Question

If we have an expression as $x=a\cos \theta$ ,$y=b\sin \theta$, then $\dfrac{{{d}^{3}}y}{d{{x}^{3}}}$ is equal to:
(1) $\dfrac{-3b}{{{a}^{3}}}{{\csc }^{4}}\theta {{\cot }^{4}}\theta$
(2) $\dfrac{-3b}{{{a}^{3}}}{{\csc }^{4}}\theta {{\cot }^{3}}\theta$
(3) $\dfrac{-3b}{{{a}^{3}}}{{\csc }^{4}}\theta \cot \theta$
(4) None of these

Answer 1

Here in this question we have been asked to find the value of $\dfrac{{{d}^{3}}y}{d{{x}^{3}}}$ given that $x=a\cos \theta$ ,$y=b\sin \theta$. From the basic concepts of differentiation we have been taught the chain rule which states that $\dfrac{dy}{dx}=\dfrac{dy}{d\theta }\times \dfrac{d\theta }{dx}$ . We will use this in order to answer the question.

Complete step by step solution:
Now considering from the question we have been asked to find the value of $\dfrac{{{d}^{3}}y}{d{{x}^{3}}}$ given that $x=a\cos \theta$ ,$y=b\sin \theta$.
From the basic concepts of differentiation we have been taught the chain rule which states that $\dfrac{dy}{dx}=\dfrac{dy}{d\theta }\times \dfrac{d\theta }{dx}$ .
We also know that $\dfrac{d}{dx}\sin x=\cos x$ and $\dfrac{d}{dx}\cos x=-\sin x$ .
Hence we can say that $\dfrac{dy}{d\theta }=b\left( \cos \theta \right)$ and $\dfrac{dx}{d\theta }=a\left( -\sin \theta \right)$ .
Now we can say that
$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=b\left( \cos \theta \right)\times \dfrac{1}{a\left( -\sin \theta \right)} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{-b}{a}\cos \theta \csc \theta \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{-b}{a}\cot \theta \\\ \end{aligned}$ .
Now by differentiating it again we will have $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{-b}{a}\cot \theta \right)$ .
By simplifying it further we will have $\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \dfrac{b}{a}{{\csc }^{2}}\theta \right)\left( \dfrac{d\theta }{dx} \right)$ since $\dfrac{d}{dx}\cot x=-{{\csc }^{2}}x$ .
By simplifying it we will have $\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \dfrac{b}{a}{{\csc }^{2}}\theta \right)\left( \dfrac{1}{-a\sin \theta } \right)$ .
Now again simplifying it further we will get $\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\left( \dfrac{-b}{{{a}^{2}}}{{\csc }^{3}}\theta \right)$ .
Now by differentiating it further we will get $\Rightarrow \dfrac{{{d}^{3}}y}{d{{x}^{3}}}=\dfrac{d}{dx}\left( \dfrac{-b}{{{a}^{2}}}{{\csc }^{3}}\theta \right)$ .
By simplifying it further we will get $\Rightarrow \dfrac{{{d}^{3}}y}{d{{x}^{3}}}=\left( \dfrac{-b}{{{a}^{2}}} \right)\dfrac{d}{dx}\left( {{\csc }^{3}}\theta \right)$ .
We know that $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ and $\dfrac{d}{dx}\csc x=-\csc x\cot x$ by using these formulae in the above expression we will have $\Rightarrow \dfrac{{{d}^{3}}y}{d{{x}^{3}}}=\left( \dfrac{-b}{{{a}^{2}}} \right)\left( 3{{\csc }^{2}}\theta \right)\left( -\csc \theta \cot \theta \right)\dfrac{d\theta }{dx}$ .
Now by simplifying this further we will have
$\begin{aligned} & \Rightarrow \dfrac{{{d}^{3}}y}{d{{x}^{3}}}=\left( \dfrac{3b}{{{a}^{2}}} \right)\left( {{\csc }^{3}}\theta \cot \theta \right)\left( \dfrac{-1}{a\sin \theta } \right) \\\ & \Rightarrow \dfrac{{{d}^{3}}y}{d{{x}^{3}}}=\left( \dfrac{-3b}{{{a}^{3}}} \right)\left( {{\csc }^{4}}\theta \cot \theta \right) \\\ \end{aligned}$ .
Therefore we can conclude that when it is given that $x=a\cos \theta$ and $y=b\sin \theta$ then the value of $\dfrac{{{d}^{3}}y}{d{{x}^{3}}}$ will be given as $\left( \dfrac{-3b}{{{a}^{3}}} \right)\left( {{\csc }^{4}}\theta \cot \theta \right)$ .
Hence we will mark the option “3” as correct.

Note: While answering questions of this type, we should be sure with the concepts that we are going to apply and the calculations that we are going to perform in between the steps. If someone had confused and forgot consider $\dfrac{d\theta }{dx}$ then we will have the resulting answer as $\dfrac{2b}{a}\cot \theta {{\csc }^{2}}\theta$ which is a wrong answer.