Cycloid Curve

• A cycloid curve is a curve that is the path traced by a circle rolling.
• 3Blue1Brown video on this is good

Derivation of cycloid curve from calculus of variations and Euler-Lagrange Equation

The speed of the particle

$$V=\frac{ds}{dt}\to dt=\frac{ds}{V}$$

The total time is the integral over $dt$

$$\tau =\underset{A}{\overset{B}{\int }}\frac{ds}{V}$$

What is $ds$

$$ds=\sqrt{d{x}^2+d{y}^2}=dx\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}=dx\sqrt{1+y^{\prime 2}}$$ $$V\to T=\frac{1}{2m{v}^2}$$ $$u=-mgx$$

$T+U$ is conserved because gravity is conservative

$$\frac{1}{2}m{V}^2-mgx=\text{constant}=0$$ $$\frac{1}{2}m{V}^2=mgx⟹\frac{1}{2v^2}=gx$$ $$V=\sqrt{2gx}$$ $$\tau =\underset{A}{\overset{B}{\int }}\frac{ds}{V}=\underset{0}{\overset{x_B}{\int }}\frac{\sqrt{1+y^{\prime 2}}dx}{\sqrt{2gx}}dx$$

Compare to $J=\int F(y(x,y^{\prime }(x),x))dx$

$$F=\sqrt{\frac{1+y^{\prime 2}}{2gx}}$$

Use Euler-Lagrange Equation

$$\frac{d}{dx}\frac{\partial F}{\partial y^{\prime }}-\frac{\partial F}{\partial y}=0$$ $$\frac{\partial F}{\partial y}=0;\frac{d}{dx}\frac{\partial F}{\partial y^{\prime }}=0$$ $$\frac{\partial F}{\partial y^{\prime }}\text{ must be constant}$$ $$\frac{\partial F}{\partial y^{\prime }}=\frac{d}{d{y}^{\prime }}\left(\frac{1+y^{\prime 2}}{2gx}\right)=\frac{1}{\sqrt{2g}}\frac{d}{d{y}^{\prime }}{\left(\frac{1+y^{\prime 2}}{x}\right)}^{1/2}$$ $$=\frac{1}{2}{\left(\frac{1+y^{\prime 2}}{x}\right)}^{-1/2}\left(\frac{2y^{\prime }}{x}\right)$$ $$={\left(\frac{x}{1+y^{\prime 2}}\right)}^{1/2}\left(\frac{y^{\prime }}{x}\right)=\text{constant}$$

Square both sides

$$\frac{x}{1+y^{\prime 2}}\frac{y^{\prime 2}}{x^2}={\text{constant}}^2=\frac{1}{2a}$$ $$2a{y}^{\prime 2}=(1+y^{\prime 2})x=x+x{y}^2$$

2ay’^{2}-xy’^{2} = x

y^{2}(2a-x) = x

y’^{2} = \frac{x}{2a-x} \left( \frac{x}{x} \right) = \frac{x^{2}}{2ax-x^{2}}

\frac{dy}{dx} = \frac{x}{\sqrt{ 2ax-x^{2} }}

\int  \, dy  = \int \frac{x}{\sqrt{ 2ax-x^{2} }} \, dx 

x = a(1-\cos(\theta)) ; \ dx = a\sin \theta d\theta

y (x) = \int \frac{a(1-\cos \theta)a\sin\theta}{\sqrt{ 2a^{2}(1-\cos\theta)-a^{2}(1-\cos\theta)^{2} }} \, d\theta

$$Denominatorbecomes$$

2a^{2} - 2a^{2}\cos\theta -a^{2} + 2a^{2}\cos\theta-a^{2}\cos ^{2}\theta

a^{2} - a^{2}\cos ^{2}\theta

y(x) = \int \frac{a^{2}\sin\theta(1-\cos\theta)}{\sqrt{ a^{2}-a^{2}\cos \theta }} , d\theta= \int a(1-\cos(\theta)) , d\theta

= y(x) = a(\theta-\sin\theta)+ \text{constant}

x=a(1-\cos(\theta))

$$>Thisisthepathofacycloidcurve!$$