# Brachistochrone Problem

## Brachistochrone Problem Solution

Brachistochrone Problem :
If a particle falls from rest under the influence of gravity from a higher to lower point in the minimum time, what is the curve that the particle will follow?

### Solution

Suppose $v$ is the speed of the particle along the curve, then in traversing ds portion of the curve time spent would be $\frac{{ds}}{v}$ so that total time taken by the particle in moving from highest point 1 to lowest point 2 will be

${t_{12}} = \int\limits_1^2 {\frac{{ds}}{v}}$

Suppose vertical distance of fall upto point 2 be x, then from the principle of conservation of energy of the particle we find that
$\begin{array}{*{20}{l}} {\frac{1}{2}m{v^2} = mgx}\\ {or}\\ {v = \sqrt {2gx} }\\ {then}\\ {{t_{12}} = \int\limits_1^2 {\frac{{\sqrt {d{x^2} + d{y^2}} }}{{\sqrt {2gx} }}} dx}\\ {{\rm{ = }}\int\limits_1^2 {\frac{{\sqrt {1 + {{\dot y}^2}} }}{{\sqrt {2gx} }}} dx}\\ {{\rm{ = }}\int\limits_1^2 {fdx} } \end{array}$

Where,

$f=\left ( \frac{1+\dot{y}^{2}}{2gx} \right )^{\frac{1}{2}}$

For ${t_{12}}$ to be minimum equation

$\frac{d}{{dx}}\left( {\frac{{\partial f}}{{\partial \dot y}}} \right) – \frac{{\partial f}}{{\partial y}} = 0$

must be satisfied. From expression for $f$ we find that

$\begin{array}{l}\frac{{\partial f}}{{\partial y}} = 0\\\frac{{\partial f}}{{\partial \dot y}} = \frac{{\dot y}}{{\sqrt {2gx} \sqrt {1 + {{\dot y}^2}} }}\\\frac{d}{{dx}}\left( {\frac{{\dot y}}{{\sqrt {2gx} \sqrt {1 + {{\dot y}^2}} }}} \right) = 0\\or\\\frac{{{{\dot y}^2}}}{{2gx(1 + {{\dot y}^2})}} = c’\end{array}$

Where c’ is the constant of integration.
Since c’ is a constant we can also write

$\frac{{{{\dot y}^2}}}{{x(1 + {{\dot y}^2})}} = c$

where c is also a constant. On integrating above equation we find

$\begin{array}{l}\frac{{{{\dot y}^2}}}{c} = x(1 + {{\dot y}^2})\\or,\\{{\dot y}^2}\left( {\frac{1}{c} – x} \right) = x\\{{\dot y}^2}\left( {\frac{x}{c} – {x^2}} \right) = {x^2}\\\dot y = \frac{x}{{\sqrt {\frac{x}{c} – {x^2}} }}\end{array}$

Putting $\frac{1}{c} = 2a$ , and on integration we get

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

$y=acos^{-1}(1-\frac{x}{a})-(2ax-x^{2})^{1/2}+c”$

where c” is new constant of integration.
In case c” is zero then y will be zero for x=0.
As such the equation

$y=acos^{-1}(1-\frac{x}{a})-(2ax-x^{2})^{1/2}$

And this y represents an inverted cycloid with its base along y-axis and cusp at the origin and is the curve that particle will follow. 