{"id":5591,"date":"2021-09-22T15:16:48","date_gmt":"2021-09-22T09:46:48","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5591"},"modified":"2021-11-13T23:39:03","modified_gmt":"2021-11-13T18:09:03","slug":"slopes-of-tangent-and-normal-to-the-curve","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/slopes-of-tangent-and-normal-to-the-curve\/","title":{"rendered":"Slopes of Tangent and Normal to the Curve"},"content":{"rendered":"
Here you will learn slopes of tangent and normal to the curve with examples.<\/p>\n
Let’s begin –<\/p>\n
Let y = f(x) be a continuous curve, and let \\(P(x_1, y_1)\\) be a point on it. Then,\u00a0<\/p>\n
\n\\(({dy\\over dx})_P\\) is the tangent to the curve y = f(x) at point P.<\/p>\n
i.e. \\(({dy\\over dx})_P\\) = tan \\(\\psi\\) = Slope of the tangent at P,<\/p>\n<\/blockquote>\n
where \\(\\psi\\) is the angle which the tangent at \\(P(x_1, y_1)\\) makes with the positive direction of x-axis.<\/p>\n
If the tangent at P is parallel to x-axis<\/strong>, then<\/p>\n
\\(\\psi\\) = 0 \\(\\implies\\) tan \\(\\psi\\) = 0 \\(\\implies\\) Slope = 0 \\(\\implies\\) \\(({dy\\over dx})_P\\) = 0<\/p>\n
If the tangent at P is perpendicular to x-axis, or parallel to y-axis<\/strong>, then<\/p>\n
\\(\\psi\\) = \\(\\pi\\over 2\\) \\(\\implies\\) cot \\(\\psi\\) = 0 \\(\\implies\\) \\(1\\over tan \\psi\\) = 0 \\(\\implies\\) \\(({dx\\over dy})_P\\) = 0<\/p>\n
(b) Slopes of Normal<\/h3>\n
The normal to the curve at \\(P(x_1, y_1)\\) is a line perpendicular to the tangent at P and passing through P.<\/p>\n
\n\\(\\therefore\\)\u00a0 Slope of the normal at P = \\(-1\\over Slope of the tangent at P\\) = \\(-({dx\\over dy})_P\\)<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : find the slopes of the tangent and the normal to the curve \\(x^2 + 3y + y^2\\) = 5 at (1, 1).<\/p>\n
Solution<\/strong><\/span> : The equation of the curve is \\(x^2 + 3y + y^2\\) = 5<\/p>\n
Differentiating with respect to x, we get<\/p>\n
2x + 3\\(dy\\over dx\\) + 2y\\(dy\\over dx\\) = 0<\/p>\n
\\(\\implies\\) \\(dy\\over dx\\) = \\(-2x\\over 2y + 3\\)<\/p>\n
\\(\\implies\\) \\(({dy\\over dx})_{(1, 1)}\\) = -(\\(2\\over 2 + 3\\)) = -\\(2\\over 5\\)<\/p>\n
\\(\\therefore\\)\u00a0 Slope of the tangent at (1, 1) = -\\(2\\over 5\\)<\/p>\n
and, Slope of normal at (1, 1) = \\(-1\\over slope of tangent at (1, 1)\\) = \\(5\\over 2\\)<\/p>\n
\nRelated Questions<\/h3>\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