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# Find $\dfrac{{dy}}{{dx}}$ for parametric equation$x = a\cos \theta$, $y = a\sin \theta$

Last updated date: 20th Jun 2024
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Hint: To find derivative of any parametric equations, we proceed the calculations in two steps. In the first step we first calculate the derivative of x w.r.t. $\theta$and derivative of y w.r.t. $\theta$. After that we divide the result of the derivative of y with the result of the derivative of x, obtained in the first step to obtain the derivative of y w.r.t. x (i.e. $\dfrac{{dy}}{{dx}}$)

Complete step-by-step solution:
Given, parametric equations are $x = a\cos \theta$and $y = a\sin \theta$
Differentiating x w.r.t. $\theta$we have,
$\dfrac{{dx}}{{d\theta }} = a\left( { - \sin \theta } \right)$
Differentiating y w.r.t. $\theta$we have,
$\dfrac{{dy}}{{d\theta }} = a\left( {\cos \theta } \right)$
Now, to find $\dfrac{{dy}}{{dx}}$ we divide the derivative of y equation by the derivative of x equation.
i.e. $\dfrac{{dy}}{{dx}} = \dfrac{{\left( {dy/d\theta } \right)}}{{\left( {dx/d\theta } \right)}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{a\left( {\cos \theta } \right)}}{{ - a\left( {\sin \theta } \right)}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\cos \theta }}{{\sin \theta }}$
$\Rightarrow \dfrac{{dy}}{{dx}} = - \cot \theta$
Hence, from above we see that $\dfrac{{dy}}{{dx}}$ of parametric equations $x = a\cos \theta ,\,y = a\sin \theta$ is$- \cot \theta$.

Note: In finding derivative of parametric equations we must differential equations separately and then we should divide derivative of y term with derivative of x term to obtain$\dfrac{{dy}}{{dx}}$. Which is the required derivative of the given parametric equations.