If x = ecos2t and y = esin2t, prove that dy/ dx = – y log x/ x log y.


If ecos2t and esin2t, prove that dy/ dx = – y log x/ x log y.



ecos2t and esin2t

So, cos 2t = log x and sin 2t = log y

Now, differentiating both the parameter functions w.r.t t, we have

$\frac{d x}{d t}=e^{\cos 2 t} \cdot \frac{d}{d t}(\cos 2 t)=e^{\cos 2 t}(-\sin 2 t) \cdot \frac{d}{d t}(2 t)$

$=-e^{\cos 2 t} \cdot \sin 2 t \cdot 2=-2 e^{\cos 2 t} \cdot \sin 2 t$

Now, $\quad y=e^{\sin 2 t}$

$\frac{d y}{d t}=e^{\sin 2 t} \cdot \frac{d}{d t}(\sin 2 t)=e^{\sin 2 t} \cdot \cos 2 t \cdot \frac{d}{d t}(2 t)$

$=e^{\sin 2 t} \cdot \cos 2 t \cdot 2=2 e^{\sin 2 t} \cdot \cos 2 t$

$\therefore \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{2 e^{\sin 2 t} \cdot \cos 2 t}{-2 e^{\cos 2 t} \cdot \sin 2 t}=\frac{e^{\sin 2 t} \cdot \cos 2 t}{-e^{\cos 2 t} \cdot \sin 2 t}=\frac{y \cos 2 t}{-x \sin 2 t}$

$=\frac{y \log x}{-x \log y}$ $\left[\begin{array}{r}\because \cos 2 t=\log x \\ \sin 2 t=\log y\end{array}\right]$

Thus,  $\frac{d y}{d x}=-\frac{y \log x}{x \log y}$

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