Question

If \[x = 3 \cos t - 2 \cos^3 t, y = 3\sin t - 2 \sin^3 t,\] find \[\frac{d^2 y}{d x^2} \] ?

Answer 1

We have,

\[x = 3\cos t - 2 \cos^3 t\]

\[ \Rightarrow \frac{dx}{dt} = 3\left( - \sin t \right) - 6 \cos^2 t\left( - \sin t \right)\]

\[ = - 3\sin t + 6\sin t \cos^2 t\]

Also,

\[y = 3\sin t - 2 \sin^3 t\]

\[ \Rightarrow \frac{dy}{dt} = 3\cos t - 6 \sin^2 t \cos t\]

Now,

\[\frac{dy}{dx} = \frac{\left( \frac{dy}{dt} \right)}{\left( \frac{dx}{dt} \right)}\]

\[ = \frac{3\cos t - 6 \sin^2 t \cos t}{- 3\sin t + 6\sin t \cos^2 t}\]

\[ = \frac{3\cos t\left( 1 - 2 \sin^2 t \right)}{3\sin t\left( - 1 + 2 \cos^2 t \right)}\]

\[ = \frac{\cot t\left( \cos2t \right)}{\left( \cos2t \right)}\]

\[ = \cot t\]

\[So, \frac{d^2 y}{d x^2} = \frac{d}{dx}\left( \frac{dy}{dx} \right)\]

\[ = \frac{d}{dx}\left( \cot t \right)\]

\[ = - {cosec}^2 t \frac{dt}{dx}\]

\[ = \frac{- {cosec}^2 t}{\left( \frac{dx}{dt} \right)}\]

\[ = \frac{- {cosec}^2 t}{- 3\sin t + 6\sin t \cos^2 t}\]

\[ = \frac{- {cosec}^2 t}{-3 \sin t\left( 1 - 2 \cos^2 t \right)}\]

\[ = \frac{{cosec}^3 t}{\left( - 3\cos 2t \right)}\]

\[ = \frac{- {cosec}^3 t}{3\cos 2t}\]