TITLE: Eliminating a parameter from 2 equations
QUESTION [0 upvotes]: The question given to me was actually of parametric differentiation, and the equations were:
$$x = \dfrac{\sin^3 t}{\sqrt{\cos2t}}\ , \ \  \ \ y = \dfrac{\cos^3 t}{\sqrt{\cos2t}}$$
and we had to find the $\dfrac{dy}{dx}$
One way of calculating could be calculating $\dfrac{dy}{dt}$ and dividing it by $\dfrac{dx}{dt}$.
But this method was very long and tiring, and takes time to solve. I was thinking of eliminating the parameter, but couldn't do so. Expressing $\cos t$ as $\sin(\frac{\pi}{2}-t)$ also doesn't help!
Is there a way by which we can eliminate the parameter $t$ ?

REPLY [2 votes]: \begin{align*}
  \dot{x} &=
  \frac{3\sin^{2} t \cos t}{\sqrt{\cos 2t}}+
  \frac{\sin^{3} t \sin 2t}{(\cos 2t)^{3/2}} \\
  &=\frac{\sin^{2} t}{(\cos 2t)^{3/2}}
    ( 3\cos t \cos 2t+\sin t \sin 2t) \\
  &=\frac{\sin^{2} t}{(\cos 2t)^{3/2}}
    (3\cos t \cos 2t+2\sin^{2} t \cos t) \\
  &=\frac{\sin^{2} t \cos t}{(\cos 2t)^{3/2}}
    (3\cos 2t+2\sin^{2} t) \\
  \dot{y} &=-
  \frac{3\cos^{2} t \sin t}{\sqrt{\cos 2t}}+
  \frac{\cos^{3} t \sin 2t}{(\cos 2t)^{3/2}} \\
  &=\frac{\cos^{2} t}{(\cos 2t)^{3/2}}
    (-3\sin t \cos 2t+\cos t \sin 2t) \\
  &=\frac{\cos^{2} t}{(\cos 2t)^{3/2}}
    (-3\sin t \cos 2t+2\cos^{2} t \sin t) \\
  &=\frac{\cos^{2} t \sin t}{(\cos 2t)^{3/2}}
    (-3\cos 2t+2\cos^{2} t) \\
  \frac{dy}{dx} &=
  \frac{\cos t(2\cos^{2} t-3\cos 2t)}{\sin t(2\sin^{2} t+3\cos 2t)} \\
  &=\frac{\cos t[2\cos^{2} t-3(2\cos^{2} t-1)]}
         {\sin t[2\sin^{2} t+3(1-2\sin^{2} t)]} \\
  &=-\frac{4\cos^{3} t-3\cos t}{3\sin t-4\sin^{3} t} \\
  &=-\frac{\cos 3t}{\sin 3t} \\
  &=-\cot 3t
\end{align*}
Further points to be noticed
\begin{align*}
  y^{2}-x^{2} &= \frac{7+\cos 4t}{8} \\
  x^{2/3}+y^{2/3} &= \frac{1}{\sqrt[3]{\cos 2t}} \\
  y^{2}-x^{2} &=
  \frac{1}{4} \left[ 3+\frac{1}{(x^{2/3}+y^{2/3})^{6}} \right] \\
  \frac{dy}{dx} &=-\frac{3\cot t-\cot^{3} t}{1-3\cot^{2} t} \\
  &=\frac{xy^{3}-3\sqrt[3]{x^{5}y}}{x^{2}-3y^{2}}
\end{align*}
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