{"id":8105,"date":"2021-11-13T23:05:00","date_gmt":"2021-11-13T17:35:00","guid":{"rendered":"https:\/\/mathemerize.com\/?p=8105"},"modified":"2021-11-14T00:14:59","modified_gmt":"2021-11-13T18:44:59","slug":"show-that-the-tangents-to-the-curve-y-2x3-3-at-the-points-where-x-2-and-x-2-are-parallel","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/show-that-the-tangents-to-the-curve-y-2x3-3-at-the-points-where-x-2-and-x-2-are-parallel\/","title":{"rendered":"Show that the tangents to the curve y = \\(2x^3 – 3\\) at the points where x =2 and x = -2 are parallel."},"content":{"rendered":"

Solution :<\/h2>\n

The equation of the curve is y = \\(2x^3 – 3\\)<\/p>\n

Differentiating with respect to x, we get<\/p>\n

\\(dy\\over dx\\) = \\(6x^2\\)<\/p>\n

Now, \\(m_1\\) = (Slope of the tangent at x = 2) = \\(({dy\\over dx})_{x = 2}\\) = \\(6 \\times (2)^2\\) = 24<\/p>\n

and, \\(m_2\\) = (Slope of the tangent at x = -2) = \\(({dy\\over dx})_{x = -2}\\) = \\(6 \\times (-2)^2\\) = 24<\/p>\n

Clearly \\(m_1\\) = \\(m_2\\).<\/p>\n

Thus, the tangents to the given curve at the points where x = 2 and x = -2 are parallel.<\/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

Find the slope of the normal to the curve x = \\(a cos^3\\theta\\), y = \\(a sin^3\\theta\\) at \\(\\theta\\) = \\(\\pi\\over 4\\).<\/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 : The equation of the curve is y = \\(2x^3 – 3\\) Differentiating with respect to x, we get \\(dy\\over dx\\) = \\(6x^2\\) Now, \\(m_1\\) = (Slope of the tangent at x = 2) = \\(({dy\\over dx})_{x = 2}\\) = \\(6 \\times (2)^2\\) = 24 and, \\(m_2\\) = (Slope of the tangent at x …<\/p>\n

Show that the tangents to the curve y = \\(2x^3 – 3\\) at the points where x =2 and x = -2 are parallel.<\/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":"\nShow that the tangents to the curve y = \\(2x^3 - 3\\) at the points where x =2 and x = -2 are parallel.<\/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\/show-that-the-tangents-to-the-curve-y-2x3-3-at-the-points-where-x-2-and-x-2-are-parallel\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Show that the tangents to the curve y = \\(2x^3 - 3\\) at the points where x =2 and x = -2 are parallel.\" \/>\n<meta property=\"og:description\" content=\"Solution : The equation of the curve is y = (2x^3 – 3) Differentiating with respect to x, we get (dyover dx) = (6x^2) Now, (m_1) = (Slope of the tangent at x = 2) = (({dyover dx})_{x = 2}) = (6 times (2)^2) = 24 and, (m_2) = (Slope of the tangent at x … Show that the tangents to the curve y = (2x^3 – 3) at the points where x =2 and x = -2 are parallel. 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