The Notes: How can we prove that the circumference of a circle is proportional to its diameter?

Tuesday, May 15, 2012

How can we prove that the circumference of a circle is proportional to its diameter?

The parametric equation of a circle of radius $\displaystyle{r}$ is given by

$\displaystyle{x}={r}{\cos{{\left({t}\right.}}}$ $\displaystyle{\)}$

$\displaystyle{y}={r}{\sin{{\left({t}\right)}}}$

where $\displaystyle{t}={\left[{0},\omega\right]}$ and $\displaystyle\omega$ is the total radians or degrees in a circle. We assume that given any circle the total degrees is always the same, therefore $\displaystyle\omega$ is constant.

To measure the length of a curve, in particular the length of the circumference of our circle, we use the following formula:

$\displaystyle{C}={\int_{{0}}^{\omega}}\sqrt{{{{\left(\frac{{\left.{d}{x}}}{{\left.{d}{t}}}\right)}}^{{2}}+{{\left(\frac{{\left.{d}{y}}}{{\left.{d}{t}}}\right)}}^{{2}}}}{\left.{d}{t}\right.}$

The derivatives $\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{t}}}$ and $\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{t}}}$ are simple to evaluate. We have $\displaystyle\frac{{\left.{d}{x}}}{{\left.{d}{t}}}=-{r}{\sin{{\left({t}\right)}}}$ and $\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{t}}}={r}{\cos{{\left({t}\right)}}}$ . Plugging in the derivatives into the integral,

$\displaystyle{C}={\int_{{0}}^{\omega}}\sqrt{{{{\left(-{r}{\sin{{\left({t}\right)}}}\right)}}^{{2}}+{{\left({r}{\cos{{\left({t}\right)}}}\right)}}^{{2}}}}{\left.{d}{t}\right.}$

$\displaystyle={\int_{{0}}^{\omega}}\sqrt{{{{r}}^{{2}}{{\sin}}^{{2}}{\left({t}\right)}+{{r}}^{{2}}{{\cos}}^{{2}}{\left({t}\right)}}}{\left.{d}{t}\right.}$

By the pythagorean theorem $\displaystyle{{\sin}}^{{2}}{\left({t}\right)}+{{\cos}}^{{2}}{\left({t}\right)}={1}$ . Therefore

$\displaystyle{C}={\int_{{0}}^{\omega}}\sqrt{{{{r}}^{{2}}}}{\left.{d}{t}\right.}={\int_{{0}}^{\omega}}{r}{\left.{d}{t}\right.}={r}{\int_{{0}}^{\omega}}{1}{\left.{d}{t}\right.}={r}{\left[\omega-{0}\right]}={r}\omega$

This shows that $\displaystyle{C}={r}\omega={2}{r}\frac{\omega}{{2}}$ , the ratio between the circumference and the diameter of the circle is $\displaystyle\frac{\omega}{{2}}$ regardless the size of the circle.

The Notes

Tuesday, May 15, 2012

How can we prove that the circumference of a circle is proportional to its diameter?

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