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 $\frac{dy}{dx}$ 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.

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

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