Question

If we have $x=a\sin t$ and $y=a\left( \cos t+\log \left( \tan \dfrac{t}{2} \right) \right)$ then find $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$.

Answer 1

We find the first order derivative and again differentiate the first order derivative to find the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$. As the given equation is function of $t$ and we have to find the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$, we can write as $\dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}$. We differentiate the given equations with respect to $t$ and substitute the values to obtain a desired answer. We also use the chain rule to solve this question. The following derivatives we use in this question:
$\dfrac{d}{dt}\sin t=\cos t$
$\dfrac{d}{dt}\tan t={{\sec }^{2}}t$
$\dfrac{d}{dt}\log t=\dfrac{1}{t}$
$\dfrac{d}{dt}\cos t=-\sin t$

Complete step by step answer:
We have given that $x=a\sin t$ and $y=a\left( \cos t+\log \left( \tan \dfrac{t}{2} \right) \right)$
We have to find $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$.
As we know that $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ is the second order derivative, then first we have to calculate $\dfrac{dy}{dx}$.
Now, $\dfrac{dy}{dx}=\dfrac{dy}{dx}\times \dfrac{dt}{dt}$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}$ as the given equations are functions of $t$ .
Now, consider the equation $x=a\sin t$.
When we differentiate the equation with respect to $t$, we get
$\begin{aligned} & x=a\sin t \\\ & \dfrac{dx}{dt}=\dfrac{d}{dt}\left( a\sin t \right) \\\ \end{aligned}$
Taking constant term out, we get
$\dfrac{dx}{dt}=a\dfrac{d}{dt}\left( \sin t \right)$
Now, we know that $\dfrac{d}{dt}\sin t=\cos t$
Then $\dfrac{dx}{dt}=a\cos t...........(i)$
Now, consider equation $y=a\left( \cos t+\log \left( \tan \dfrac{t}{2} \right) \right)$
When we differentiate the equation with respect to $t$, we get

& y=a\left( \cos t+\log \left( \tan \dfrac{t}{2} \right) \right) \\\ & \Rightarrow \dfrac{dy}{dt}=a\dfrac{d}{dt}\left( \cos t+\log \left( \tan \dfrac{t}{2} \right) \right) \\\ & \Rightarrow \dfrac{dy}{dt}=a\left( \dfrac{d}{dt}\cos t+\dfrac{d}{dt}\log \left( \tan \dfrac{t}{2} \right) \right) \\\ \end{aligned}$$ Now, we know that $\dfrac{d}{dt}\log t=\dfrac{1}{t}$ and $\dfrac{d}{dt}\cos t=-\sin t$ We know that $\dfrac{d}{dt}\tan t={{\sec }^{2}}t$ Also, we use chain rule to solve $$\dfrac{d}{dt}\log \left( \tan \dfrac{t}{2} \right)$$, which is as follows: $$\begin{aligned} & \Rightarrow \dfrac{1}{\left( \tan \dfrac{t}{2} \right)}\dfrac{d}{dt}\left( \tan \dfrac{t}{2} \right) \\\ & \Rightarrow \dfrac{1}{\left( \tan \dfrac{t}{2} \right)}\left( {{\sec }^{2}}\dfrac{t}{2} \right)\dfrac{d}{dt}\left( \dfrac{t}{2} \right) \\\ & \Rightarrow \dfrac{1}{\left( \tan \dfrac{t}{2} \right)}\left( {{\sec }^{2}}\dfrac{t}{2} \right)\dfrac{1}{2} \\\ \end{aligned}$$ When we substitute the values, we get $$\Rightarrow \dfrac{dy}{dt}=a\left( -\sin t+\dfrac{1}{\left( \tan \dfrac{t}{2} \right)}.{{\sec }^{2}}\dfrac{t}{2}.\dfrac{1}{2} \right)$$ Now, we know that $\dfrac{1}{\tan \dfrac{t}{2}}=\cot \dfrac{t}{2}=\dfrac{\cos \dfrac{t}{2}}{\sin \dfrac{t}{2}}$ and $\sec \dfrac{t}{2}=\dfrac{1}{\cos \dfrac{t}{2}}$ . Now, substitute the values, we get $$\begin{aligned} & \Rightarrow \dfrac{dy}{dt}=a\left( -\sin t+\dfrac{\cos \dfrac{t}{2}}{\sin \dfrac{t}{2}}.\dfrac{1}{{{\cos }^{2}}\dfrac{t}{2}}.\dfrac{1}{2} \right) \\\ & \Rightarrow \dfrac{dy}{dt}=a\left( -\sin t+\dfrac{1}{\sin \dfrac{t}{2}}.\dfrac{1}{\cos \dfrac{t}{2}}.\dfrac{1}{2} \right) \\\ & \Rightarrow \dfrac{dy}{dt}=a\left( -\sin t+\dfrac{1}{2\sin \dfrac{t}{2}\cos \dfrac{t}{2}} \right) \\\ \end{aligned}$$ Now, we know that $2\sin A\cos A=\sin 2A$ So, $$\begin{aligned} & \Rightarrow \dfrac{dy}{dt}=a\left( -\sin t+\dfrac{1}{\sin 2.\dfrac{t}{2}} \right) \\\ & \Rightarrow \dfrac{dy}{dt}=a\left( -\sin t+\dfrac{1}{\sin t} \right) \\\ \end{aligned}$$ Now, simplify further we get $$\Rightarrow \dfrac{dy}{dt}=a\left( \dfrac{1-{{\sin }^{2}}t}{\sin t} \right)$$ Now, we know that $$1-{{\sin }^{2}}t={{\cos }^{2}}t$$ Now, we get $$\Rightarrow \dfrac{dy}{dt}=a\left( \dfrac{{{\cos }^{2}}t}{\sin t} \right)...................(ii)$$ Now, we have $\Rightarrow \dfrac{dy}{dx}=\dfrac{dy}{dt}\times \dfrac{dt}{dx}$, substitute the values from equation (i) and (ii), we get $$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=a\left( \dfrac{{{\cos }^{2}}t}{\sin t} \right)\times \dfrac{1}{a\cos t} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{\cos t}{\sin t} \\\ & \Rightarrow \dfrac{dy}{dx}=\cot t \\\ \end{aligned}$$ Now, to find $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$, we again differentiate the above equation with respect to $x$ $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \cot t \right)$ Now we know that $\dfrac{d}{dx}\left( \cot x \right)=-cose{{c}^{2}}x$ So, we get $\begin{aligned} & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-cose{{c}^{2}}t\dfrac{dt}{dx} \\\ & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-cose{{c}^{2}}t\times \dfrac{1}{a\cos t} \\\ \end{aligned}$ $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-cose{{c}^{2}}t}{a\cos t}$ **So, the correct answer is “Option A”.** **Note:** To solve such a type of question we need to have knowledge about the differentiation formulas. As this question has a complex calculation while solving differentiation so try to solve step by step. As we stop the calculation at $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-cose{{c}^{2}}t}{a\cos t}$, if you want to solve further, use identities. $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-cose{{c}^{2}}t}{a\cos t}$ As we know that $\cos ec\theta =\dfrac{1}{\sin \theta }$ $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{a{{\sin }^{2}}t\cos t}$