Question

If \[x=a\cos 2t,y=b{{\sin }^{2}}t\], then at \[(-a,b),\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] is equal to
(a) $1$ (b) \[\dfrac{1}{2}\]
(c) $0$ (d) \[\dfrac{1}{\sqrt{2}}\]

Answer 1

Hint: Use parametric equations and chain rule.
As per the given information, \[x=a\cos 2t,y=b{{\sin }^{2}}t\]
For this problem, first we shall find \[\dfrac{dy}{dt}\And \dfrac{dx}{dt}.\]
So, consider, \[y=b{{\sin }^{2}}t\]
Now take differentiation on both sides with respect to ‘t’, we get
\[\dfrac{dy}{dt}=\dfrac{d}{dt}(b{{\sin }^{2}}t)\]
Taking out the constant term, we get
\[\begin{align}
  & \dfrac{dy}{dt}=b.\dfrac{d}{dt}({{\sin }^{2}}t) \\
 & \dfrac{dy}{dt}=b(2\sin t)\dfrac{d}{dt}(\sin t) \\
\end{align}\]
We know derivative of $sinx$ is $\cos x$ , so we get
\[\dfrac{dy}{dt}=b(2\sin t)(cost)\]
But we know, $\sin 2t=2\sin t\cos t$ , so above equation becomes,
\[\dfrac{dy}{dt}=b\sin 2t........(i)\]
Now consider,\[x=a\cos 2t\]
Now take differentiation on both sides with respect to ‘t’, we get
\[\dfrac{dx}{dt}=\dfrac{d}{dt}(a\cos 2t)\]
Taking out the constant term and derivative of $\cos x$ is $-\sin x$ , so we get
\[\begin{align}
  & \dfrac{dx}{dt}=a(-\sin 2t)\dfrac{d}{dt}(2t) \\
 & \dfrac{dx}{dt}=-2a.\sin 2t..........(ii) \\
\end{align}\]
Now dividing equations (ii) by (i), we have
\[\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\dfrac{b\sin 2t}{-2a.\sin 2t}\]
Cancelling like terms, we get
\[\dfrac{dy}{dx}=\dfrac{-b}{2a}\]
Now as this is free from the ‘t’ term, we can directly differentiate to find the second derivative.
Now take differentiation on both sides with respect to ‘t’, we get
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{-b}{2a} \right)\]
We know derivative of constant term is zero, so we have
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=0\]
As the second derivative is free of variable terms, so its value is zero at any point.
Hence the correct answer is option (c).

Note: Instead of deriving with respect to ‘t’ and then dividing, we can find the value of ‘t’ from the given values of ‘x’.
We know,
$\cos 2t=1-{{\sin }^{2}}t$
Substituting this in value of ‘x’, we get
\[\begin{align}
  & x=a\cos 2t \\
 & \Rightarrow x=a(1-{{\sin }^{2}}t) \\
 & \Rightarrow \dfrac{x}{a}+1={{\sin }^{2}}t \\
\end{align}\]
Now substituting this value in the value of ‘y’, we get
\[\begin{align}
  & y=b{{\sin }^{2}}t \\
 & \Rightarrow y=b\left( \dfrac{x}{a}+1 \right) \\
\end{align}\]
Now we can see that this can be differentiated directly with respect to ‘x’.
But this is a complicated process.