Differentiation of Parametric Functions

Here you will learn differentiation of parametric functions with example.

Let’s begin –

Differentiation of Parametric Functions

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  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

\(dy\over dx\) = \(dy/dt\over dx/dt\)

Example : find \(dy\over dx\) where x = a{cos t + \({1\over 2} log tan^2 {t\over 2}\)} and y = a sin t

Solution : We have,

x = a{cos t + \({1\over 2} log tan^2 {t\over 2}\)} and y = a sin t

\(\implies\) x = a{cos t + \({1\over 2} \times 2 log tan{t\over 2}\)} and y = a sin t

\(\implies\) x = a{cos t + {\(log tan{t\over 2}\)} and y = a sin t

Differentiating with respect to t, we get

\(dx\over dt\) = a{-sin t + \({1\over tan t/2}(sec^2{t\over 2})\times {1\over 2}\)} and \(dy\over dt\) = a cost

\(dx\over dt\) = a{-sin t + \(1\over 2 sin (t/2) cos (t/2)\)} and \(dy\over dt\) = a cost

\(\implies\)  \(dx\over dt\) = a{-sin t + \(1\over sint \)} and \(dy\over dt\) = a cost

\(\implies\)  \(dx\over dt\) = a{\(-sin^2t + 1\over sint \)} and \(dy\over dt\) = a cost

\(dx\over dt\) = a{\(cos^2t\over sint \)} and \(dy\over dt\) = a cost

\(\therefore\)   \(dy\over dx\) = \(dy/dt\over dx/dt\) = \(acost\over {acos^2t\over sint}\) 

\(dy\over dx\) = tan t

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