Question

If we have the value of x and y as $x=2\cos t-\cos 2t,y=2\sin t-\sin 2t,$ then the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ at $t=\dfrac{\pi }{2}$ is:
(A). $\dfrac{1}{2}$
(B). $2$
(C). $\dfrac{3}{2}$
(D). $-\dfrac{3}{2}$

Answer 1

Hint: First of all do the single derivative of x with respect to t and single derivative of y with respect to t. Then do the derivative of the single derivate of x with respect to t and derivate the single derivate of y with respect to t. Now, divide this double derivative of y with respect to t by double derivative of x with respect to t and then put the value of t in the obtained expression as $\dfrac{\pi }{2}$.

Complete step-by-step solution -
The value of x and y given in the expression in terms of t is:
$\begin{align}
  & x=2\cos t-\cos 2t \\
 & y=2\sin t-\sin 2t \\
\end{align}$
Taking the derivative of x with respect to t we get,
$\begin{align}
  & x=2\cos t-\cos 2t \\
 & \Rightarrow \dfrac{dx}{dt}=-2\sin t+2\sin 2t \\
\end{align}$
In the above equation, we have used the relation that the derivative of $\cos t$ with respect to t is $-\sin t$.
Taking the derivative of y with respect to t we get,
$\begin{align}
  & y=2\sin t-\sin 2t \\
 & \Rightarrow \dfrac{dy}{dt}=2\cos t-2\cos 2t \\
\end{align}$
We are going to take the derivative of $\dfrac{dx}{dt}$ with respect to t.
$\begin{align}
  & \dfrac{dx}{dt}=-2\sin t+2\sin 2t \\
 & \Rightarrow \dfrac{{{d}^{2}}x}{d{{t}^{2}}}=-2\cos t+4\cos 2t \\
\end{align}$
We are going to take the derivative of $\dfrac{dy}{dt}$ with respect to t.
$\begin{align}
  & \dfrac{dy}{dt}=2\cos t-2\cos 2t \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{t}^{2}}}=-2\sin t+4\sin 2t \\
\end{align}$
Now, we are going to write $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ as follows:
$\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\dfrac{{{d}^{2}}y}{d{{t}^{2}}}}{\dfrac{{{d}^{2}}x}{d{{t}^{2}}}} \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-2\sin t+4\sin 2t}{-2\cos t+4\cos 2t} \\
\end{align}$
Taking -2 as common from the numerator and denominator in the above equation we get.
$\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-2\sin t+4\sin 2t}{-2\cos t+4\cos 2t} \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\sin t-2\sin 2t}{\cos t-2\cos 2t} \\
\end{align}$
Substituting the value of t as $\dfrac{\pi }{2}$ in the above equation we get,
$\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\sin t-2\sin 2t}{\cos t-2\cos 2t} \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\sin \dfrac{\pi }{2}-2\sin \pi }{\cos \dfrac{\pi }{2}-2\cos \pi } \\
\end{align}$
We know from the trigonometric ratios that the value of:
$\begin{align}
  & \sin \dfrac{\pi }{2}=1,\sin \pi =0 \\
 & \cos \dfrac{\pi }{2}=0,\cos \pi =-1 \\
\end{align}$
Plugging these values in the above double derivative equation we get,
$\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\sin \dfrac{\pi }{2}-2\sin \pi }{\cos \dfrac{\pi }{2}-2\cos \pi } \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1-0}{0-2\left( -1 \right)} \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{2} \\
\end{align}$
From the above solution, we have found that the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ at $t=\dfrac{\pi }{2}$ is $\dfrac{1}{2}$.
Hence, the correct option is (a).

Note: You might think of how we have taken the derivative of $\sin 2t$ with respect to t.
Let us assume $y=\sin 2t$. Taking derivative on both the sides we get,
$dy=d\left( \sin 2t \right)$
The answer is let us assume $2t=u$. Take the derivative on both the sides we get $2dt=du$.
Now, write u in place of 2t in $\sin 2t$ then it will look like $\sin u$.
$\begin{align}
  & dy=d\left( \sin u \right) \\
 & \Rightarrow dy=\cos udu \\
\end{align}$
In the above, we have taken the differentiation of $\sin u$ with respect to u.
Substituting $2dt=du$ and $u=2t$ in the above equation we get,
$\begin{align}
  & dy=\cos 2t\left( 2dt \right) \\
 & \Rightarrow dy=2\cos 2t dt \\
\end{align}$
Dividing the above equation by dt we get,
$\dfrac{dy}{dt}=2\cos 2t$
Hence, we have shown the derivative of $\sin 2t$ with respect to t.