Question

We have given if $x=a\left( \cos t+\log \left( \tan \dfrac{t}{2} \right) \right),y=a\sin t$ and we have to find the following expressions:
$\dfrac{{{d}^{2}}y}{d{{t}^{2}}},\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$

Answer 1

Hint: We have given x and y in terms of “t”. To find the value of this expression $\dfrac{{{d}^{2}}y}{d{{t}^{2}}}$ we have to first find single derivative of y with respect to t then again derivate it with respect to t. Now, to find $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ we want to first differentiate x with respect to t. The find $\dfrac{dy}{dx}$ by dividing $\dfrac{dy}{dt}$ by $\dfrac{dx}{dt}$ and then differentiate $\dfrac{dy}{dx}$ with respect to x.

Complete step by step answer:
In the above problem, we have given x and y as:
$x=a\left( \cos t+\log \left( \tan \dfrac{t}{2} \right) \right),y=a\sin t$
And we are asked to find $\dfrac{{{d}^{2}}y}{d{{t}^{2}}},\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$.
First of all, we are finding $\dfrac{{{d}^{2}}y}{d{{t}^{2}}}$ which we are going to do by first differentiating y with respect to t then again we are differentiating with respect to t.
$y=a\sin t$
Differentiating with respect to t on both the sides we get,
$\dfrac{dy}{dt}=a\cos t$ ……… Eq. (1)
Now, again differentiating the above equation we with respect to t we get,
$\dfrac{{{d}^{2}}y}{d{{t}^{2}}}=-a\sin t$
Hence, we have got the value of $\dfrac{{{d}^{2}}y}{d{{t}^{2}}}=-a\sin t$.
Now, we are finding the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ which are going to do as follows:
First of all, we are going to find $\dfrac{dy}{dx}$ by finding $\dfrac{dy}{dt}$ then $\dfrac{dx}{dt}$ and dividing both of them,
We have already calculated $\dfrac{dy}{dt}$ and $\dfrac{dx}{dt}$ we are about to calculate as follows:
$x=a\left( \cos t+\log \left( \tan \dfrac{t}{2} \right) \right)$
Differentiating on both the sides with respect to “t” we get,
$\begin{align}
  & \dfrac{dx}{dt}=a\left( -\sin t+\dfrac{1}{\tan \dfrac{t}{2}}\left( \dfrac{1}{2} \right){{\sec }^{2}}\dfrac{t}{2} \right) \\
 & \Rightarrow \dfrac{dx}{dt}=a\left( -\sin t+\dfrac{\cos \dfrac{t}{2}}{\sin \dfrac{t}{2}}\left( \dfrac{1}{2{{\cos }^{2}}\dfrac{t}{2}} \right) \right) \\
\end{align}$
One $\cos \dfrac{t}{2}$ will be cancelled out from the numerator and the denominator.
$\dfrac{dx}{dt}=a\left( -\sin t+\dfrac{1}{2\sin \dfrac{t}{2}}\left( \dfrac{1}{\cos \dfrac{t}{2}} \right) \right)$
We know that, $\sin t=2\sin \dfrac{t}{2}\cos \dfrac{t}{2}$ using this relation in the above equation we get,
$\begin{align}
  & \dfrac{dx}{dt}=a\left( -\sin t+\dfrac{1}{\sin t} \right) \\
 & \Rightarrow \dfrac{dx}{dt}=a\left( \dfrac{1-{{\sin }^{2}}t}{\sin t} \right) \\
\end{align}$
We know that, $1-{{\sin }^{2}}t={{\cos }^{2}}t$ using this relation in the above equation we get,
$\dfrac{dx}{dt}=a\left( \dfrac{{{\cos }^{2}}t}{\sin t} \right)$…………. Eq. (2)
Now, dividing eq. (1) by eq. (2) we get,
\[\begin{align}
  & \dfrac{dy}{dx}=\dfrac{a\cos t}{a\left( \dfrac{{{\cos }^{2}}t}{\sin t} \right)} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{\cos t\left( \sin t \right)}{{{\cos }^{2}}t} \\
\end{align}\]
One $\cos t$ will be cancelled out in the above equation we get,
\[\dfrac{dy}{dx}=\dfrac{\left( \sin t \right)}{\cos t}=\tan t\]
Now, differentiating the above equation with respect to x on both the sides we get,
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}={{\sec }^{2}}t\dfrac{dt}{dx}............Eq.(3)$
Substituting the value of $\dfrac{dt}{dx}$ using the eq. (2) in the above equation we get,
Eq. (2) is shown below:
$\dfrac{dx}{dt}=a\left( \dfrac{{{\cos }^{2}}t}{\sin t} \right)$
Reciprocating the above equation we get,
$\dfrac{dt}{dx}=\dfrac{1}{a}\left( \dfrac{\sin t}{{{\cos }^{2}}t} \right)$
Using the above value of $\dfrac{dt}{dx}$ in eq. (3) we get,
$\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}={{\sec }^{2}}t\dfrac{1}{a}\left( \dfrac{\sin t}{{{\cos }^{2}}t} \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{a}\sin t{{\sec }^{4}}t \\
\end{align}$

Hence, we have got the value of $\dfrac{{{d}^{2}}y}{d{{t}^{2}}},\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ as follows:
$\dfrac{{{d}^{2}}y}{d{{t}^{2}}}=-a\sin t$
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{a}\sin t{{\sec }^{4}}t$

Note: The most plausible mistake that could happen in this problem is that just like we found the value of $\dfrac{dy}{dx}$ we can also find the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ which we have shown below:
You might have written $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ as follows:
$\dfrac{\dfrac{{{d}^{2}}y}{d{{t}^{2}}}}{\dfrac{{{d}^{2}}x}{d{{t}^{2}}}}$
And then find $\dfrac{{{d}^{2}}y}{d{{t}^{2}}}\And \dfrac{{{d}^{2}}x}{d{{t}^{2}}}$ and then divide both of them to find the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$. This is the wrong way to find $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ so make sure you won’t make this blunder.