{"id":8079,"date":"2021-11-13T21:28:50","date_gmt":"2021-11-13T15:58:50","guid":{"rendered":"https:\/\/mathemerize.com\/?p=8079"},"modified":"2021-11-14T00:56:33","modified_gmt":"2021-11-13T19:26:33","slug":"find-the-equations-of-the-tangent-and-the-normal-at-the-point-t-on-the-curve-x-a-sin3-t-y-b-cos3-t","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/find-the-equations-of-the-tangent-and-the-normal-at-the-point-t-on-the-curve-x-a-sin3-t-y-b-cos3-t\/","title":{"rendered":"Find the equations of the tangent and the normal at the point ‘t’ on the curve x = \\(a sin^3 t\\), y = \\(b cos^3 t\\)."},"content":{"rendered":"

Solution :<\/h2>\n

We have, x = \\(a sin^3 t\\), y = \\(b cos^3 t\\)<\/p>\n

\\(\\implies\\)\u00a0 \\(dx\\over dt\\) = \\(3a sin^2t cos t\\)\u00a0 and, \\(dy\\over dt\\)\u00a0 = \\(-3b cos^2t sin t\\)<\/p>\n

\\(\\therefore\\)\u00a0 \u00a0\\(dy\\over dx\\) = \\(dy\/dt\\over dx\/dt\\) = \\(-b\\over a\\) \\(cos t\\over sin t\\)<\/p>\n

So, the equation of the tangent<\/a> at the point ‘t’ is<\/p>\n

y – \\(b cos^3 t\\) = (\\(dy\\over dx\\))(x – \\(a sin^3 t\\))<\/p>\n

or, y – \\(b cos^3 t\\) = \\(-b\\over a\\) \\(cos t\\over sin t\\)(x – \\(a sin^3 t\\))<\/p>\n

or, bx cos t + ay sin t = ab sin t cos t<\/p>\n

The equation of the normal<\/a> at the point ‘t’ is<\/p>\n

y – \\(b cos^3 t\\) = \\((-1\\over ({dy\\over dx})\\)(x – \\(a sin^3 t\\))<\/p>\n

or, y – \\(b cos^3 t\\) = \\((-1\\over ({-b\\over a}{cos t\\over sin t})\\)(x – \\(a sin^3 t\\))<\/p>\n

or, ax sin t – by cos t = \\(a^2 sin^4 t – b^2 cos^4 t\\)<\/p>\n

\n

Similar Questions<\/h3>\n

Find the equation of the tangent to curve y = \\(-5x^2 + 6x + 7\\)\u00a0 at the point (1\/2, 35\/4).<\/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

Check the orthogonality of the curves \\(y^2\\) = x and \\(x^2\\) = y.<\/a><\/p>\n

The angle of intersection between the curve \\(x^2\\) = 32y and \\(y^2\\) = 4x at point (16, 8) is<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : We have, x = \\(a sin^3 t\\), y = \\(b cos^3 t\\) \\(\\implies\\)\u00a0 \\(dx\\over dt\\) = \\(3a sin^2t cos t\\)\u00a0 and, \\(dy\\over dt\\)\u00a0 = \\(-3b cos^2t sin t\\) \\(\\therefore\\)\u00a0 \u00a0\\(dy\\over dx\\) = \\(dy\/dt\\over dx\/dt\\) = \\(-b\\over a\\) \\(cos t\\over sin t\\) So, the equation of the tangent at the point ‘t’ is y …<\/p>\n

Find the equations of the tangent and the normal at the point ‘t’ on the curve x = \\(a sin^3 t\\), y = \\(b cos^3 t\\).<\/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 equations of the tangent and the normal at the point 't' on the curve x = \\(a sin^3 t\\), y = \\(b cos^3 t\\).<\/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-equations-of-the-tangent-and-the-normal-at-the-point-t-on-the-curve-x-a-sin3-t-y-b-cos3-t\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Find the equations of the tangent and the normal at the point 't' on the curve x = \\(a sin^3 t\\), y = \\(b cos^3 t\\).\" \/>\n<meta property=\"og:description\" content=\"Solution : We have, x = (a sin^3 t), y = (b cos^3 t) (implies)\u00a0 (dxover dt) = (3a sin^2t cos t)\u00a0 and, (dyover dt)\u00a0 = (-3b cos^2t sin t) (therefore)\u00a0 \u00a0(dyover dx) = (dy\/dtover dx\/dt) = (-bover a) (cos tover sin t) So, the equation of the tangent at the point ‘t’ is y … Find the equations of the tangent and the normal at the point ‘t’ on the curve x = (a sin^3 t), y = (b cos^3 t). 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