# Perimeter of a Circle is $$2\pi r$$

From parametric or vector form (easiest)

$$\begin{pmatrix}x \\y \end{pmatrix}=\begin{pmatrix} r\cos{t} \\r \sin{t} \end{pmatrix} \\ s=\int_{0}^{2\pi}\sqrt{\left ( \frac{dx}{dt} \right )^2+\left ( \frac{dy}{dt} \right )^2} dt \\ =\int_{0}^{2\pi}\sqrt{\left (-r\sin{t} \right)^2+\left ( r\cos{t} \right)^2} dt \\ =\int_{0}^{2\pi}\sqrt{r\cos^2{t}+r\sin^2{t}} dt \\ =\int_{0}^{2\pi}\sqrt{r} dt \\ =\left [ rt\right ]_0^{2\pi} \\ =r2\pi-0 \\ \therefore s=2\pi r$$

From Cartesian Form

$$x^2+y^2=r^2 \\ \Rightarrow y=\pm sqrt(r^2-x^2)$$

If restricted to only the positive square root, this plots a semi-circle that is always positive and vice versa if it were the negative square root. So applying the arc length formula for one will only give half the circumference of a circle.

$$\frac{s}{2}=\int_{-r}^{r}\sqrt{1+\left ( \frac{dy}{dx} \right )^2} dx \\ =\int_{-r}^{r}\sqrt{1+\left ( \frac{-x}{\sqrt{r^2-x^2}} \right )^2} dx \\ =\int_{-r}^{r}\sqrt{1+\frac{x^2}{r^2-x^2}} dx \\ =\int_{-r}^{r}\sqrt{\frac{r^2-x^2}{r^2-x^2}+\frac{x^2}{r^2-x^2}} dx \\ =\int_{-r}^{r}\sqrt{\frac{r^2}{r^2-x^2}} dx \\ =r\int_{-r}^{r}\sqrt{\frac{1}}{\sqrt{r^2-x^2}} dx \\ =r\left[\arcsin{\frac{1}{r}}\right]_{-r}^r \\ =r \arcsin{(1/r)}-r \arcsin{(-1/r)} \\ =\frac{\pi r}{2}--\frac{\pi r}{2} \\ =\frac{\pi r}{2}+\frac{\pi r}{2} \\ \frac{s}{2}=\pi r \\$$

Double perimeter of semi-circle to get circumference of circle.

$$\therefore s=2\pi r$$

From Polar Form

$$r=r \\ s=\int_0^{2\pi} \sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2} \\ =\int_0^{2\pi} \sqrt{r^2+0^2} d\theta \\ =\int_0^{2\pi} \sqrt{r^2} d\theta \\ =\int_0^{2\pi} r d\theta \\ =\left [ r\theta \right ]_0^{2\pi} \\ 2\pi r- 0 \\ \therefore s=2\pi r$$

Note by Jack Han
3 years, 5 months ago

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