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

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.