Question

If $x=a\left( \sin \theta -\theta \cos \theta \right)$ $y=a\left( \cos \theta +\theta \sin \theta \right)$, then show that $\dfrac{dy}{dx}=\cot \theta$ .

Answer 1

In this question we have been given that $x=a\left( \sin \theta -\theta \cos \theta \right)$ and $y=a\left( \cos \theta +\theta \sin \theta \right)$ we need to show that $\dfrac{dy}{dx}=\cot \theta$ . For that we will use chain rule given as $\dfrac{dy}{dx}=\dfrac{dy}{d\theta }\times \dfrac{d\theta }{dx}\Rightarrow \dfrac{\left( \dfrac{dy}{d\theta } \right)}{\left( \dfrac{dx}{d\theta } \right)}$ and product rule given as $\dfrac{d}{dx}\left( u\times v \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ .

Complete step by step solution:
Now considering from the question we have been given that $x=a\left( \sin \theta -\theta \cos \theta \right)$ and $y=a\left( \cos \theta +\theta \sin \theta \right)$ we need to show that $\dfrac{dy}{dx}=\cot \theta$ .
From the basic concepts we know that chain rule given as $\dfrac{dy}{dx}=\dfrac{dy}{d\theta }\times \dfrac{d\theta }{dx}\Rightarrow \dfrac{\left( \dfrac{dy}{d\theta } \right)}{\left( \dfrac{dx}{d\theta } \right)}$ and product rule given as $\dfrac{d}{dx}\left( u\times v \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ . We will use them and simplify and answer this question.
We also know the differential formula given as $\dfrac{d}{d\theta }\left( \sin \theta \right)=\cos \theta$ and $\dfrac{d}{d\theta }\left( \cos \theta \right)=-\sin \theta$ .
So by applying chain rule we will have $\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{d\theta } \right)}{\left( \dfrac{dx}{d\theta } \right)}$ . Now we need to evaluate the values of $\dfrac{dy}{d\theta }$ and $\dfrac{dx}{d\theta }$ .
We will evaluate $\dfrac{dy}{d\theta }$ as $\Rightarrow \dfrac{dy}{d\theta }=\dfrac{d}{d\theta }\left( a\left( \cos \theta +\theta \sin \theta \right) \right)$.
We will further simplify this using the product rule for differentiating the part $\theta \sin \theta$ and we will have
$\begin{aligned} & \Rightarrow \dfrac{dy}{d\theta }=a\left( -\sin \theta +\sin \theta +\theta \cos \theta \right) \\\ & \Rightarrow \dfrac{dy}{d\theta }=a\theta \cos \theta \\\ \end{aligned}$
We will evaluate $\dfrac{dx}{d\theta }$ as $\Rightarrow \dfrac{dx}{d\theta }=\dfrac{d}{d\theta }\left( a\left( \sin \theta -\theta \cos \theta \right) \right)$.
We will further simplify this using the product rule for differentiating the part $\theta \cos \theta$ and we will have
$\begin{aligned} & \Rightarrow \dfrac{dx}{d\theta }=a\left( \cos \theta -\cos \theta +\theta \sin \theta \right) \\\ & \Rightarrow \dfrac{dx}{d\theta }=a\theta \sin \theta \\\ \end{aligned}$
Hence we can say that when $x=a\left( \sin \theta -\theta \cos \theta \right)$ and $y=a\left( \cos \theta +\theta \sin \theta \right)$ then $\dfrac{dy}{dx}=\cot \theta$ .

Note: While answering questions of this type we should be sure with our differential concepts. This question can be answered easily and in a short span of time and very few mistakes are possible in it. We have many other differential formulae similarly for example $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ , $\dfrac{d}{dx}\log x=\dfrac{1}{x}$ . This formula can be used in the questions to solve them. We have quotient rule of differentiation similar to the product rule given as $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{vu'-uv'}{{{v}^{2}}}$ .