{"id":1794,"date":"2020-10-10T20:34:04","date_gmt":"2020-10-10T20:34:04","guid":{"rendered":"http:\/\/physicscatalyst.com\/article\/?p=1794"},"modified":"2021-06-05T13:45:19","modified_gmt":"2021-06-05T13:45:19","slug":"brachistochrone-problem","status":"publish","type":"post","link":"https:\/\/physicscatalyst.com\/graduation\/brachistochrone-problem\/","title":{"rendered":"Brachistochrone Problem"},"content":{"rendered":"\n

Brachistochrone Problem Solution<\/h2>\n\n\n\n

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?<\/p>\n\n\n\n

Solution<\/strong><\/h3>\n\n\n\n

Suppose $v$ is the speed<\/a> 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<\/p>\n\n\n\n

\\({t_{12}} = \\int\\limits_1^2 {\\frac{{ds}}{v}} \\)<\/p>\n\n\n\n

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}\\]<\/p>\n\n\n\n

Where,<\/p>\n\n\n\n

\\( f=\\left ( \\frac{1+\\dot{y}^{2}}{2gx} \\right )^{\\frac{1}{2}} \\)<\/p>\n\n\n\n

For \\({t_{12}}\\) to be minimum equation<\/p>\n\n\n\n

\\(\\frac{d}{{dx}}\\left( {\\frac{{\\partial f}}{{\\partial \\dot y}}} \\right) – \\frac{{\\partial f}}{{\\partial y}} = 0\\)<\/p>\n\n\n\n

must be satisfied. From expression for \\(f\\) we find that<\/p>\n\n\n\n

\\(\\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}\\)<\/p>\n\n\n\n

Where c\u2019 is the constant of integration.
Since c\u2019 is a constant we can also write<\/p>\n\n\n\n

\\(\\frac{{{{\\dot y}^2}}}{{x(1 + {{\\dot y}^2})}} = c\\)<\/p>\n\n\n\n

where c is also a constant. On integrating above equation we find<\/p>\n\n\n\n

\\(\\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}\\)<\/p>\n\n\n\n

Putting \\(\\frac{1}{c} = 2a\\) , and on integration we get<\/p>\n\n\n\n

\\(\\int dy=\\int \\frac{x}{\\sqrt{2ax-x^{2}}}dx\\)<\/p>\n\n\n\n

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

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

\\( y=acos^{-1}(1-\\frac{x}{a})-(2ax-x^{2})^{1\/2}\\)<\/p>\n\n\n\n

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.<\/p>\n\n\n\n

\"Brachistochrone\"<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Question If a particle falls from rest under the influence of gravity from 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<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[18],"tags":[],"class_list":["post-1794","post","type-post","status-publish","format-standard","hentry","category-classical-mechanics"],"yoast_head":"\nBrachistochrone Problem - Learn about education and B.Sc. 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