Question:

\(x\) and \(y\) are given as follows:

\[  x = e^t \cos t  \]

\[  y = e^t \sin t  \]

Express \(\frac{d^2y}{dx^2}\) in terms of \(t\).

Answer:

Take the derivative with respect to \(t\).

\[  \frac{dx}{dt} = e^t (\cos t -\sin t)  \]

\[  \frac{dy}{dt} = e^t (\cos t +\sin t)  \]

Therefore,

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

\[              = \frac{\cos t + \sin t}{\cos t – \sin t}\]

Now, \(\frac{d^2y}{dx^2}\) is a little tricky.

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\frac{dy}{dx} = \frac{d}{dt}\frac{dy}{dx}\frac{dt}{dx} \]

Therefore,

\[  \frac{d^2y}{dx^2} = \frac{d}{dt}\left(\frac{\cos t + \sin t}{\cos t – \sin t}\right)\frac{1}{e^t (\cos t -\sin t)}  \]

\[                                = \frac{(-\sin t + \cos t)(\cos t – \sin t)-(\cos t + \sin t)(-\sin t – \cos t)}{e^t(\cos t – \sin t)^3}  \]

\[                                = \frac{\cos^2 t – 2\cos t\sin t + \sin^2 t + \cos^2 t + 2\cos t\sin t + \sin^2 t}{e^t(\cos t – \sin t)^3}  \]

\[                                = \frac{2(\cos^2 t + \sin^2 t)}{e^t(\cos t – \sin t)^3}    \]

\[                                = \frac{2}{e^t(\cos t – \sin t)^3}    \]