{"id":233081,"date":"2025-06-12T14:17:48","date_gmt":"2025-06-12T14:17:48","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=233081"},"modified":"2025-06-12T14:17:51","modified_gmt":"2025-06-12T14:17:51","slug":"find-dy-dx-and-d2y-dx2-without-eliminating-the-parameter","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/12\/find-dy-dx-and-d2y-dx2-without-eliminating-the-parameter\/","title":{"rendered":"Find dy\/dx and d2y\/dx2 without eliminating the parameter."},"content":{"rendered":"\n

Find dy\/dx and d2y\/dx2 without eliminating the parameter. x=1-t^(2), y=1+t<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Given the parametric equations:x=1\u2212t2,y=1+tx = 1 – t^2,\\quad y = 1 + tx=1\u2212t2,y=1+t<\/p>\n\n\n\n

We are to find:<\/p>\n\n\n\n

    \n
  1. dydx\\frac{dy}{dx}dxdy\u200b \u2014 the first derivative of yyy with respect to xxx,<\/li>\n\n\n\n
  2. d2ydx2\\frac{d^2y}{dx^2}dx2d2y\u200b \u2014 the second derivative of yyy with respect to xxx,<\/li>\n<\/ol>\n\n\n\n

    without eliminating the parameter<\/strong>.<\/p>\n\n\n\n

    \n\n\n\n

    Step 1: First Derivative dydx\\frac{dy}{dx}dxdy\u200b<\/h3>\n\n\n\n

    Using parametric differentiation, the first derivative is given by:dydx=dydtdxdt\\frac{dy}{dx} = \\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}dxdy\u200b=dtdx\u200bdtdy\u200b\u200b<\/p>\n\n\n\n

    Differentiate xxx and yyy with respect to ttt:<\/p>\n\n\n\n

      \n
    • dxdt=ddt(1\u2212t2)=\u22122t\\frac{dx}{dt} = \\frac{d}{dt}(1 – t^2) = -2tdtdx\u200b=dtd\u200b(1\u2212t2)=\u22122t<\/li>\n\n\n\n
    • dydt=ddt(1+t)=1\\frac{dy}{dt} = \\frac{d}{dt}(1 + t) = 1dtdy\u200b=dtd\u200b(1+t)=1<\/li>\n<\/ul>\n\n\n\n

      So,dydx=1\u22122t=\u221212t\\frac{dy}{dx} = \\frac{1}{-2t} = -\\frac{1}{2t}dxdy\u200b=\u22122t1\u200b=\u22122t1\u200b<\/p>\n\n\n\n

      \n\n\n\n

      Step 2: Second Derivative d2ydx2\\frac{d^2y}{dx^2}dx2d2y\u200b<\/h3>\n\n\n\n

      The second derivative in parametric form is given by:d2ydx2=ddt(dydx)\u00f7dxdt\\frac{d^2y}{dx^2} = \\frac{d}{dt}\\left( \\frac{dy}{dx} \\right) \\div \\frac{dx}{dt}dx2d2y\u200b=dtd\u200b(dxdy\u200b)\u00f7dtdx\u200b<\/p>\n\n\n\n

      We already have dydx=\u221212t\\frac{dy}{dx} = -\\frac{1}{2t}dxdy\u200b=\u22122t1\u200b, so compute its derivative with respect to ttt:ddt(\u221212t)=12t2\\frac{d}{dt}\\left( -\\frac{1}{2t} \\right) = \\frac{1}{2t^2}dtd\u200b(\u22122t1\u200b)=2t21\u200b<\/p>\n\n\n\n

      Now divide by dxdt=\u22122t\\frac{dx}{dt} = -2tdtdx\u200b=\u22122t:d2ydx2=12t2\u22122t=\u221214t3\\frac{d^2y}{dx^2} = \\frac{\\frac{1}{2t^2}}{-2t} = -\\frac{1}{4t^3}dx2d2y\u200b=\u22122t2t21\u200b\u200b=\u22124t31\u200b<\/p>\n\n\n\n

      \n\n\n\n

      Final Answers:<\/h3>\n\n\n\n

      dydx=\u221212t,d2ydx2=\u221214t3\\frac{dy}{dx} = -\\frac{1}{2t},\\quad \\frac{d^2y}{dx^2} = -\\frac{1}{4t^3}dxdy\u200b=\u22122t1\u200b,dx2d2y\u200b=\u22124t31\u200b<\/p>\n\n\n\n

      \n\n\n\n

      Explanation<\/h3>\n\n\n\n

      When working with parametric equations, such as x=1\u2212t2x = 1 – t^2x=1\u2212t2 and y=1+ty = 1 + ty=1+t, we treat both xxx and yyy as functions of a third variable ttt, known as the parameter. Instead of expressing yyy directly in terms of xxx, we differentiate each with respect to ttt, and use these derivatives to find the derivatives of yyy with respect to xxx.<\/p>\n\n\n\n

      To find the first derivative dydx\\frac{dy}{dx}dxdy\u200b, we apply the chain rule of calculus. Since dy\/dx=(dy\/dt)\/(dx\/dt)dy\/dx = (dy\/dt)\/(dx\/dt)dy\/dx=(dy\/dt)\/(dx\/dt), we compute the derivatives:<\/p>\n\n\n\n

        \n
      • dy\/dt=1dy\/dt = 1dy\/dt=1, since y=1+ty = 1 + ty=1+t,<\/li>\n\n\n\n
      • dx\/dt=\u22122tdx\/dt = -2tdx\/dt=\u22122t, because x=1\u2212t2x = 1 – t^2x=1\u2212t2.<\/li>\n<\/ul>\n\n\n\n

        Substituting, we get dydx=1\u22122t=\u221212t\\frac{dy}{dx} = \\frac{1}{-2t} = -\\frac{1}{2t}dxdy\u200b=\u22122t1\u200b=\u22122t1\u200b.<\/p>\n\n\n\n

        To find the second derivative d2ydx2\\frac{d^2y}{dx^2}dx2d2y\u200b, we use the formula:d2ydx2=ddt(dydx)\u00f7dxdt\\frac{d^2y}{dx^2} = \\frac{d}{dt}\\left( \\frac{dy}{dx} \\right) \\div \\frac{dx}{dt}dx2d2y\u200b=dtd\u200b(dxdy\u200b)\u00f7dtdx\u200b<\/p>\n\n\n\n

        This involves differentiating \u221212t-\\frac{1}{2t}\u22122t1\u200b with respect to ttt, which gives 12t2\\frac{1}{2t^2}2t21\u200b, and then dividing this result by \u22122t-2t\u22122t, the derivative of xxx. Thus:d2ydx2=12t2\u22122t=\u221214t3\\frac{d^2y}{dx^2} = \\frac{\\frac{1}{2t^2}}{-2t} = -\\frac{1}{4t^3}dx2d2y\u200b=\u22122t2t21\u200b\u200b=\u22124t31\u200b<\/p>\n\n\n\n

        This process allows us to compute the derivatives directly from the parametric equations without needing to eliminate the parameter.<\/p>\n\n\n\n

        \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

        Find dy\/dx and d2y\/dx2 without eliminating the parameter. x=1-t^(2), y=1+t The Correct Answer and Explanation is: Given the parametric equations:x=1\u2212t2,y=1+tx = 1 – t^2,\\quad y = 1 + tx=1\u2212t2,y=1+t We are to find: without eliminating the parameter. Step 1: First Derivative dydx\\frac{dy}{dx}dxdy\u200b Using parametric differentiation, the first derivative is given by:dydx=dydtdxdt\\frac{dy}{dx} = \\frac{\\frac{dy}{dt}}{\\frac{dx}{dt}}dxdy\u200b=dtdx\u200bdtdy\u200b\u200b Differentiate xxx […]<\/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-233081","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233081","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=233081"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233081\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=233081"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=233081"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=233081"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}