Ucale

# Differentiation of parametric functions

Instead of one equation relating say, x and y, we have two equations, one relating x with the parameter, and one relating y with the parameter.

Sometimes x and y are given as functions of a single variable, e.g., $\displaystyle x=\phi \left( t \right) \quad y=\psi \left( t \right)$ are two functions and t is a variable. In such a case x and y are called parametric functions or parametric equations and t is called the parameter.

To find $\displaystyle \frac { dy }{ dx }$ in case of parametric function , we first obtain the relationship  between x and y  by eliminating the parameter t and then we differentiate at with respect to x . But every time it is not convenient to eliminate the parameter . Therefore $\displaystyle \frac { dy }{ dx }$  acn also be obtained by following formula $\displaystyle \frac { dy }{ dx } =\frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } }$

To prove it , let $\displaystyle \Delta x\quad and\quad \Delta y$  be the changes in x and y respectively corresponding to a small change $\displaystyle \Delta t\quad in\quad t\\ \because \frac { \Delta y }{ \Delta x } =\frac { \frac { \Delta y }{ \Delta t } }{ \frac { \Delta x }{ \Delta t } } \\ \therefore \frac { dy }{ dx } =\lim _{ \Delta x\longrightarrow 0 }{ \frac { \Delta y }{ \Delta x } } \\ =\frac { \lim _{ \Delta t\longrightarrow 0 }{ \frac { \Delta y }{ \Delta t } } }{ \lim _{ \Delta t\longrightarrow 0 }{ \frac { \Delta x }{ \Delta t } } } \\ =\frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } }$

### Example $\displaystyle Find\quad \frac { dy }{ dx } \quad x=a\left( \theta -\sin { \theta } \right) \quad and\quad y=a\left( 1-\cos { \theta } \right)$

Solution: $\displaystyle we\quad have,x=a\left( \theta -\sin { \theta } \right) \quad and\quad y=a\left( 1-\cos { \theta } \right)$

Differentiating w.r.t. θ , we get $\displaystyle \frac { dx }{ d\theta } =a\left( 1-\cos { \theta } \right) \quad and\quad \frac { dy }{ d\theta } =a\sin { \theta } \\ \therefore \frac { dy }{ dx } =\frac { \frac { dy }{ d\theta } }{ \frac { dx }{ d\theta } } \\ =\frac { a\sin { \theta } }{ a\left( 1-\cos { \theta } \right) } \\ =\frac { 2\sin { \frac { \theta }{ 2 } } \cos { \frac { \theta }{ 2 } } }{ 2\sin ^{ 2 }{ \frac { \theta }{ 2 } } } \\ =\cot { \frac { \theta }{ 2 } }$

April 18, 2019
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