{"id":8109,"date":"2021-11-13T23:27:17","date_gmt":"2021-11-13T17:57:17","guid":{"rendered":"https:\/\/mathemerize.com\/?p=8109"},"modified":"2021-11-14T00:33:41","modified_gmt":"2021-11-13T19:03:41","slug":"find-the-slope-of-the-normal-to-the-curve-x-a-cos3theta-y-a-sin3theta-at-theta-piover-4","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/find-the-slope-of-the-normal-to-the-curve-x-a-cos3theta-y-a-sin3theta-at-theta-piover-4\/","title":{"rendered":"Find the slope of the normal to the curve x = \\(a cos^3\\theta\\), y = \\(a sin^3\\theta\\) at \\(\\theta\\) = \\(\\pi\\over 4\\)."},"content":{"rendered":"

Solution :<\/h2>\n

We have,\u00a0x = \\(a cos^3\\theta\\), y = \\(a sin^3\\theta\\)<\/p>\n

\\(\\implies\\)\u00a0 \\(dx\\over d\\theta\\) = \\(-3 a cos^2\\theta sin\\theta\\),\u00a0 \\(dy\\over d\\theta\\) = \\(3 a sin^2\\theta cos\\theta\\)<\/p>\n

Now, \\(dy\\over dx\\)\u00a0 = \\(dy\/d\\theta\\over dx\/d\\theta\\)<\/p>\n

\\(\\implies\\)\u00a0 \\(dy\\over dx\\) = -\\(tan\\theta\\)<\/p>\n

\\(\\therefore\\)\u00a0 Slope of the normal<\/a> at any point on the curve = \\(-1\\over dy\/dx\\) = \\(cot\\theta\\)<\/p>\n

Hence, the slope of the normal at \\(\\theta\\) = \\(\\pi\\over 4\\) = \\(cot\\pi\\over 4\\) = 1.<\/p>\n

\n

Similar Questions<\/h3>\n

Find the slope of normal to the curve x = 1 \u2013 \\(a sin\\theta\\), y = \\(b cos^2\\theta\\) at \\(\\theta\\) = \\(\\pi\\over 2\\).<\/a><\/p>\n

Show that the tangents to the curve y = \\(2x^3 \u2013 3\\) at the points where x =2 and x = -2 are parallel.<\/a><\/p>\n

Find the equations of the tangent and the normal at the point \u2018t\u2019 on the curve x = \\(a sin^3 t\\), y = \\(b cos^3 t\\).<\/a><\/p>\n

Find the equation of the normal to the curve y = \\(2x^2 + 3 sin x\\) at x = 0.<\/a><\/p>\n

Find the angle between the curves xy = 6 and \\(x^2 y\\) =12.<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : We have,\u00a0x = \\(a cos^3\\theta\\), y = \\(a sin^3\\theta\\) \\(\\implies\\)\u00a0 \\(dx\\over d\\theta\\) = \\(-3 a cos^2\\theta sin\\theta\\),\u00a0 \\(dy\\over d\\theta\\) = \\(3 a sin^2\\theta cos\\theta\\) Now, \\(dy\\over dx\\)\u00a0 = \\(dy\/d\\theta\\over dx\/d\\theta\\) \\(\\implies\\)\u00a0 \\(dy\\over dx\\) = -\\(tan\\theta\\) \\(\\therefore\\)\u00a0 Slope of the normal at any point on the curve = \\(-1\\over dy\/dx\\) = \\(cot\\theta\\) Hence, the …<\/p>\n

Find the slope of the normal to the curve x = \\(a cos^3\\theta\\), y = \\(a sin^3\\theta\\) at \\(\\theta\\) = \\(\\pi\\over 4\\).<\/span> Read More »<\/a><\/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":"default","ast-global-header-display":"","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":""},"categories":[66,43],"tags":[],"yoast_head":"\nFind the slope of the normal to the curve x = \\(a cos^3\\theta\\), y = \\(a sin^3\\theta\\) at \\(\\theta\\) = \\(\\pi\\over 4\\).<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/find-the-slope-of-the-normal-to-the-curve-x-a-cos3theta-y-a-sin3theta-at-theta-piover-4\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Find the slope of the normal to the curve x = \\(a cos^3\\theta\\), y = \\(a sin^3\\theta\\) at \\(\\theta\\) = \\(\\pi\\over 4\\).\" \/>\n<meta property=\"og:description\" content=\"Solution : We have,\u00a0x = (a cos^3theta), y = (a sin^3theta) (implies)\u00a0 (dxover dtheta) = (-3 a cos^2theta sintheta),\u00a0 (dyover dtheta) = (3 a sin^2theta costheta) Now, (dyover dx)\u00a0 = (dy\/dthetaover dx\/dtheta) (implies)\u00a0 (dyover dx) = -(tantheta) (therefore)\u00a0 Slope of the normal at any point on the curve = (-1over dy\/dx) = (cottheta) Hence, the … Find the slope of the normal to the curve x = (a cos^3theta), y = (a sin^3theta) at (theta) = (piover 4). 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