joined: Thu Jul 07, 2011 2:08 pm

### Re: derivative

This actually still works with tau.

$$A=\frac{1}{2}\tau r^2$$
$$\frac{dA}{dr}=\tau r$$

and the Circumference is $$C=\tau r$$

It should work regardless of which circle constant is used
Math - It's in you to give.

SpikedMath

Posts: 133
Joined: Mon Feb 07, 2011 1:31 am

### Re: derivative

SpikedMath wrote:This actually still works with tau.

$$A=\frac{1}{2}\tau r^2$$
$$\frac{dA}{dr}=\tau r$$

and the Circumference is $$C=\tau r$$

It should work regardless of which circle constant is used

If you take the inverse of the derivative on the circumference, you will realise why $$\tau$$ is the better one to use in the case of area.

$$\int x y \ dx = {x y^2 \over 2}$$

Filling for $$x = \tau$$ and $$y = r$$,

$$A = \int C(r) \ dr = \int \tau r \ dr = {\tau r^2 \over 2} = \pi r^2$$

$$\pi$$ does not take the meat pie when it comes to the area containing all that juicy meat. Indeed, $$\tau$$ is victorious.
rdococ
Elementary School

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Joined: Tue Sep 10, 2013 11:23 am

### Re: derivative

Area comes from circumference, not the other way around, in the same way that two dimensions come from one dimension. Archimedes' method can be used to find circumference, and area is found by integrating circumference with respect to radius, yielding 1/2 C r or by splitting the polygons used in Archimedes' method into TRIANGLES, which have an area of 1/2 bh. Notice that finding the areas of the polygons using this method results in 1/2 Pa, which for a circle is 1/2 Cr. This also shows again that area comes from perimeter, since you have to find the perimeter of a polygon first before you measure its area.
τ>π
1=0
Mathlete

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