Find the Parametric equation of the tangent vector of curve r(t) = t^2 i + 2t^3 j + 3t k at t = 1.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To find the parametric equation of the tangent vector<\/strong> of the curve The velocity vector is the derivative of (\\mathbf{r}(t)):<\/p>\n\n\n\n [ Differentiating component-wise:<\/p>\n\n\n\n Thus, the velocity (tangent) vector is:<\/p>\n\n\n\n [ Substituting ( t = 1 ):<\/p>\n\n\n\n [ [ The tangent line at ( t = 1 ) passes through the point ( \\mathbf{r}(1) ) with direction ( \\mathbf{v}(1) ).<\/p>\n\n\n\n First, find ( \\mathbf{r}(1) ):<\/p>\n\n\n\n [ The parametric equations of the tangent line:<\/p>\n\n\n\n [ where ( s ) is a parameter along the tangent direction.<\/p>\n\n\n\n A parametric equation of the tangent vector<\/strong> provides a way to describe a tangent line at a specific point on a curve. In vector form, the curve is given by ( \\mathbf{r}(t) ), and the tangent vector at any point is obtained by differentiating ( \\mathbf{r}(t) ). This derivative, ( \\mathbf{v}(t) ), represents the instantaneous rate of change of ( \\mathbf{r}(t) ) and is thus tangent to the curve at each ( t ).<\/p>\n\n\n\n For this problem, we first computed ( \\mathbf{v}(t) ) by differentiating each component of ( \\mathbf{r}(t) ). This gives a vector function ( \\mathbf{v}(t) = (2t, 6t^2, 3) ), which describes how the position changes over time.<\/p>\n\n\n\n To find the specific tangent vector at ( t = 1 ), we evaluated ( \\mathbf{v}(1) ), yielding ( (2,6,3) ). This vector represents the direction of the tangent line.<\/p>\n\n\n\n Next, we found the position of the curve at ( t = 1 ), which is ( \\mathbf{r}(1) = (1,2,3) ). Using the point-direction equation of a line<\/strong>, we wrote parametric equations for the tangent line using the form:<\/p>\n\n\n\n [ Substituting ( (1,2,3) ) as the point and ( (2,6,3) ) as the direction, we derived the parametric equations ( x = 1 + 2s ), ( y = 2 + 6s ), ( z = 3 + 3s ). These describe all points along the tangent line at ( t = 1 ).<\/p>\n","protected":false},"excerpt":{"rendered":" Find the Parametric equation of the tangent vector of curve r(t) = t^2 i + 2t^3 j + 3t k at t = 1. The Correct Answer and Explanation is : To find the parametric equation of the tangent vector of the curve[\\mathbf{r}(t) = t^2 \\mathbf{i} + 2t^3 \\mathbf{j} + 3t \\mathbf{k}]at ( t = […]<\/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-187117","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187117","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=187117"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187117\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187117"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187117"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187117"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
[
\\mathbf{r}(t) = t^2 \\mathbf{i} + 2t^3 \\mathbf{j} + 3t \\mathbf{k}
]
at ( t = 1 ), follow these steps:<\/p>\n\n\n\nStep 1: Compute the Velocity Vector<\/strong><\/h3>\n\n\n\n
\\mathbf{v}(t) = \\frac{d}{dt} (t^2 \\mathbf{i} + 2t^3 \\mathbf{j} + 3t \\mathbf{k})
]<\/p>\n\n\n\n\n
\\mathbf{v}(t) = (2t) \\mathbf{i} + (6t^2) \\mathbf{j} + 3 \\mathbf{k}
]<\/p>\n\n\n\nStep 2: Evaluate at ( t = 1 )<\/strong><\/h3>\n\n\n\n
\\mathbf{v}(1) = (2(1)) \\mathbf{i} + (6(1)^2) \\mathbf{j} + 3 \\mathbf{k}
]<\/p>\n\n\n\n
= 2 \\mathbf{i} + 6 \\mathbf{j} + 3 \\mathbf{k}
]<\/p>\n\n\n\nStep 3: Parametric Equations<\/strong><\/h3>\n\n\n\n
\\mathbf{r}(1) = (1^2) \\mathbf{i} + 2(1)^3 \\mathbf{j} + 3(1) \\mathbf{k} = 1 \\mathbf{i} + 2 \\mathbf{j} + 3 \\mathbf{k}
]<\/p>\n\n\n\n
x = 1 + 2s, \\quad y = 2 + 6s, \\quad z = 3 + 3s
]<\/p>\n\n\n\nExplanation (300 Words)<\/strong><\/h3>\n\n\n\n
(x, y, z) = (x_0, y_0, z_0) + s(v_x, v_y, v_z)
]<\/p>\n\n\n\n