Question

How do you differentiate the following parametric equation: $x\left( t \right)=t\ln t$, $y\left( t \right)=\cos t-t{{\sin }^{2}}t$?

Answer 1

Hint: In this problem we need to find the differentiation of the given parametric equations. In the problem we have two parametric equations. So, we will consider them separately and calculate the differentiation separately. First, we will consider the equation $x\left( t \right)=t\ln t$ Here we will apply the $uv$ formula of differentiation. After applying the $uv$ formula we will apply basic differentiation rules and simplify the equation to get the differentiation of the first parametric equation. Now we will consider the second parametric equation. In this equation also we will apply the $uv$ formula and simplify the obtained equation to get the required solution.

Complete step-by-step solution:
Given parametric equations $x\left( t \right)=t\ln t$, $y\left( t \right)=\cos t-t{{\sin }^{2}}t$.
Considering the first parametric equation which is $x\left( t \right)=t\ln t$.
Differentiating the above equation with respect to $t$, then we will get
$\Rightarrow \dfrac{dx}{dt}=\dfrac{d}{dt}\left( t\ln t \right)$
We have the $uv$ formula for the differentiation $\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$. Applying this formula in the above equation, then we will get
$\Rightarrow \dfrac{dx}{dt}=t\dfrac{d}{dx}\left( \ln t \right)+\ln t\dfrac{d}{dt}\left( t \right)$
We have the differentiation formulas $\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}$, $\dfrac{d}{dx}\left( x \right)=1$. Substituting these values in the above equation, then we will get
$\Rightarrow \dfrac{dx}{dt}=t\times \dfrac{1}{t}+\ln t\left( 1 \right)$
Simplifying the above equation, then we will have
$\therefore \dfrac{dx}{dt}=1+\ln t$
Now considering the second parametric equation $y\left( t \right)=\cos t-t{{\sin }^{2}}t$.
Differentiating the above equation with respect to $t$, then we will get
$\Rightarrow \dfrac{dy}{dt}=\dfrac{d}{dt}\left( \cos t-t{{\sin }^{2}}t \right)$
Applying the differentiation to each term individually, then we will have
$\Rightarrow \dfrac{dy}{dt}=\dfrac{d}{dt}\left( \cos t \right)-\dfrac{d}{dt}\left( t{{\sin }^{2}}t \right)$
Again, applying the $uv$ formula for the differentiation $\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ in the above equation, then we will get
$\Rightarrow \dfrac{dy}{dt}=\dfrac{d}{dt}\left( \cos t \right)-\left[ t\dfrac{d}{dt}\left( {{\sin }^{2}}t \right)+{{\sin }^{2}}t\dfrac{d}{dt}\left( t \right) \right]$
We have the differentiation formulas $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$, $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$, $\dfrac{d}{dx}\left( x \right)=1$. From these formulas the above equation is modified as
$\Rightarrow \dfrac{dy}{dt}=-\sin x-\left[ t\times 2\sin t\dfrac{d}{dt}\left( \sin t \right)+{{\sin }^{2}}t \right]$
Again, we have the differentiation formula $\dfrac{d}{dx}\left( \sin x \right)=\cos x$, then we will get
$\Rightarrow \dfrac{dy}{dt}=-\sin t-2t\sin t\cos t-{{\sin }^{2}}t$
Taking $-\sin t$ as common in the above equation, then we will get
$\therefore \dfrac{dy}{dt}=-\sin t\left( 1+\sin t+2t\cos t \right)$

Note: In this problem they have only mentioned to calculate the derivatives of given parametric equations, so we have calculated individually. If they have asked to calculate the value of $\dfrac{dy}{dx}$, then we need to calculate the ratio of $\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$. Now this value will be our required value $\dfrac{dy}{dx}$.