Question

If it is given that $x=a\left( \cos t+\log \tan \dfrac{t}{2} \right)$, $y=a\sin t$, find $\dfrac{{{d}^{2}}y}{d{{t}^{2}}}$ and $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$.

Answer 1

Hint: In order to solve the question, we have to calculate two things first one is $\dfrac{{{d}^{2}}y}{d{{t}^{2}}}$ which is easier just differentiate $y$ two times with respect $t$. The other one is $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ which involves little bit of complexity, first calculate $\dfrac{dy}{dt}$ and $\dfrac{dx}{dt}$ then divide to get $\dfrac{dy}{dx}$ and then differentiate with respect to $x$in order to get $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ .

Complete step by step answer:
It is given that $x=a\left( \cos t+\log \tan \dfrac{t}{2} \right)$ and $y=a\sin t$.
As $y=a\sin t$
$\Rightarrow \dfrac{dy}{dt}=\dfrac{d}{dt}\left( a\sin t \right)$
As we know that the derivative of $\sin x$ is $\cos x$ i.e. $\dfrac{d}{dx}\left( \sin x \right)=\cos x$
$\Rightarrow \dfrac{dy}{dt}=a\cos t....................(1)$
We also know that the derivative of $\cos x$ is $-\sin x$ i.e. $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{t}^{2}}}=\dfrac{d}{dt}\left( \dfrac{dy}{dt} \right)=\dfrac{d}{dt}\left( a\cos t \right)=-a\sin t$
Now, we have to calculate $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ . So first we have to calculate $\dfrac{dy}{dx}$ and then differentiate with respect to $x$ and get $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$.

As there is no any direct relation between $y$ and $x$ so we cannot directly calculate $\dfrac{dy}{dx}$. So, in order to calculate $\dfrac{dy}{dx}$ we have to first calculate $\dfrac{dy}{dt}$ and $\dfrac{dx}{dt}$ then divide to get $\dfrac{dy}{dx}$ .
Now, from equation (1), we get $\dfrac{dy}{dt}=a\cos t$
As, $x=a\left( \cos t+\log \tan \dfrac{t}{2} \right)$
$\Rightarrow \dfrac{dx}{dt}=\dfrac{d}{dt}\left( a\left( \cos t+\log \tan \dfrac{t}{2} \right) \right)$
$\Rightarrow \dfrac{dx}{dt}=\dfrac{d}{dt}\left( a\cos t \right)+\dfrac{d}{dt}\left( a\log \tan \dfrac{t}{2} \right)..............(2)$
In order to get $\dfrac{d}{dt}\left( a\log \tan \dfrac{t}{2} \right)$, we have to apply chain rule which state that the derivative of \[f\left( g\left( x \right) \right)\] is \[f'\left( g\left( x \right) \right)\cdot g'\left( x \right)\]. In other words, it helps us differentiate composite functions.
Now, $\dfrac{d}{dt}\left( a\log \tan \dfrac{t}{2} \right)=\dfrac{d\left( a\log \tan \dfrac{t}{2} \right)}{d\left( \tan \dfrac{t}{2} \right)}\times \dfrac{d\left( \tan \dfrac{t}{2} \right)}{d\left( \dfrac{t}{2} \right)}\times \dfrac{d\left( \dfrac{t}{2} \right)}{dt}$
As we know that the derivative of $\log x$ is $\dfrac{1}{x}$ i.e. $\dfrac{d}{dx}(\log x)=\dfrac{1}{x}$.
Similarly, the derivative of $\tan x$ is ${{\sec }^{2}}x$ i.e. $\dfrac{d}{dx}(\tan x)={{\sec }^{2}}x$, substituting all those values in above equation, we get
$\Rightarrow \dfrac{d}{dt}\left( a\log \tan \dfrac{t}{2} \right)=\dfrac{a}{\tan \dfrac{t}{2}}\times {{\sec }^{2}}\dfrac{t}{2}\times \dfrac{1}{2}$
Substituting $\sec \dfrac{t}{2}=\dfrac{1}{\cos \dfrac{t}{2}}$ and $\tan \dfrac{t}{2}=\dfrac{\sin \dfrac{t}{2}}{\cos \dfrac{t}{2}}$ in the above equation, we get
$\Rightarrow \dfrac{d}{dt}\left( a\log \tan \dfrac{t}{2} \right)=\dfrac{a\cos \dfrac{t}{2}}{2\sin \dfrac{t}{2}{{\cos }^{2}}\dfrac{t}{2}}=\dfrac{a}{2\sin \dfrac{t}{2}\cos \dfrac{t}{2}}=\dfrac{a}{\sin t}$
Substituting the value of $\dfrac{d}{dt}\left( a\log \tan \dfrac{t}{2} \right)$ in equation (2), we get
$\Rightarrow \dfrac{dx}{dt}=-a\sin t+\dfrac{a}{\sin t}......................(3)$
Dividing equation (1) with (3), we get
$\Rightarrow \dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{a\cos t}{-a\sin t+\dfrac{a}{\sin t}}...............(4)$
We have to differentiate it again with respect to $x$. We will use the quotient rule for the same. It is given by $\dfrac{d}{dx}\left[ \dfrac{f\left( x \right)}{g\left( x \right)} \right]=\dfrac{f'\left( x \right).g\left( x \right)-f\left( x \right).g'\left( x \right)}{{{\left[ g\left( x \right) \right]}^{2}}}$ . Now, using it, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\dfrac{d}{dx}\left( a\cos t \right)\left( -a\sin t+\dfrac{a}{\sin t} \right)-a\cos t\dfrac{d}{dx}\left( -a\sin t+\dfrac{a}{\sin t} \right)}{{{\left( -a\sin t+\dfrac{a}{\sin t} \right)}^{2}}}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-a\sin t\times \dfrac{dt}{dx}\left( -a\sin t+\dfrac{a}{\sin t} \right)-a\cos t\left( -a\cos t\dfrac{dt}{dx}-\dfrac{a}{{{\sin }^{2}}t}\times \cos t\dfrac{dt}{dx} \right)}{{{\left( -a\sin t+\dfrac{a}{\sin t} \right)}^{2}}}$
We know that from equation (3), $\dfrac{dx}{dt}=-a\sin t+\dfrac{a}{\sin t}\Rightarrow \dfrac{dt}{dx}=\dfrac{1}{-a\sin t+\dfrac{a}{\sin t}}$
So putting the value of $\dfrac{dt}{dx}$ in above equation, we get
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-a\sin t\times \dfrac{1}{\left( -a\sin t+\dfrac{a}{\sin t} \right)}\left( -a\sin t+\dfrac{a}{\sin t} \right)+{{a}^{2}}{{\cos }^{2}}t\left( 1+\dfrac{1}{{{\sin }^{2}}t} \right)\times \dfrac{1}{\left( -a\sin t+\dfrac{a}{\sin t} \right)}}{{{\left( -a\sin t+\dfrac{a}{\sin t} \right)}^{2}}}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-a\sin t+{{a}^{2}}{{\cos }^{2}}t\left( 1+\dfrac{1}{{{\sin }^{2}}t} \right)\times \dfrac{\sin t}{a{{\cos }^{2}}t}}{{{\left( -a\sin t+\dfrac{a}{\sin t} \right)}^{2}}}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-a\sin t+\left( a\sin t+\dfrac{a}{{{\sin }^{2}}t} \right)}{{{\left( -a\sin t+\dfrac{a}{\sin t} \right)}^{2}}}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\dfrac{a}{{{\sin }^{2}}t}}{{{a}^{2}}\dfrac{{{\left( 1-{{\sin }^{2}}t \right)}^{2}}}{{{\sin }^{2}}t}}=\dfrac{1}{a{{\cos }^{4}}t}=\dfrac{{{\sec }^{4}}t}{a}............(5)$

Hence, $\dfrac{{{d}^{2}}y}{d{{t}^{2}}}=-a\sin t$ and $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{{{\sec }^{4}}t}{a}$ is the correct answer.

Note: In these types of questions when we don’t have the direct relation between $y$ and $x$ then we first calculate $\dfrac{dy}{dt}$ and $\dfrac{dx}{dt}$ then divide to get $\dfrac{dy}{dx}$. Also, the main crux is to use the chain rule properly and substitute the value of derivatives in the expressions and get the desired value. There are a lot of computations and care must be taken not to make any silly mistakes.