{"id":5137,"date":"2021-09-08T20:42:42","date_gmt":"2021-09-08T15:12:42","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5137"},"modified":"2021-11-22T20:43:56","modified_gmt":"2021-11-22T15:13:56","slug":"differentiation-of-parametric-functions","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/differentiation-of-parametric-functions\/","title":{"rendered":"Differentiation of Parametric Functions"},"content":{"rendered":"
Here you will learn differentiation of parametric functions with example.<\/p>\n
Let’s begin –<\/p>\n
Sometimes x and y are given as functions of a single variable e.g. x = \\(\\phi\\)(t), y = \\(\\psi\\)(t) are two functions of a single variable. In such a case x and y are called parametric functions or parametric equations and it is called the parameter. To find \\(dy\\over dx\\) in case of parametric functions, we first obtain the relationship between x and y\u00a0 by eliminating the parameter t and then we differentiate it with respect to x. But, it is not always convenient to eliminate the parameter. Therefore, \\(dy\\over dx\\) can also be obtained by the formula<\/p>\n
\n\\(dy\\over dx\\) = \\(dy\/dt\\over dx\/dt\\)<\/p>\n<\/blockquote>\n
Example<\/span><\/strong>\u00a0: find \\(dy\\over dx\\) where x = a{cos t + \\({1\\over 2} log tan^2 {t\\over 2}\\)} and y = a sin t<\/p>\n
Solution <\/span><\/strong>: We have,<\/p>\n
x = a{cos t + \\({1\\over 2} log tan^2 {t\\over 2}\\)} and y = a sin t<\/p>\n
\\(\\implies\\) x = a{cos t + \\({1\\over 2} \\times 2 log tan{t\\over 2}\\)} and y = a sin t<\/p>\n
\\(\\implies\\) x = a{cos t + {\\(log tan{t\\over 2}\\)} and y = a sin t<\/p>\n
Differentiating with respect to t, we get<\/p>\n
\\(dx\\over dt\\) = a{-sin t + \\({1\\over tan t\/2}(sec^2{t\\over 2})\\times {1\\over 2}\\)} and \\(dy\\over dt\\) = a cost<\/p>\n
\\(dx\\over dt\\) = a{-sin t + \\(1\\over 2 sin (t\/2) cos (t\/2)\\)} and \\(dy\\over dt\\) = a cost<\/p>\n
\\(\\implies\\)\u00a0 \\(dx\\over dt\\) = a{-sin t + \\(1\\over sint \\)} and \\(dy\\over dt\\) = a cost<\/p>\n
\\(\\implies\\)\u00a0 \\(dx\\over dt\\) = a{\\(-sin^2t + 1\\over sint \\)} and \\(dy\\over dt\\) = a cost<\/p>\n
\\(dx\\over dt\\) = a{\\(cos^2t\\over sint \\)} and \\(dy\\over dt\\) = a cost<\/p>\n
\\(\\therefore\\)\u00a0 \u00a0\\(dy\\over dx\\) = \\(dy\/dt\\over dx\/dt\\) = \\(acost\\over {acos^2t\\over sint}\\)\u00a0<\/p>\n
\\(dy\\over dx\\) = tan t<\/p>\n\n\n