## Introduction: Circumference of a Circle

In this instructable we will have some with the almighty and omnipresent pi! In effect we will explore the use of pi in finding the circumference of a circle.

## Step 1: The Equation

To find the circumference we can either use the diameter or the radius of the circle, (see picture) the radius being half the length of the diameter.

Using equation

C=d X �

Or substituting the diameter for the radius

C= 2� X r or C= � X 2r

Using equation

C=d X �

Or substituting the diameter for the radius

C= 2� X r or C= � X 2r

## Step 2: Getting Fancy

If you really want to go all out you could get extremely pancy and throw in some calculus, though if you want the teacher to know you are a mathematical genious/show off just make sure you show your working.

Excerpts from [http://wikipedia www.wikipedia.org]:

The upper half of a circle centered at the origin is the graph of the function f(x) = \sqrt{r

Thus the circle circumference can be calculated as dara:)

c = 2 \int_{-r}

The antiderivative needed to solve this definite integral is the arcsine function:

c = 2r \left[\arcsin\left(\frac{x}{r}\right) \right]_{-r}

Pi (�) is the ratio of the circumference of a circle to its diameter.

Excerpts from [http://wikipedia www.wikipedia.org]:

The upper half of a circle centered at the origin is the graph of the function f(x) = \sqrt{r

^{2-x}2}, where x runs from -r to +r. The circumference (c) of the entire circle can be represented as twice the sum of the lengths of the infinitesimal arcs that make up this half circle. The length of a single infinitesimal part of the arc can be calculated using the Pythagorean formula for the length of the hypotenuse of a rectangular triangle with side lengths dx and f'(x)dx, which gives us \sqrt{(dx)^{2+(f'(x)dx)}2} = \left( \sqrt{1+f'^{2(x)} \right) dx.}Thus the circle circumference can be calculated as dara:)

c = 2 \int_{-r}

^{r \sqrt{1+f'}2(x)}dx = 2 \int_{-r}^{r \sqrt{1+\frac{x}2}{r^{2-x}2}}dx = 2 \int_{-r}^{r \sqrt{\frac{1}{1-\frac{{x}}2}{{r}^{2}}}dx}The antiderivative needed to solve this definite integral is the arcsine function:

c = 2r \left[\arcsin\left(\frac{x}{r}\right) \right]_{-r}

^{{r} = 2r \left[\arcsin(1)-\arcsin(-1) \right] = 2r(\tfrac{\pi}{2}-(-\tfrac{\pi}{2})) = 2\pi r.}Pi (�) is the ratio of the circumference of a circle to its diameter.

## Step 3: Thanks

Have fun learning maths! And please vote, if you do a puppy will suddenly be very happy and will also be polite and cease to urinate in unpleasant places..... Think of the puppies.