Question

If \[x=a\cos \theta ,y=b\sin \theta \], then \[\dfrac{{{d}^{3}}y}{d{{x}^{3}}}\]is equal to
(A) \[\left( \dfrac{-3b}{{{a}^{3}}} \right)\text{cose}{{\text{c}}^{4}}\theta {{\cot }^{4}}\theta \]
(B) \[\left( \dfrac{3b}{{{a}^{3}}} \right)\text{cose}{{\text{c}}^{4}}\theta \cot \theta \]
(C) \[\left( \dfrac{-3b}{{{a}^{3}}} \right)\text{cose}{{\text{c}}^{4}}\theta \cot \theta \]
(D) None of the above

Answer 1

Hint: If \[x=f(\theta )\] and\[y=g(\theta )\] then this represents the parametric form of x and y in terms of $\theta $ , and its differentiation is done accordingly by \[\dfrac{dy}{dx}=\dfrac{dy}{d\theta }\times \dfrac{d\theta }{dx}\] , then \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{d\theta }(\dfrac{dy}{dx})\times \dfrac{d\theta }{dx}\]and finally \[\dfrac{{{d}^{3}}y}{d{{x}^{3}}}=\dfrac{d}{d\theta }(\dfrac{{{d}^{2}}y}{d{{x}^{2}}})\times \dfrac{d\theta }{dx}\]. Then, using this method we will find the solution of \[\dfrac{{{d}^{3}}y}{d{{x}^{3}}}\] for \[x=a\cos \theta ,y=b\sin \theta \].

Complete step by step answer:
We are given the expressions \[x=a\cos \theta \] and \[y=b\sin \theta \]. It can clearly be seen that \[x\] and \[y\]are in parametric form and \[\theta \] is the parameter.
Hence , in order to find the value of the derivative \[\dfrac{dy}{dx}\], first we have to differentiate \[x\] and \[y\]separately with respect to \[\theta \].
i.e. \[\dfrac{dy}{dx}=\dfrac{dy}{d\theta }\times \dfrac{d\theta }{dx}\]
Now, we will differentiate \[x\] and \[y\] separately with respect to \[\theta \].
On differentiating \[y\] with respect to \[\theta \], we get
 \[\dfrac{dy}{d\theta }=\dfrac{d}{d\theta }\left( b\sin \theta \right)=b\cos \theta \]
And, on differentiating \[x\] with respect to \[\theta \], we get
\[\dfrac{dx}{d\theta }=\dfrac{d}{d\theta }\left( a\cos \theta \right)=-a\sin \theta \]
Now , we know inverse function theorem of differentiation says that if \[x=f(\theta )\] and \[\dfrac{dx}{d\theta }=x'\] then , \[\dfrac{d\theta }{dx}=\theta '=\dfrac{1}{x'}\] .
So , we can write \[\dfrac{d\theta }{dx}=\dfrac{1}{\dfrac{dx}{d\theta }}=\dfrac{1}{-a\sin \theta }\].

Now, we will substitute the values of \[\dfrac{dy}{d\theta }\] and \[\dfrac{d\theta }{dx}\] in the expression of \[\dfrac{dy}{dx}\].
So , on substituting the values of \[\dfrac{dy}{d\theta }\] and \[\dfrac{d\theta }{dx}\] in the expression of \[\dfrac{dy}{dx}\], we get \[\dfrac{dy}{dx}=\dfrac{b\cos \theta }{-a\sin \theta }=-\dfrac{b}{a}\cot \theta \]
Again , we can see that \[\dfrac{dy}{dx}\]is a function of \[\theta \].
So , we can say \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{d\theta }\left( \dfrac{dy}{dx} \right).\dfrac{d\theta }{dx}\]
Now , we will substitute the values of \[\dfrac{dy}{dx}\] and \[\dfrac{d\theta }{dx}\] in the expression of \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\].
So , on substituting the values of \[\dfrac{dy}{dx}\] and \[\dfrac{d\theta }{dx}\] in the expression of \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\], we get ,
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-b}{a}\text{(}-\text{cose}{{\text{c}}^{2}}\theta )\left( \dfrac{-1}{a\sin \theta } \right)\] , as $\dfrac{d}{dx}(\cot \theta )=-\cos e{{c}^{2}}\theta $
On simplifying, we get
\[=\dfrac{-b}{a}\text{(}-\text{cose}{{\text{c}}^{2}}\theta )\left( \dfrac{-\cos ec\theta }{a} \right)\]
\[=\dfrac{-b}{{{a}^{2}}}\text{cose}{{\text{c}}^{3}}\theta \]
Again , we can see that \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] is a function of\[\theta \].
So , \[\dfrac{{{d}^{3}}y}{d{{x}^{3}}}=\dfrac{d}{d\theta }\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right).\dfrac{d\theta }{dx}\]
Now , we will substitute the values of \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] and \[\dfrac{d\theta }{dx}\] in the expression of \[\dfrac{{{d}^{3}}y}{d{{x}^{3}}}\].
So , on substituting the values of \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] and \[\dfrac{d\theta }{dx}\] in the expression of \[\dfrac{{{d}^{3}}y}{d{{x}^{3}}}\], we get ,
\[\dfrac{{{d}^{3}}y}{d{{x}^{3}}}=\dfrac{d}{d\theta }\left( \dfrac{-b}{{{a}^{2}}}\text{cose}{{\text{c}}^{3}}\theta \right).\left( \dfrac{-1}{a\sin \theta } \right)\]
\[=\dfrac{-b}{{{a}^{2}}}\left( 3\text{cose}{{\text{c}}^{2}}\theta \right).\left( -\text{cosec}\theta \cot \theta \right).\left( \dfrac{-1}{a\sin \theta } \right)\] as $\dfrac{d}{dx}(\operatorname{cosec}\theta )=-\cos ec\theta \cot \theta $
On simplifying, we get
\[=\dfrac{-3b}{{{a}^{3}}}\text{cose}{{\text{c}}^{4}}\theta \cot \theta \]
So , \[\dfrac{{{d}^{3}}y}{d{{x}^{3}}}=\dfrac{-3b}{{{a}^{3}}}\text{cose}{{\text{c}}^{4}}\theta \cot \theta \]

So, the correct answer is “Option C”.

Note: A common mistake made in such question is writing \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\dfrac{{{d}^{2}}y}{d{{\theta }^{2}}}}{\dfrac{{{d}^{2}}x}{d{{\theta }^{2}}}}\] which is wrong.
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{d\theta }\left( \dfrac{dy}{dx} \right).\left( \dfrac{d\theta }{dx} \right)\]. Such mistakes should be avoided. These mistakes can lead to incorrect values and the student will end up getting a wrong answer.