Find the orthogonal trajectories of the family of curves. Sketch several members of each family<\/p>\n\n\n\n
y = mx<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To find the orthogonal trajectories of the family of curves given by the equation:<\/p>\n\n\n\n [ The equation ( y = mx ) represents a family of straight lines with different slopes ( m ). For each value of ( m ), you get a different line that passes through the origin. The slope of the line is represented by ( m ), where ( m ) can be any constant.<\/p>\n\n\n\n The given equation can be written as:<\/p>\n\n\n\n [ This represents a family of lines with slope ( m ), which vary depending on the value of ( m ).<\/p>\n\n\n\n To find the orthogonal trajectories, we need to determine the slope of the given family of curves. We differentiate the equation ( y = mx ) implicitly with respect to ( x ). Since ( m ) is constant, the derivative of ( y = mx ) with respect to ( x ) is:<\/p>\n\n\n\n [ So, the slope of the given family of curves at any point on the line is ( m ).<\/p>\n\n\n\n The orthogonal trajectories of a family of curves have slopes that are the negative reciprocal of the slopes of the given curves. Therefore, if the slope of the given curve is ( m ), the slope of the orthogonal trajectory will be:<\/p>\n\n\n\n [ To find the equation of the orthogonal trajectories, we replace ( m ) with ( -\\frac{1}{m} ). However, instead of directly using ( m ), we know that the relationship between ( x ) and ( y ) for the orthogonal trajectory will be of the form ( y = -\\frac{1}{m}x + C ), where ( C ) is a constant.<\/p>\n\n\n\n The family of orthogonal curves is a family of curves with slope ( -1\/m ). These are families of lines with slopes that are the negative reciprocals of the slopes of the original family of lines<\/strong>. The equation of the orthogonal trajectory will thus be:<\/p>\n\n\n\n [ The family of curves ( y = mx ) consists of straight lines through the origin with varying slopes, and the family of orthogonal trajectories consists of hyperbolas or curves with decreasing and increasing slopes. Both families intersect at right angles, and each member of the orthogonal trajectory family intersects a member of the original family at a 90-degree angle.<\/p>\n\n\n\n The original family is a set of lines passing through the origin, and the orthogonal family is a set of curves that tend to be perpendicular to these lines, resulting in a set of hyperbolic curves.<\/p>\n\n\n\n In conclusion, the orthogonal trajectories to the family ( y = mx ) are represented by the family of curves:<\/p>\n\n\n\n [ where ( C ) is a constant. These curves are hyperbolas that are perpendicular to the lines ( y = mx ).<\/p>\n","protected":false},"excerpt":{"rendered":" Find the orthogonal trajectories of the family of curves. Sketch several members of each family y = mx The correct answer and explanation is : To find the orthogonal trajectories of the family of curves given by the equation: [y = mx] Step 1: Understanding the family of curves The equation ( y = mx […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","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":"","ast-disable-related-posts":"","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":[25],"tags":[],"class_list":["post-203014","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/203014","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=203014"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/203014\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=203014"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=203014"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=203014"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
y = mx
]<\/p>\n\n\n\nStep 1: Understanding the family of curves<\/h3>\n\n\n\n
Step 2: Equation of the family of curves<\/h3>\n\n\n\n
y = mx
]<\/p>\n\n\n\nStep 3: Differentiating to find the slope<\/h3>\n\n\n\n
\\frac{dy}{dx} = m
]<\/p>\n\n\n\nStep 4: Orthogonal slope<\/h3>\n\n\n\n
\\text{slope of orthogonal trajectory} = -\\frac{1}{m}
]<\/p>\n\n\n\nStep 5: Equation of the orthogonal trajectory<\/h3>\n\n\n\n
Step 6: Solution<\/h3>\n\n\n\n
y = -\\frac{1}{x} + C
]<\/p>\n\n\n\nStep 7: Sketching the families<\/h3>\n\n\n\n
y = -\\frac{1}{x} + C
]<\/p>\n\n\n\n