Differentiation of Function in Parametric Form - Practice Questions & MCQ

Differentiation of Function in Parametric Form

Sometimes, $x$ and $y$ are given as functions of a single variable, i.e., $x=g(t)$ and $y=f(t)$ are two functions and $t$ is a variable. In such cases $x$ and $y$ are called parametric functions or parametric equations and $t$ is called the parameter.

To find $\frac{d y}{d x}$ in such cases, first find the relationship between x and y by eliminating the parameter t and then differentiate concerning $t$.

But sometimes it is not possible to eliminate $t$, then in that case use.

$
\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{f^{\prime}(t)}{g^{\prime}(t)}
$
For example
If $x=a(1-\cos \theta)$ and $y=a(\theta-\sin \theta)$, then $d y / d x$ is
Solution.

$
\begin{aligned}
& \text { Given } \mathrm{x}=\mathrm{a}(1-\cos \theta) \text { and } \mathrm{y}=\mathrm{a}(\theta+\sin \theta) \\
& \Rightarrow \frac{d x}{d \theta}=a(\sin \theta) \text { and } \frac{d y}{d \theta}=a(1+\cos \theta) \\
& \Rightarrow \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{1+\cos \theta}{\sin \theta}
\end{aligned}
$